Free math worksheets

- Read the constant rate from the vertical axis. - Identify the length of the time interval. - Multiply the rate by the elapsed time.

1. The outflow rate is constant at \(200\,\text{L/min}\). 2. The accumulated outflow is the rectangular area under the rate graph: \((200\,\text{L/min})(2.5\,\text{min})=500\,\text{L}\).

\(500\,\text{L}\)

- Read the constant rate from the vertical axis. - Find the altitude change shown on the horizontal axis. - Multiply the rate by the altitude interval.

1. The magnitude of the decrease rate is constant at \(5\,\text{K/km}\). 2. Over an altitude change of \(4\,\text{km}\), the temperature decrease is \(5\,\frac{\text{K}}{\text{km}}\cdot4\,\text{km}=20\,\text{K}\).

The temperature decreases by \(20\,\text{K}\).

- Integrating a growth rate gives accumulated growth. - Find an antiderivative of the rate. - Use the initial population to determine the constant. - Check the units of the integral.

1. An antiderivative of \(r(t)\) is \(R(t)=t^3+20t^2\). 2. \(\int_0^4r(t)\,\text{d}t=[t^3+20t^2]_0^4=384\). 3. The integral is the increase in the number of bacteria during the first \(4\) hours. 4. Since \(N'(t)=r(t)\), \(N(t)=t^3+20t^2+C\). Using \(N(0)=1000\) gives \(C=1000\). 5. Thus, \(N(t)=t^3+20t^2+1000\).

a) \(384\) bacteria; this is the increase during the first \(4\) hours. b) \(N(t)=t^3+20t^2+1000\)

- Integrate the flow rate over the pumping interval. - Use \(0\) and \(12\) as the limits of integration. - Convert liters to cubic meters at the end.

1. The total volume in liters is \(\int_0^{12}(120t-10t^2)\,\mathrm{d}t\). 2. An antiderivative is \(60t^2-\frac{10}{3}t^3\). 3. Therefore, \(\left[60t^2-\frac{10}{3}t^3\right]_0^{12} =8640-5760 =2880\,\text{L}\). 4. Since \(1000\,\text{L}=1\,\text{m}^3\), \(2880\,\text{L}=2.88\,\text{m}^3\).

The pumps remove \(2.88\,\text{m}^3\) of water.

- Use Hooke's law to determine the spring constant. - Convert all distances to meters before calculating work. - Work is the definite integral of force with respect to displacement. - Choose integration limits that match each stretch interval.

1. a) Convert \(6\,\text{cm}\) to \(0.06\,\text{m}\). Then \(k=\frac{15}{0.06}=250\,\text{N/m}\). 2. b) The work is \(W=\int_0^{0.1}250s\,\text{d}s=\left[125s^2\right]_0^{0.1}=1.25\,\text{J}\). 3. c) The additional work is \(\int_{0.1}^{0.2}250s\,\text{d}s=125(0.2^2-0.1^2)=3.75\,\text{J}\).

a) \(250\,\text{N/m}\) b) \(1.25\,\text{J}\) c) \(3.75\,\text{J}\)

- Write work as the integral of force with respect to displacement. - Use the power rule for integration. - When substituting \(3a\), cube the entire expression. - Simplify the resulting terms.

1. Work is the definite integral of force: \(W=\int_a^{3a}cs^2\,\text{d}s\). 2. Evaluate the integral: \(W=\left[\frac{c}{3}s^3\right]_a^{3a}=\frac{c}{3}(27a^3-a^3)=\frac{26}{3}ca^3\).

\(W=\frac{26}{3}ca^3\)

- Integrate the production rate to obtain the total-production function. - Use the initial completed quantity to determine the constant. - Substitute the requested time into the function.

1. Integrating the production rate gives \(N(t)=-0.2t^3+3t^2+4t+C\). 2. Since \(N(0)=50\), \(C=50\). Thus, \(N(t)=-0.2t^3+3t^2+4t+50\). 3. \(N(5)=-0.2(5)^3+3(5)^2+4(5)+50=120\) components.

a) \(N(t)=-0.2t^3+3t^2+4t+50\) b) \(120\) components

- Interpret the shaded area as energy produced during the first \(6\) hours. - Integrating power over time gives energy. - Use the average-value formula over the full \(12\)-hour interval. - Track the difference between kilowatts and kilowatt-hours.

1. Energy is the integral of power: \(\int_0^6(-t^2+12t)\,\text{d}t=\left[-\frac{1}{3}t^3+6t^2\right]_0^6=-72+216=144\,\text{kWh}\). 2. The average power is \(\frac{1}{12}\int_0^{12}(-t^2+12t)\,\text{d}t=\frac{288}{12}=24\,\text{kW}\).

1. \(144\,\text{kWh}\) 2. \(24\,\text{kW}\)

- Integrate the polynomial term by term. - Apply the Fundamental Theorem of Calculus at \(t=1\) and \(t=7\). - An integral of a rate over time gives accumulated volume. - Identify the corresponding area on the graph.

1. Integrating term by term gives \(F(t)=-\frac{1}{6}t^3+2t^2+2t+C\). 2. The accumulated volume is \(\int_1^7f(t)\,\mathrm{d}t=F(7)-F(1)\). Now, \(F(7)=\frac{329}{6}\) and \(F(1)=\frac{23}{6}\). Therefore, \(\int_1^7f(t)\,\mathrm{d}t =\frac{329}{6}-\frac{23}{6} =51\). 3. The integral is the total volume entering the basin from hour \(1\) through hour \(7\). On the graph, it is the area under \(f\) and above the \(t\)-axis between \(t=1\) and \(t=7\).

a) \(F(t)=-\frac{1}{6}t^3+2t^2+2t+C\) b) \(51\,\text{m}^3\) c) It is the accumulated inflow from \(t=1\) to \(t=7\), represented by the area under the rate graph on that interval.

- Multiply the horizontal and vertical units to determine the integral's units. - Break the region into simple geometric shapes. - Add the areas of the two triangles and the rectangle.

1. Mathematically, the integral is the area under the graph of \(f\) from \(t=0\) to \(t=10\). In context, it is the total number of visitors who enter during the first \(10\) minutes. 2. Divide the region into two triangles and one rectangle: \(\frac{1}{2}\cdot4\cdot20=40\), \(4\cdot20=80\), and \(\frac{1}{2}\cdot2\cdot20=20\). Therefore, \(\int_0^{10}f(t)\,\text{d}t=140\) visitors.

a) It is the area under the rate graph and represents the total number of visitors entering during the first \(10\) minutes. b) \(140\) visitors

- Check separately whether each proposed rate is nonnegative on the whole interval. - Integrate each rate over \([0,2]\) to find the delivered volume. - A correct model must satisfy both the physical condition and the stated total-volume condition.

1. A valid inflow model must be nonnegative throughout the interval, and its integral over \([0,2]\) must equal \(1\,\text{L}\). 2. For \(f\), the minimum on \([0,2]\) is \(f(2)=0.25>0\), so the rate is nonnegative. However, \(\int_0^2(-0.5t+1.25)\,\text{d}t=1.5\,\text{L}\), not \(1\,\text{L}\). Therefore, the claim is false for \(f\). 3. For \(g\), \(g(0)=-1<0\), so it cannot represent inflow throughout the interval. Although \(\int_0^2(1.5t-1)\,\text{d}t=1\,\text{L}\), the claim is also false for \(g\).

Neither model satisfies both conditions. The function \(f\) is nonnegative but delivers \(1.5\,\text{L}\), while \(g\) has integral \(1\,\text{L}\) but is negative on part of the interval.

- Volume is represented by area under the flow-rate graph. - Break each requested region into rectangles or trapezoids. - For part c), first read the rate at \(t=4\) from the decreasing line segment. - Add adjacent interval volumes for the larger interval.

1. a) The area over \([0,1]\) is a rectangle: \((1)(1)=1\,\text{L}\). 2. b) The area over \([1,3]\) is a trapezoid: \(\frac{1+4}{2}(2)=5\,\text{L}\). 3. c) The line decreases from \((3,4)\) to \((5,0)\), so \(q(4)=2\). The trapezoid over \([3,4]\) has area \(\frac{4+2}{2}(1)=3\,\text{L}\). 4. d) Add the adjacent volumes: \(5+3=8\,\text{L}\).

a) \(1\,\text{L}\) b) \(5\,\text{L}\) c) \(3\,\text{L}\) d) \(8\,\text{L}\)

- Identify the geometric shape between the line and the horizontal axis. - Read the growth rates at the beginning and end of the interval. - Use the area formula for a trapezoid.

1. The accumulated increase in diameter is the area under the rate graph from \(t=0\) to \(t=3\). 2. The region is a trapezoid with parallel sides \(2\,\text{mm/week}\) and \(8\,\text{mm/week}\) and width \(3\,\text{weeks}\). 3. Its area is \(\frac{2+8}{2}(3)=15\,\text{mm}\).

The diameter increases by \(15\,\text{mm}\).

- Interpret the area under the rate graph as accumulated volume. - Identify the base and height of the triangular region. - Use the area formula for a triangle.

1. The accumulated outflow is the area under the rate graph from \(t=0\) to \(t=2\). 2. This region is a right triangle with base \(2\,\text{min}\) and height \(400\,\text{L/min}\). 3. The volume is \(\frac{1}{2}(2)(400)=400\,\text{L}\).

\(400\,\text{L}\)

- Interpret accumulated production as area under the rate graph. - Use a triangle for the full interval and the first hour. - Split the interval in part 3 at \(t=2\), where the rate formula changes. - Use trapezoid areas on the two resulting subintervals.

1. The full region is a triangle with base \(3\) and height \(16\), so \(\frac12(3)(16)=24\) boards are processed. 2. The area from \(0\) to \(1\) is a triangle with base \(1\) and height \(8\): \(\frac12(1)(8)=4\) boards. 3. From \(1.5\) to \(2\), the trapezoid area is \(\frac{12+16}{2}(0.5)=7\). From \(2\) to \(2.5\), the trapezoid area is \(\frac{16+8}{2}(0.5)=6\). Thus, \(7+6=13\) boards are processed during that one-hour interval.

1. \(24\) boards 2. \(4\) boards 3. \(13\) boards are processed from \(t=1.5\) to \(t=2.5\).

- Volume is area under the flow-rate graph. - Break the requested interval into a trapezoid and a rectangle. - For the full interval, use a triangle and a rectangle.

1. a) From \(1\) to \(2\), the area is a trapezoid: \(\frac{2+4}{2}(1)=3\,\text{L}\). From \(2\) to \(4\), the area is a rectangle: \((2)(4)=8\,\text{L}\). Therefore, the volume is \(3+8=11\,\text{L}\). 2. b) The area from \(0\) to \(2\) is a triangle with area \(\frac12(2)(4)=4\,\text{L}\), and the area from \(2\) to \(6\) is a rectangle with area \((4)(4)=16\,\text{L}\). The total is \(20\,\text{L}\).

a) \(11\,\text{L}\) b) The total volume is \(4+16=20\,\text{L}\).

- Accumulated mass is area under the deposition-rate graph. - Use the grid to identify the first triangle’s dimensions. - For part b), first determine the graph’s height at \(t=3\). - Then use the area formula for a triangle.

1. a) The area from \(0\) to \(1\) is a triangle with base \(1\) and height \(4\): \(\frac12(1)(4)=2\,\text{g}\). 2. b) The line from \((1,4)\) to \((5,0)\) has equation \(r(t)=-t+5\), so \(r(3)=2\,\text{g/min}\). The region from \(3\) to \(5\) is a triangle with base \(2\) and height \(2\): \(\frac12(2)(2)=2\,\text{g}\).

a) \(2\,\text{g}\) b) \(2\,\text{g}\)

- Integrate the rate over the requested time interval. - Use the initial volume to determine the constant in the accumulation function. - Read the variable limits as a moving \(5\)-minute interval. - Track the units from liters per minute to liters.

1. An antiderivative is \(F(t)=t^{3/2}+20t\). 2. \(\int_{10}^{40}f(t)\,\text{d}t=[t^{3/2}+20t]_{10}^{40}\approx821.36\,\text{L}\). This is the water that enters between minute \(10\) and minute \(40\). 3. Since \(V(0)=500\), \(V(t)=t^{3/2}+20t+500=t\sqrt{t}+20t+500\). 4. The equation in part c asks for a starting time \(t\) such that exactly \(150\,\text{L}\) enters during the next \(5\) minutes.

a) Approximately \(821.36\,\text{L}\), the amount entering from minute \(10\) through minute \(40\) b) \(V(t)=t\sqrt{t}+20t+500\) c) It asks when a \(5\)-minute interval begins during which exactly \(150\,\text{L}\) enters.

- Integrate the exponential rate. - Distinguish accumulated increase from total population. - Use the initial population to determine the constant. - Interpret the variable limits as a \(2\)-hour interval.

1. An antiderivative is \(W(t)=2000e^{0.1t}\). 2. The increase is \(\int_0^{10}200e^{0.1t}\,\text{d}t=2000(e-1)\approx3436.56\), or about \(3437\) bacteria. 3. A total-population function has the form \(N(t)=2000e^{0.1t}+C\). Since \(N(0)=5000\), \(C=3000\). 4. Thus, \(N(t)=2000e^{0.1t}+3000\). 5. The equation asks for the starting time of a \(2\)-hour interval during which the population grows by exactly \(2000\) bacteria.

a) \(2000(e-1)\approx3436.56\) bacteria, or about \(3437\) bacteria b) \(N(t)=2000e^{0.1t}+3000\) c) “At what time does a \(2\)-hour period begin during which the culture grows by exactly \(2000\) bacteria?”

- Integrate the rate over the full shift to find the mass captured per square meter. - Find the combined area of all filter elements. - Convert square centimeters to square meters and milligrams to grams.

1. The mass captured per square meter is \(\int_{-4}^{4}500\left(1-\frac{t^2}{16}\right)\,\mathrm{d}t =\left[500t-\frac{500}{48}t^3\right]_{-4}^{4} =\frac{8000}{3}\,\frac{\text{mg}}{\text{m}^2}\). 2. The total filter area is \(120\cdot2500\,\text{cm}^2 =300{,}000\,\text{cm}^2 =30\,\text{m}^2\). 3. Therefore, the total mass is \(30\,\text{m}^2\cdot\frac{8000}{3}\,\frac{\text{mg}}{\text{m}^2} =80{,}000\,\text{mg} =80\,\text{g}\).

The system captures \(80\,\text{g}\) of particulate matter.

- A population increases when its rate of change is positive. - Add accumulated change to the initial population. - Keep track of the “thousands of bacteria” unit. - Interpret each side of the equation in part c as a population.

1. The population increases where \(f(t)>0\). The zeros are \(t=2\) and \(t=6\), and the parabola opens downward, so \(f(t)>0\) on \((2,6)\). 2. In thousands of bacteria, \(B(6)=10+\int_0^6(-0.1t^2+0.8t-1.2)\,\text{d}t\). 3. The integral equals \(\left[-\frac{t^3}{30}+0.4t^2-1.2t\right]_0^6=0\). Therefore, \(B(6)=10\), or \(10{,}000\) bacteria. 4. The equation in part c asks for all times \(k\) when the population is \(9.5\) thousand, or \(9500\), bacteria.

a) \((2,6)\) hours b) \(10{,}000\) bacteria c) “At what times \(k\) does the culture contain \(9500\) bacteria?”

- A maximum of the accumulated quantity occurs when its rate changes from positive to negative. - Integrate the rate to find net change. - A zero accumulated change means equal starting and ending volumes.

1. The volume is greatest when the rate changes from positive to negative. Solve \(4e^{-0.5t}-2=0\): \(t=2\ln2\approx1.39\,\text{h}\). 2. The net change is \(\int_0^4(4e^{-0.5t}-2)\,\text{d}t=[-8e^{-0.5t}-2t]_0^4=-8e^{-2}\approx-1.08\,\text{m}^3\). 3. A solution would mean that the water volume at time \(x\) equals the volume at \(t=2\). No such solution exists for \(2<x\le5\), because \(h(t)<0\) throughout that interval.

a) \(t=2\ln2\approx1.39\,\text{h}\) b) \(-8e^{-2}\,\text{m}^3\approx-1.08\,\text{m}^3\), a decrease c) It would mean the volume at time \(x\) equals the volume at \(t=2\); no such \(x\) exists in \((2,5]\).

- Accumulated volume is the integral of flow rate over time. - Split the interval where the rate formula changes. - Add the contributions from the two subintervals.

1. Split the accumulation at \(t=2\), where the rate formula changes. 2. From \(t=1\) to \(t=2\), the delivered volume is \(\int_1^2 0.2t\,\text{d}t=\left[0.1t^2\right]_1^2=0.3\,\text{L}\). 3. From \(t=2\) to \(t=3\), the delivered volume is \(\int_2^3 0.4\,\text{d}t=0.4\,\text{L}\). 4. The total delivered volume is \(0.3+0.4=0.7\,\text{L}\).

\(0.7\,\text{L}\)

- Integrate the rate to find an accumulated amount of data. - Split an interval wherever the rate formula changes. - Keep megabytes and megabytes per second distinct. - An instantaneous rate is found by evaluating \(r(t)\), not by integrating.

1. The data received during the first second is \(\int_0^1 0.4t\,\text{d}t=\left[0.2t^2\right]_0^1=0.2\,\text{MB}\). 2. The data received during the first \(3\) seconds is \(0.2+\int_1^3(-0.1t+0.5)\,\text{d}t=0.8\,\text{MB}\). 3. The data received from \(0.5\) to \(2\) is \(\int_{0.5}^1 0.4t\,\text{d}t+\int_1^2(-0.1t+0.5)\,\text{d}t=0.15+0.35=0.5\,\text{MB}\). 4. The data received from \(4\) to \(5\) is \(\int_4^5(-0.1t+0.5)\,\text{d}t=0.05\,\text{MB}\). 5. The instantaneous rate is \(r(2)=-0.1(2)+0.5=0.3\,\text{MB/s}\). This is a rate, not an accumulated amount.

1. \(0.2\,\text{MB}\) 2. \(0.8\,\text{MB}\) 3. \(0.5\,\text{MB}\) 4. \(0.05\,\text{MB}\) 5. \(0.3\,\text{MB/s}\)

- Use the chain rule to differentiate the exponential function. - What happens to \(e^{-z^2/2}\) as \(|z|\) becomes very large? - What does \(\int x f(x)\,\text{d}x\) represent for a continuous random variable? - Examine how \(\sigma\) appears in the leading factor of a general normal density.

1. a) By the chain rule, \(\frac{\text{d}}{\text{d}z}\left(-e^{-z^2/2}\right)=-e^{-z^2/2}(-z)=ze^{-z^2/2}=h(z)\). 2. b) \(\int_{-\infty}^{\infty}z\phi(z)\,\text{d}z=\frac{1}{\sqrt{2\pi}}\left[-e^{-z^2/2}\right]_{-\infty}^{\infty}=0\), because \(e^{-z^2/2}\to0\) as \(z\to\pm\infty\). This integral is the mean, so the standard normal distribution has \(\mu=0\). 3. c) \(\phi(0)=\frac{1}{\sqrt{2\pi}}\approx0.3989\). For a normal density with standard deviation \(\sigma\), the maximum is \(\frac{1}{\sigma\sqrt{2\pi}}\), so increasing \(\sigma\) lowers the maximum.

a) \(\frac{\text{d}}{\text{d}z}\left(-e^{-z^2/2}\right)=ze^{-z^2/2}\) b) The integral equals \(0\); it is the mean \(\mu\) of the standard normal distribution. c) The maximum is \(\frac{1}{\sqrt{2\pi}}\approx0.3989\). A larger standard deviation produces a lower maximum.

- Check both nonnegativity and total area when verifying a density. - Translate “longer than two years” into the limits of an integral. - Check whether your mean is reasonable within the interval \([0, 3]\).

1. a) On \([0, 3]\), both \(t^2\) and \(3-t\) are nonnegative. Also, \(\int_0^3\frac{4}{27}(3t^2-t^3)\,\text{d}t=\frac{4}{27}\left[t^3-\frac{t^4}{4}\right]_0^3=1\). Therefore, \(g\) is a density. 2. b) \(E(T)=\int_0^3t g(t)\,\text{d}t=\frac{4}{27}\left[\frac{3t^4}{4}-\frac{t^5}{5}\right]_0^3=1.8\) years. 3. c) \(P(T>2)=\int_2^3\frac{4}{27}(3t^2-t^3)\,\text{d}t=\frac{4}{27}\left[t^3-\frac{t^4}{4}\right]_2^3=\frac{11}{27}\approx0.4074\).

a) \(g(t)\ge0\) and \(\int_0^3g(t)\,\text{d}t=1\) b) \(E(T)=1.8\) years c) \(P(T>2)=\frac{11}{27}\approx0.4074\), or about \(40.74\%\)

- Evaluate \(r(t)\) for an instantaneous rate. - Integrate \(r(t)\) to find an accumulated volume. - Use the requested times as the limits of integration. - Keep liters per hour distinct from liters.

1. a) The instantaneous rate is \(r(0.5)=6(0.5-0.25)=1.5\,\text{L/h}\). 2. An antiderivative of \(6(t-t^2)\) is \(R(t)=3t^2-2t^3\). 3. b) The volume delivered during the first \(0.5\) hour is \(R(0.5)-R(0)=0.5\,\text{L}\). 4. c) The volume delivered from \(0.2\) to \(0.8\) is \(R(0.8)-R(0.2)=0.896-0.104=0.792\,\text{L}\). 5. d) The volume delivered from \(t=0.75\) to \(t=1\) is \(R(1)-R(0.75)=1-0.84375=0.15625\,\text{L}\).

a) \(1.5\,\text{L/h}\) b) \(0.5\,\text{L}\) c) \(0.792\,\text{L}\) d) \(0.15625\,\text{L}\)

- The integral of flow rate over the full cycle must equal the required volume. - After finding \(k\), use a definite integral for each requested time interval. - Check that the subinterval volumes are smaller than the full-cycle volume.

1. a) The full-cycle volume must satisfy \(\int_0^4 k(4-t)\,\text{d}t=1\). Evaluating gives \(8k=1\), so \(k=\frac{1}{8}\). 2. An accumulation function is \(R(t)=\frac{1}{8}\left(4t-\frac{t^2}{2}\right)\). 3. b) The volume from \(t=2\) to \(t=4\) is \(R(4)-R(2)=1-\frac{6}{8}=0.25\,\text{L}\). 4. c) The volume from \(t=1\) to \(t=3\) is \(R(3)-R(1)=0.9375-0.4375=0.5\,\text{L}\).

a) \(k=\frac{1}{8}=0.125\) b) \(0.25\,\text{L}\) c) \(0.5\,\text{L}\)

- Integrate the rate to find the net inventory change. - Divide the accumulated change by the interval length to find an average rate. - Account for the scale factor of \(100\) in the rate units. - Use the sign of the rate to determine when inventory decreases and increases.

1. a) The net inventory change is \(100\int_0^4f(t)\,\text{d}t\). An antiderivative is \(-0.025t^4+0.2t^3-0.4t^2\), and evaluation from \(0\) to \(4\) gives \(0\). Thus, the net inventory change is \(0\) packages. 2. b) The average model rate on \([0,2]\) is \(\frac{1}{2}\int_0^2f(t)\,\text{d}t=-0.2\). Since the model is measured in hundreds of packages per hour, the average net inventory rate is \(-20\) packages per hour. 3. c) Factor \(f(t)=-0.1t(t-2)(t-4)\). The rate is negative on \((0,2)\) and positive on \((2,4)\), so inventory decreases until \(t=2\) and increases afterward. Therefore, the minimum occurs at \(t=2\,\text{h}\).

a) \(0\) packages b) \(-20\) packages per hour c) \(t=2\,\text{h}\)

- Integrate the emission rate to find accumulated mass. - Divide the total mass by \(10\) hours to find the average rate. - Evaluate the definite integral separately on \([0,5]\) and \([5,10]\).

1. An antiderivative is \(F(t)=-0.0025t^4+0.04t^3+0.5t\). 2. a) \(\int_0^{10}E(t)\,\text{d}t=F(10)-F(0)=-25+40+5=20\,\text{kg}\). 3. b) The average rate is \(\frac{1}{10}\int_0^{10}E(t)\,\text{d}t=2\,\text{kg/h}\). 4. c) During the first half, \(\int_0^5E(t)\,\text{d}t=\frac{95}{16}=5.9375\,\text{kg}\). During the second half, \(\int_5^{10}E(t)\,\text{d}t=\frac{225}{16}=14.0625\,\text{kg}\). Therefore, substantially more pollutant is released during the second half.

a) \(20\,\text{kg}\) b) \(2\,\text{kg/h}\) c) First \(5\) hours: \(5.9375\,\text{kg}\) Final \(5\) hours: \(14.0625\,\text{kg}\)

- Integrate the transfer rate to find the accumulated data. - Account for the inner derivative when integrating the sine function. - Add the accumulated data to the initial stored amount for part b).

1. An antiderivative of the data-transfer rate is \(G(t)=-\frac{160}{\pi}\cos\left(\frac{\pi}{4}t\right)+50t\). 2. a) \(\int_0^4g(t)\,\text{d}t=\frac{320}{\pi}+200\approx301.86\,\text{MB}\). 3. b) \(\int_0^6g(t)\,\text{d}t=300+\frac{160}{\pi}\approx350.93\,\text{MB}\). Adding the initial amount gives \(200+350.93\approx550.93\,\text{MB}\).

a) \(200+\frac{320}{\pi}\,\text{MB}\approx301.86\,\text{MB}\) b) \(500+\frac{160}{\pi}\,\text{MB}\approx550.93\,\text{MB}\)

- The sign of a rate indicates whether the quantity is increasing or decreasing. - Integrate the rate to find the net change. - Use radian measure when evaluating the trigonometric functions.

1. At \(t=3\), \(v(3)=20\cos\left(\frac{3\pi}{4}\right) =-10\sqrt{2} \approx-14.14\). The negative sign means the volume in the upper reservoir is decreasing at that time. 2. An antiderivative is \(\frac{80}{\pi}\sin\left(\frac{\pi}{4}t\right)\). Therefore, \(\int_2^6v(t)\,\mathrm{d}t =\frac{80}{\pi}\left(\sin\left(\frac{3\pi}{2}\right)-\sin\left(\frac{\pi}{2}\right)\right) =-\frac{160}{\pi} \approx-50.93\). The reservoir loses approximately \(50.93\times10^3\,\text{m}^3\) of water from \(t=2\) to \(t=6\).

a) \(v(3)=-10\sqrt{2}\approx-14.14\) thousand \(\text{m}^3/\text{h}\); the reservoir volume is decreasing. b) \(\int_2^6v(t)\,\mathrm{d}t=-\frac{160}{\pi}\approx-50.93\) thousand \(\text{m}^3\); the reservoir loses approximately \(50.93\times10^3\,\text{m}^3\).

- Interpret a negative rate as a decrease. - Integrate the rate over the full time interval to find the net change. - Compare the cosine values at the two endpoints.

1. At \(t=5\), \(k(5)=-0.5\sin\left(\frac{\pi}{2}\right)=-0.5\). The concentration is decreasing at a rate of \(0.5\,\text{mol}/(\text{L}\cdot\text{min})\). 2. An antiderivative is \(\frac{5}{\pi}\cos\left(\frac{\pi}{10}t\right)\). Therefore, \(\int_0^{20}k(t)\,\mathrm{d}t =\frac{5}{\pi}\left(\cos(2\pi)-\cos(0)\right) =0\). The net change in concentration over the \(20\) minutes is zero, so the final concentration equals the initial concentration.

a) \(k(5)=-0.5\,\text{mol}/(\text{L}\cdot\text{min})\); the concentration is decreasing. b) \(\int_0^{20}k(t)\,\mathrm{d}t=0\); the final concentration equals the initial concentration.

- Use the zeros and sign of the rate function. - A definite integral of a rate gives net change. - Add accumulated change to the initial amount. - A positive-to-negative rate change produces a local maximum of the accumulated quantity.

1. The zeros of \(f(t)\) are \(t=2\) and \(t=10\). Since the parabola opens downward, \(f(t)<0\) on \([0,2)\) and \((10,12]\), and \(f(t)>0\) on \((2,10)\). 2. The integral from \(4\) to \(9\) is the net change in the grain amount, in metric tons, during that time interval. 3. \(M(10)=100+\int_0^{10}(-t^2+12t-20)\,\text{d}t\). 4. The integral is \(\left[-\frac{t^3}{3}+6t^2-20t\right]_0^{10}=\frac{200}{3}\), so \(M(10)=\frac{500}{3}\approx166.67\) metric tons. 5. At \(t=10\), the rate changes from positive to negative, so the amount changes from increasing to decreasing and has a local maximum.

a) Removed on \([0,2)\) and \((10,12]\); added on \((2,10)\) b) The net change in the grain amount from hour \(4\) to hour \(9\) c) \(\frac{500}{3}\) metric tons, or approximately \(166.67\) metric tons d) The rate changes from positive to negative at \(t=10\).

- Factor the rate to make a sign chart. - Add the accumulated rate to the initial volume. - The sign of the total integral determines whether the ending volume is above or below the initial volume.

1. Factor the rate: \(v(t)=0.2t(t-8)(t-10)\). The sign is positive on \((0,8)\) and negative on \((8,10)\). 2. \(V(8)=500+\int_0^80.2(t^3-18t^2+80t)\,\text{d}t\). 3. An antiderivative is \(0.2\left(\frac{t^4}{4}-6t^3+40t^2\right)\). Its value at \(8\) is \(102.4\), so \(V(8)=602.4\,\text{m}^3\). 4. The net change from \(0\) to \(10\) is \(100\,\text{m}^3\), so \(V(10)=600\,\text{m}^3\), greater than the initial volume.

a) \(v(t)=0.2t(t-8)(t-10)\), which has the stated signs b) \(602.4\,\text{m}^3\) c) Greater; \(V(10)=600\,\text{m}^3\)

- Use the maximum value of sine. - Energy is the integral of power over time. - Compare the interval length with the period. - Scale one-system, one-shift energy by the numbers of systems and shifts.

1. The sine term is greatest when its argument is \(\frac{\pi}{2}\). This gives \(t=6\), and \(P(6)=35\,\text{kW}\). 2. The interval length \(12\) is one full period, so the sinusoidal term has integral zero. Therefore, \(\int_0^{12}P(t)\,\text{d}t=20(12)=240\,\text{kWh}\). 3. For \(8\) systems and \(2\) shifts, the total is \(240\cdot8\cdot2=3840\,\text{kWh}\).

a) \(35\,\text{kW}\) at \(t=6\,\text{h}\) b) \(240\,\text{kWh}\) c) \(3840\,\text{kWh}\)

- Integrate the inflow rate and add the initial reservoir volume. - Use the derivative of the rate to locate its maximum. - Factor the rate function to find when it is zero.

1. a) The accumulated inflow through day \(12\) is \(\int_0^{12}(-0.25t^4+4t^3)\,\text{d}t=8294.4\,\text{m}^3\). Therefore, the reservoir contains \(5000+8294.4=13{,}294.4\,\text{m}^3\). 2. b) Differentiate the rate: \(f'(t)=t^2(12-t)\). The derivative is positive before \(t=12\) and negative after \(t=12\), so the greatest inflow rate occurs at \(t=12\). The maximum rate is \(f(12)=1728\,\text{m}^3/\text{day}\). 3. c) Factor \(f(t)=0.25t^3(16-t)\). Besides \(t=0\), the rate is zero at \(t=16\). At that time, the modeled inflow stops.

a) \(13{,}294.4\,\text{m}^3\) b) At \(t=12\) days; \(1728\,\text{m}^3/\text{day}\) c) At \(t=16\) days; the modeled inflow stops.

- An integral of a rate gives the net change. - Compare the numerator with the derivative of the denominator. - Interpret a negative accumulated change as a loss from the tank.

1. Since the numerator is a constant multiple of the derivative of the denominator, an antiderivative is \(-30\ln(t^2+36)\). 2. Therefore, \(\int_0^8f(t)\,\mathrm{d}t =\left[-30\ln(t^2+36)\right]_0^8 =-30\ln\left(\frac{25}{9}\right) \approx-30.65\). The volume in the tank decreases by approximately \(30.65\,\text{L}\) during the first \(8\) hours. 3. The inequality states that the net volume change is less than \(-40\,\text{L}\), meaning more than \(40\,\text{L}\) has leaked out.

a) \(-30\ln\left(\frac{25}{9}\right)\approx-30.65\,\text{L}\); approximately \(30.65\,\text{L}\) leaks out. b) “After how many hours has the tank lost more than \(40\,\text{L}\) of coolant?”

- Integrate the inflow rate to find accumulated volume. - Compare the numerator with the derivative of the denominator. - Use the requested time interval as the limits of integration.

1. Because the numerator is a constant multiple of the derivative of the denominator, an antiderivative is \(H(t)=50\ln\left((t+2)^2+16\right)\). 2. For the first \(6\) hours, \(\int_0^6h(t)\,\mathrm{d}t =50\ln(80)-50\ln(20) =50\ln4 \approx69.31\). Thus approximately \(69.31\,\text{m}^3\) flows into the basin. 3. From \(t=2\) to \(t=4\), \(\int_2^4h(t)\,\mathrm{d}t =50\ln(52)-50\ln(32) =50\ln\left(\frac{13}{8}\right) \approx24.28\). This is the volume that flows into the basin between the second and fourth hours.

a) \(50\ln4\approx69.31\,\text{m}^3\) b) \(50\ln\left(\frac{13}{8}\right)\approx24.28\,\text{m}^3\); this is the inflow from \(t=2\) to \(t=4\).

- Express the given work as a definite integral containing the unknown spring constant. - Convert centimeters to meters. - After finding the spring constant, substitute the requested stretch into Hooke's law.

1. a) Convert the limits to meters. The work equation is \(0.72=\int_{0.04}^{0.08}ks\,\text{d}s=\frac{k}{2}(0.08^2-0.04^2)\). 2. Since \(\frac{1}{2}(0.08^2-0.04^2)=0.0024\), \(k=\frac{0.72}{0.0024}=300\,\text{N/m}\). 3. b) At \(s=0.15\,\text{m}\), \(F(0.15)=300(0.15)=45\,\text{N}\).

a) \(k=300\,\text{N/m}\) b) \(45\,\text{N}\)

- Use the force at \(s=0\) to determine one constant immediately. - The area under a force-versus-displacement graph represents work. - Set the definite integral equal to the given work. - Solve the resulting linear equation.

1. Since \(F(0)=d=10\), \(d=10\,\text{N}\). 2. Use the work condition: \(80=\int_0^4(ks+10)\,\text{d}s=\left[\frac{k}{2}s^2+10s\right]_0^4=8k+40\). 3. Solving \(8k+40=80\) gives \(k=5\,\text{N/m}\).

\(k=5\,\text{N/m}\) and \(d=10\,\text{N}\)

- Determine the starting and ending distances from the Moon's center. - Work is the integral of the position-dependent force. - Convert all distances to meters. - Use the antiderivative of \(r^{-2}\).

1. The distance from the Moon's center changes from \(r_1=1.738\times10^6\,\text{m}\) to \(r_2=3.476\times10^6\,\text{m}\). 2. The required work is \(W=\int_{r_1}^{r_2}\frac{GMm}{r^2}\,\text{d}r=GMm\left[ -\frac{1}{r}\right]_{r_1}^{r_2}=GMm\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\). 3. Since \(r_2=2r_1\), \(W=\frac{GMm}{2r_M}\). 4. Substitution gives \(W\approx5.643\times10^8\,\text{J}\), or approximately \(564.3\,\text{MJ}\).

\(W\approx5.643\times10^8\,\text{J}\), or \(564.3\,\text{MJ}\)

- Integrate the rate using a negative exponent. - Use the initial condition first. - Use the one-hour measurement to determine \(k\). - Substitute the constant back into the mass function.

1. Integrating the rate gives \(m(t)=-\frac{k}{t+1}+C\). 2. Since \(m(0)=0\), \(-k+C=0\), so \(C=k\). 3. Since \(m(1)=2\), \(-\frac{k}{2}+k=2\), so \(k=4\). 4. Therefore, \(m(t)=4-\frac{4}{t+1}\).

\(k=4\) and \(m(t)=4-\frac{4}{t+1}\)

- Multiply the units represented by a grid square's width and height. - The rate's greatest and least values indicate the fastest increase and decrease. - Use the sign of the rate to determine when inventory is increasing. - Integrate the rate to find the net inventory change.

1. Each grid square is \(1\,\text{h}\) wide and \(2\) pallets per hour high, so it represents \(2\) pallets. 2. Inventory increases fastest when \(f\) is greatest. The vertex occurs at \(t=5\), where \(f(5)=9\). Inventory decreases fastest when \(f\) is least. On the interval, the minimum value is \(f(0)=f(10)=-16\), so this occurs at \(t=0\) and \(t=10\). Because \(f(t)>0\) for \(2<t<8\), inventory increases throughout that interval and is greatest at \(t=8\). 3. Expanding gives \(f(t)=-t^2+10t-16\), with antiderivative \(F(t)=-\frac{1}{3}t^3+5t^2-16t\). Therefore, \(\int_2^8f(t)\,\text{d}t=36\). The warehouse gains a net total of \(36\) pallets from \(t=2\) to \(t=8\).

a) \(2\) pallets b) Increasing fastest at \(t=5\); decreasing fastest at \(t=0\) and \(t=10\); greatest inventory on \([2,8]\) at \(t=8\) c) \(36\) pallets, the net inventory gain from \(t=2\) to \(t=8\)

- The integral of power over time gives energy. - Determine the energy represented by one grid square. - Shift the variable by \(6\) when applying the given antiderivative formula. - Compare the exact decimal value with your graphical estimate.

1. The integral is the electrical energy generated from hour \(4\) through hour \(8\), measured in kilowatt-hours. 2. One reasonable graphical estimate is about \(250\,\text{kWh}\). For example, counting approximately \(25\) grid squares at \(10\,\text{kWh}\) per square gives this estimate. 3. An antiderivative is \(P(t)=160\arctan\left(0.5(t-6)\right)\). Therefore, \(\int_4^8p(t)\,\text{d}t=160\left(\arctan(1)-\arctan(-1)\right)=80\pi\,\text{kWh}\approx251.33\,\text{kWh}\), which is close to the graphical estimate.

a) The energy generated from \(t=4\) to \(t=8\), in kilowatt-hours b) A reasonable estimate is approximately \(250\,\text{kWh}\). c) \(80\pi\,\text{kWh}\approx251.33\,\text{kWh}\), which agrees closely with the estimate.

- Integrating an arrival rate gives the accumulated number of arrivals. - Use the Fundamental Theorem of Calculus for the first \(2\) hours. - Set the accumulation function equal to \(2160\). - Compare the arrival rate with the processing capacity.

1. \(I_0(x)\) is the total number of visitors who arrive from opening until time \(x\). 2. An antiderivative is \(-10t^3+120t^2\). Thus, \(I_0(2)=\left[-10t^3+120t^2\right]_0^2=400\). 3. Solve \(-10x^3+120x^2=2160\). The solution in \(0\le x\le8\) is \(x=6\). 4. Set the arrival rate equal to capacity: \(-30t^2+240t=360\). This gives \((t-2)(t-6)=0\), so the arrival rate exceeds capacity from \(t=2\) to \(t=6\). After \(t=6\), the arrival rate drops below capacity, so the waiting line can begin to shrink.

a) The total number of visitors arriving from opening through time \(x\) b) \(400\) visitors c) \(x=6\) hours d) After \(t=6\) hours

- Accumulated production and shipping are areas under their rate graphs. - Break the regions into triangles and rectangles. - Subtract accumulated shipping from accumulated production. - Include the initial inventory of \(20\) machines.

1. From \(t=0\) to \(t=4\), production is the triangular area \(\frac{1}{2}(4)(8)=16\) machines, and shipping is \(\frac{1}{2}(4)(2)=4\) machines. Thus, inventory is \(20+16-4=32\) machines. 2. From \(t=0\) to \(t=8\), production is \(16+(4)(8)=48\) machines, and shipping is \(\frac{1}{2}(8)(4)=16\) machines. Thus, inventory is \(20+48-16=52\) machines. 3. From \(t=0\) to \(t=12\), production is \(48+\frac{1}{2}(4)(8)=64\) machines, and shipping is \(\frac{1}{2}(12)(6)=36\) machines. Thus, inventory is \(20+64-36=48\) machines.

After \(4\) hours: \(32\) machines After \(8\) hours: \(52\) machines After \(12\) hours: \(48\) machines

- Relate the sign of power to whether stored energy is increasing or decreasing. - A minimum can occur where power changes from negative to positive. - Interpret signed area under a power graph as a change in energy. - Use both axis scales to determine the energy represented by one grid square.

1. a) The power is negative on \([0,5)\) and positive on \((5,8]\). Therefore, stored energy decreases until \(t=5\) and increases afterward, so its minimum occurs at \(t=5\,\text{h}\). 2. b) The change in stored energy is \(\int_0^5p(t)\,\text{d}t\). Each grid square represents \((1\,\text{h})(10\,\text{kW})=10\,\text{kWh}\). The region below the axis from \(0\) to \(5\) has an area of about \(10\) squares, so the energy changes by about \(-100\,\text{kWh}\). 3. Thus, the stored energy is approximately \(400-100=300\,\text{kWh}\).

a) \(t=5\,\text{h}\), because the power changes from negative to positive there. b) Approximately \(300\,\text{kWh}\)

- Interpret each accumulated number as area under the rate graph. - Break the full region into two triangles and a rectangle. - For the middle interval, subtracting the two small end regions may be efficient. - On \([0,1]\), use the first rate formula inside the accumulation integral.

1. a) The total area is two triangles and one rectangle: \(\frac12(1)(20)+(1)(20)+\frac12(1)(20)=40\). Therefore, \(40\) bicycles are returned. 2. b)(1) The first-hour region is a triangle, so \(\frac12(1)(20)=10\) bicycles. 3. b)(2) From \(1.5\) to \(2\), the area is \((0.5)(20)=10\). From \(2\) to \(3\), the triangular area is \(10\). The total is \(20\) bicycles. 4. b)(3) The two excluded end triangles each have area \(\frac12(0.5)(10)=2.5\). Thus, \(40-2(2.5)=35\) bicycles. 5. c) For \(0\le t\le1\), \(B(t)=\int_0^t20s\,\text{d}s=10t^2\).

a) \(40\) bicycles b) (1) \(10\) bicycles (2) \(20\) bicycles (3) \(35\) bicycles c) \(B(t)=10t^2\) for \(0\le t\le1\)

- Flow accumulated over time is represented by area under the graph. - Divide the region into triangles, rectangles, or trapezoids. - First find half of the total volume. - Locate the interval in which the cumulative volume reaches that value, then use the constant rate on that interval.

1. a) The total area consists of a triangle on \([0,1]\), a rectangle on \([1,2]\), and a triangle on \([2,4]\): \(\frac12(1)(40)+(1)(40)+\frac12(2)(40)=20+40+40=100\,\text{L}\). 2. b) From \(1\) to \(2\), the area is \(40\,\text{L}\). From \(2\) to \(3\), the trapezoid area is \(\frac{40+20}{2}(1)=30\,\text{L}\). Thus, the volume is \(70\,\text{L}\). 3. c) Half of the total volume is \(50\,\text{L}\). By \(t=1\), \(20\,\text{L}\) has been transferred, and by \(t=2\), \(60\,\text{L}\) has been transferred, so \(m\in[1,2]\). On this interval, the rate is \(40\,\text{L/min}\). Solve \(20+40(m-1)=50\), which gives \(m=1.75\,\text{min}\).

a) \(100\,\text{L}\) b) \(70\,\text{L}\) c) \(m=1.75\,\text{min}\)

- Use geometric area for accumulated numbers of visitors. - Keep instantaneous rate and accumulated arrivals distinct; they have different units. - Build the cumulative function separately on each interval where the rate formula changes. - Check that the cumulative formulas agree at \(t=2\) and \(t=4\).

1. a)(1) The triangular area from \(0\) to \(1\) is \(\frac12(1)(20)=10\) visitors. 2. a)(2) The triangular area from \(0\) to \(2\) is \(\frac12(2)(40)=40\) visitors. 3. a)(3) From \(2\) to \(3\), the trapezoid area is \(\frac{40+20}{2}(1)=30\). Thus, \(40+30=70\) visitors arrive during the first \(3\) hours. 4. a)(4) Subtract the first-hour accumulation from the first-three-hour accumulation: \(70-10=60\) visitors. 5. a)(5) The last-hour region is a triangle with area \(\frac12(1)(20)=10\) visitors. 6. a)(6) \(r(2)=40\,\text{people/h}\). This is a rate, not an accumulated number of people. 7. b) For \(t<0\), \(A(t)=0\). For \(0\le t<2\), \(A(t)=\int_0^t20s\,\text{d}s=10t^2\). For \(2\le t\le4\), \(A(t)=40+\int_2^t(-20s+80)\,\text{d}s=-10t^2+80t-80\). For \(t>4\), \(A(t)=80\).

a) (1) \(10\) visitors (2) \(40\) visitors (3) \(70\) visitors (4) \(60\) visitors (5) \(10\) visitors (6) \(40\,\text{people/h}\) b) \(A(t)=\begin{cases}0 & t<0\\10t^2 & 0\le t<2\\-10t^2+80t-80 & 2\le t\le4\\80 & t>4\end{cases}\)

- What two conditions must a probability density function satisfy? - An interval probability is an integral of the density. - The cumulative distribution function is the accumulated area from the left endpoint to \(x\). - For a continuous random variable, the mean is \(\int x f(x)\,dx\).

1. a) On \([0, 3]\), \(f(x)=\frac{4}{27}x^2(3-x)\ge 0\), and it is \(0\) outside that interval. Also, \(\int_0^3\left(\frac{4}{9}x^2-\frac{4}{27}x^3\right)\,dx=\left[\frac{4}{27}x^3-\frac{1}{27}x^4\right]_0^3=4-3=1\). Thus, \(f\) is a valid density. 2. b) \(P(X\le 1.5)=\left[\frac{4}{27}x^3-\frac{1}{27}x^4\right]_0^{3/2}=\frac{5}{16}=0.3125\). 3. c) For \(0\le x\le 3\), \(F(x)=\int_0^x f(t)\,dt=\frac{4}{27}x^3-\frac{1}{27}x^4\). 4. d) \(E(X)=\int_0^3 x f(x)\,dx=\left[\frac{1}{9}x^4-\frac{4}{135}x^5\right]_0^3=1.8\) hours. Since \(0.8\cdot 60=48\), this is \(1\) hour \(48\) minutes.

a) \(f(x)\ge 0\) and \(\int_{-\infty}^{\infty}f(x)\,dx=1\). b) \(P(X\le 1.5)=\frac{5}{16}=0.3125\), or \(31.25\%\) c) \(F(x)=\frac{4}{27}x^3-\frac{1}{27}x^4\) for \(0\le x\le 3\) d) \(E(X)=1.8\) hours, or \(1\) hour \(48\) minutes

- Locate the maximum and the line of symmetry to find the mean. - The inflection points are one standard deviation from the mean. - Interpret the probability as area under the density curve. - For a concave-down graph, compare the curve with the line segment joining two points on it. - Use the endpoint heights to calculate a trapezoid area.

1. The curve reaches its maximum at \(x=5\), so \(\mu=5\). Its inflection points are at \(x=3\) and \(x=7\), so \(\sigma=2\). 2. On \([5, 7]\), the density curve is concave down, so the line segment joining \((5, f(5))\) and \((7, f(7))\) lies below the curve. From the graph, \(f(5)\approx0.20\) and \(f(7)\approx0.12\). The trapezoid below the curve has area approximately \(\frac{0.20+0.12}{2}\cdot(7-5)=0.32\). Therefore, the probability is greater than about \(0.32\), and in particular \(P(5\le X\le7)>0.3\).

1. \(\mu=5\) and \(\sigma=2\) 2. Using graph heights of about \(0.20\) and \(0.12\), the trapezoid area is about \(0.32\), so \(P(5\le X\le7)>0.3\).

- Use the sign of the rate to decide whether stored mass is increasing or decreasing. - A stored quantity is greatest when its rate changes from positive to negative. - Integrate the polynomial term by term. - Compare the signed areas on either side of \(t=4\).

1. \(f(2)=6\) means that after \(2\) minutes, recyclable material is being added at an instantaneous rate of \(6\,\text{kg/min}\). 2. The rate is positive before \(t=4\) and negative after \(t=4\). Therefore, the mass increases until \(t=4\) and decreases afterward, so it is greatest at \(t=4\). 3. An antiderivative is \(F(t)=\frac{1}{16}t^4-t^3+4t^2\). Thus, \(\int_0^4f(t)\,\text{d}t=16\). The hopper gains \(16\,\text{kg}\) during the first \(4\) minutes. 4. The graph is symmetric under a \(180^\circ\) rotation about \((4,0)\). The positive signed area on \([0,4]\) and the negative signed area on \([4,8]\) cancel, so the net change is \(0\,\text{kg}\).

a) At \(t=2\), material is being added at \(6\,\text{kg/min}\). b) \(t=4\) minutes c) \(16\,\text{kg}\), the net gain during the first \(4\) minutes d) \(0\,\text{kg}\)

- Stored energy increases where net power is positive and decreases where it is negative. - Integrate the cosine function using the chain rule. - Add the accumulated energy change to the initial stored energy.

1. The power is positive on \(0<t<2\) and negative on \(2<t<6\), so stored energy changes from increasing to decreasing at \(t=2\). The later positive accumulation from \(t=6\) to \(t=8\) does not recover the earlier loss, so the greatest stored energy occurs at \(t=2\). 2. An antiderivative is \(P(t)=3\sin\left(\frac{\pi}{4}t\right)\). Therefore, \(\int_2^6p(t)\,\text{d}t=-6\). The storage unit has a net energy loss of \(6\,\text{kWh}\) from \(t=2\) to \(t=6\). 3. The accumulated change from \(t=0\) to \(t=6\) is \(-3\,\text{kWh}\). Thus, the stored energy after \(6\) hours is \(15-3=12\,\text{kWh}\).

a) \(t=2\) hours b) \(-6\,\text{kWh}\), a net energy loss of \(6\,\text{kWh}\) c) \(12\,\text{kWh}\)

- Use the sign of \(f(t)\) to decide when attendance is increasing or decreasing. - The accumulated change in attendance is a definite integral of the net entry rate. - Add the initial attendance after finding the accumulated change. - For an absolute maximum on a closed interval, compare critical-point and endpoint values.

1. a) Since \(f(t)=-3(t-2)(t-10)\), the rate is negative on \([0,2)\) and \((10,12]\). Therefore, net departures occur from noon to 2:00 p.m. and from 10:00 p.m. to midnight. 2. b) The accumulated attendance change by \(t=6\) is \(\int_0^6(-3t^2+36t-60)\,\text{d}t=72\) people. Thus, attendance is \(200+72=272\) people. 3. c) The rate is negative before \(t=2\), positive on \((2,10)\), and negative after \(t=10\). Therefore, attendance decreases, then increases, then decreases, so a maximum occurs at \(t=10\). 4. The relevant values are \(A(0)=200\), \(A(10)=400\), and \(A(12)=344\). Therefore, the absolute maximum is \(400\) attendees.

a) From noon to 2:00 p.m. and from 10:00 p.m. to midnight b) \(272\) attendees c) The rate changes from positive to negative at 10:00 p.m.; the absolute maximum is \(400\) attendees.

- Distinguish between a rate decreasing and a stored amount decreasing. - Total supplied volume is the area under a flow-rate curve. - Compare complete areas, not only the rates near the end. - Evaluate or estimate the integral for pipe \(B\).

1. Lucas’s statement is false. All three flow rates are nonnegative on \(0\le t\le60\), so the amount of water supplied continues to increase. A decreasing curve means only that water is entering more slowly. 2. Mia’s statement is true. Total volume is the area under the flow-rate curve. Curve \(B\) has the greatest area. 3. Jonah’s statement is false. The total volumes are \(\int_0^{60}a(t)\,\mathrm{d}t=\frac{896}{9}\approx99.56\,\text{m}^3\) and \(\int_0^{60}c(t)\,\mathrm{d}t=\frac{256}{3}\approx85.33\,\text{m}^3\). Thus pipe \(A\) supplies more water overall. 4. Emma’s statement is true: \(\int_0^{60}0.005t(60-t)\,\mathrm{d}t=180\,\text{m}^3\).

1. False; a positive but decreasing rate still adds water. 2. True; curve \(B\) has the greatest area under it. 3. False; \(A\) supplies approximately \(99.56\,\text{m}^3\), while \(C\) supplies approximately \(85.33\,\text{m}^3\). 4. True; pipe \(B\) supplies exactly \(180\,\text{m}^3\).

- A negative net flow rate makes the stored volume decrease. - Possible extrema occur where the rate changes sign or at an endpoint. - Use signed geometric areas to find changes in volume. - Add the total accumulated change to the initial volume.

1. a) The volume decreases when \(v(t)<0\), which occurs on \((3,8)\). 2. b) The volume increases until \(t=3\), decreases until \(t=8\), and then increases. The positive area from \(t=0\) to \(t=3\) is \(5\,\text{L}\), while the negative area from \(t=3\) to \(t=8\) is \(-7\,\text{L}\). Therefore, the accumulated volume at \(t=8\) is below its initial value, so the absolute minimum occurs at \(t=8\,\text{min}\). 3. c) From \(0\) to \(5\), the signed areas are \(4\), \(1\), \(-1\), and \(-2\) liters. Their sum is \(4+1-1-2=2\,\text{L}\). 4. d) Over the full \(10\) minutes, the signed area is \(5-7+2=0\,\text{L}\). Thus, the final volume is \(50+0=50\,\text{L}\).

a) \((3,8)\) minutes b) \(t=8\,\text{min}\) c) An increase of \(2\,\text{L}\) d) \(50\,\text{L}\)

- Use the graph’s axis scales when reading values. - An integral of liters per minute over minutes gives liters. - Use the antiderivative of \(\sin(kt)\). - Recall the cosine values at \(0\) and \(\pi\).

1. From the graph, \(r(0)=15\,\text{L/min}\), \(r(3)=25\,\text{L/min}\), and \(r(6)=15\,\text{L/min}\). 2. The area under the rate graph represents the total volume entering the tank during the first \(6\) minutes. 3. The volume is \(\int_0^6\left(10\sin\left(\frac{\pi}{6}t\right)+15\right)\,\mathrm{d}t\). An antiderivative is \(-\frac{60}{\pi}\cos\left(\frac{\pi}{6}t\right)+15t\). Therefore, \(V=\left[-\frac{60}{\pi}\cos\left(\frac{\pi}{6}t\right)+15t\right]_0^6 =90+\frac{120}{\pi} \approx128.20\,\text{L}\).

a) \(r(0)=15\,\text{L/min}\), \(r(3)=25\,\text{L/min}\), and \(r(6)=15\,\text{L/min}\) b) The total volume entering during the first \(6\) minutes c) \(90+\frac{120}{\pi}\approx128.20\,\text{L}\)

- Use signed area under the net-rate graph. - A negative net arrival rate means the number of visitors inside is decreasing. - Compare the positive and negative geometric areas. - Distinguish net change from final attendance.

1. From \(t=0\) to \(t=6\), the net increase in the number of visitors inside is the area above the axis: \(\frac12(2)(4)+(2)(4)+\frac12(2)(4)=16\) visitors. Thus, the number inside increases by \(16\), so statement a) is false. 2. On \(6<t<10\), \(v(t)<0\), so more visitors leave than enter and the number inside decreases. Statement b) is true. 3. The area below the axis from \(t=6\) to \(t=10\) has magnitude \(\frac12(4)(2)=4\) visitors. The net change is \(16-4=12\) visitors, which is positive. Statement c) is true. 4. The integral gives the change in the number of visitors, not the final number. The initial attendance would have to be added. Statement d) is false.

a) False b) True c) True d) False

- Use the sign of the net inventory rate to determine when inventory increases or decreases. - The integral of the rate is the accumulated inventory change. - Compute the signed areas as rectangles and triangles. - After \(t=8\), use the constant shipping surplus to determine when the remaining pallets are gone.

1. a) The inventory increases while \(r(t)>0\) and decreases while \(r(t)<0\). The rate changes from positive to negative at \(t=6\), so the greatest inventory occurs at \(t=6\,\text{h}\). 2. b) The signed areas from \(0\) to \(10\) are \((4)(4)=16\), \(\frac12(2)(4)=4\), \(-\frac12(2)(4)=-4\), and \((2)(-4)=-8\), all in pallets. Thus, \(I(10)=16+4-4-8=8\) pallets. 3. c) For \(t\ge8\), the net rate is \(-4\) pallets per hour. Starting from \(I(10)=8\), \(I(t)=8-4(t-10)\). Solving \(8-4(t-10)=0\) gives \(t=12\,\text{h}\).

a) \(t=6\,\text{h}\) b) An increase of \(8\) pallets c) \(t=12\,\text{h}\)

- The stored amount is constant when the net rate is zero. - Use the signs of the rate to determine when stored data increases or decreases. - Compute the signed areas of the triangles and rectangles.

1. The amount of stored data remains constant when \(q(t)=0\), which occurs from \(t=4\) to \(t=6\). 2. The rate is positive before \(t=4\), zero from \(t=4\) to \(t=6\), and negative after \(t=6\). Therefore, the stored amount is greatest throughout \(4\le t\le6\). 3. The positive signed area from \(t=0\) to \(t=4\) is \(2\cdot2+\frac12\cdot2\cdot2=6\,\text{MB}\). The negative signed area from \(t=6\) to \(t=12\) is \(-\left(\frac12\cdot2\cdot2+2\cdot2+\frac12\cdot2\cdot2\right)=-8\,\text{MB}\). 4. The total change is \(6\,\text{MB}-8\,\text{MB}=-2\,\text{MB}\). The system stores \(2\,\text{MB}\) less data than it did initially.

a) The stored amount is constant and greatest from \(t=4\) to \(t=6\) minutes. b) \(-2\,\text{MB}\); the system stores \(2\,\text{MB}\) less than at the beginning.

- Multiply the units represented by a grid square’s width and height. - Treat area below the time axis as negative. - The stored volume is greatest when the rate changes from positive to negative.

1. One grid square is \(1\,\text{h}\) wide and \(1\,\text{m}^3/\text{h}\) high, so it represents \(1\,\text{h}\cdot1\,\frac{\text{m}^3}{\text{h}}=1\,\text{m}^3\). 2. The positive area between the zeros \(t=0.5\) and \(t=5.5\) is a trapezoid: \(\frac{1}{2}\cdot(5+2)\cdot3=10.5\,\text{m}^3\). The two negative triangles have combined area \(2\cdot\left(\frac{1}{2}\cdot0.5\cdot1\right)=0.5\,\text{m}^3\). Thus the net change is \(10.5\,\text{m}^3-0.5\,\text{m}^3=10\,\text{m}^3\). 3. The volume increases while \(f(t)>0\), from \(t=0.5\) to \(t=5.5\), and decreases afterward. Therefore, the volume is greatest at \(t=5.5\) hours.

a) \(1\,\text{m}^3\) b) \(10\,\text{m}^3\) c) \(t=5.5\) hours

- A definite integral of the inflow rate gives the accumulated volume. - Integration by parts is useful for a product of a linear term and an exponential term. - Locate the highest point of the rate graph. - Remember that the rate is measured in thousands of cubic meters per hour.

1. An antiderivative is \(F(t)=-20(t+10)e^{-0.1t}\), found by integration by parts. 2. a) \(\int_0^{10}2te^{-0.1t}\,\text{d}t=200-400e^{-1}\approx52.85\). Therefore, approximately \(52.85\times10^3\,\text{m}^3\) enters during the first \(10\) hours. 3. b) The graph reaches its highest point at \(t=10\,\text{h}\). Algebraically, \(f'(t)=(2-0.2t)e^{-0.1t}\), which is zero at \(t=10\). 4. c) \(\int_0^{20}2te^{-0.1t}\,\text{d}t=200-600e^{-2}\approx118.80\). Thus, approximately \(118.80\times10^3\,\text{m}^3\) enters during the first \(20\) hours, which is more than twice the first \(10\)-hour amount.

a) \((200-400e^{-1})\times10^3\,\text{m}^3\approx52.85\times10^3\,\text{m}^3\) b) \(t=10\,\text{h}\) c) \((200-600e^{-2})\times10^3\,\text{m}^3\approx118.80\times10^3\,\text{m}^3\), which is more than twice the amount from part a).

- The area under a rate graph represents the accumulated volume. - Use the axis scales to determine the volume represented by one grid square. - Estimate the area with grid squares or trapezoids. - Multiply the estimated volume by the treatment cost per cubic meter.

1. The incoming volume is the area under the rate graph from \(t=4\) to \(t=12\). Each grid square represents \((1\,\text{h})(1\,\text{m}^3/\text{h})=1\,\text{m}^3\). 2. Counting grid squares or using a trapezoidal estimate gives approximately \(57\,\text{m}^3\). The integral is about \(57.04\,\text{m}^3\). 3. The estimated cost is \((57.04)(\$1.50)\approx\$85.56\), so \(\$86\) is a reasonable graph-based estimate.

Approximately \(\$86\), based on about \(57\,\text{m}^3\) of water

- Recover the volume function by integrating the net flow rate. - The volume stops increasing when the rate changes from positive to negative. - Returning to the initial level means the accumulated signed area is zero. - Use the initial volume to determine the constant of integration.

1. Since \(V'(t)=f(t)\), \(V(t)=-\frac{1}{3}t^3+3t^2+C\). The condition \(V(0)=80\) gives \(V(t)=-\frac{1}{3}t^3+3t^2+80\). 2. a) \(V(3)=-9+27+80=98\,\text{m}^3\). 3. b) Since \(f(t)=t(6-t)\), the rate changes from positive to negative at \(t=6\). Therefore, the maximum occurs at \(t=6\), and \(V(6)=-72+108+80=116\,\text{m}^3\). 4. c) Set \(V(t)=80\): \(-\frac{1}{3}t^3+3t^2=0\), so \(t^2\left(3-\frac{t}{3}\right)=0\). The first positive solution is \(t=9\,\text{h}\). 5. d) \(V(10)=-\frac{1000}{3}+300+80=\frac{140}{3}\,\text{m}^3\approx46.67\,\text{m}^3\).

a) \(98\,\text{m}^3\) b) At \(t=6\,\text{h}\); \(116\,\text{m}^3\) c) \(t=9\,\text{h}\) d) \(\frac{140}{3}\,\text{m}^3\approx46.67\,\text{m}^3\)

- Read the time coordinate of the highest point on the graph. - Substitute a time value into the concentration function to find the concentration. - Evaluate the antiderivative at the upper and lower limits and subtract. - Interpret the area under a concentration-time graph and include its units.

1. The graph reaches its highest point at about \(t=2\,\text{h}\). 2. The maximum concentration is \(f(2)=4\cdot2\cdot e^{-1}=8e^{-1}\approx2.94\,\text{mg/L}\). 3. After \(6\) hours, \(f(6)=24e^{-3}\approx1.19\,\text{mg/L}\). 4. Using the given antiderivative, \(\int_0^4 f(t)\,\text{d}t=F(4)-F(0)=-48e^{-2}+16=16-48e^{-2}\approx9.50\,\text{mg}\cdot\text{h}/\text{L}\). 5. The integral is the total drug exposure, measured as area under the concentration-time curve, during the first \(4\) hours.

a) The maximum occurs at \(t=2\,\text{h}\), and the maximum concentration is about \(2.94\,\text{mg/L}\). b) About \(1.19\,\text{mg/L}\) c) \(16-48e^{-2}\approx9.50\,\text{mg}\cdot\text{h}/\text{L}\). This is the total drug exposure during the first \(4\) hours.

- Power is the rate of change of energy. - Integrate the difference between production and consumption. - Add the accumulated net energy to the initial battery charge. - The battery charges where the production graph lies above the consumption graph.

1. The battery energy is \(E(t)=5+\int_0^t(p(\tau)-c(\tau))\,\text{d}\tau\). Using the given models, \(p(t)=-\frac{1}{9}t(t-12)\) and \(c(t)=1.5\). 2. a) \(\int_0^6\left(\frac{4}{3}t-\frac{1}{9}t^2-1.5\right)\,\text{d}t=7\,\text{kWh}\), so \(E(6)=5+7=12\,\text{kWh}\). 3. Also, \(\int_0^{12}\left(\frac{4}{3}t-\frac{1}{9}t^2-1.5\right)\,\text{d}t=14\,\text{kWh}\), so \(E(12)=5+14=19\,\text{kWh}\). 4. b) The battery charges when \(p(t)>c(t)\). Solving \(-\frac{1}{9}t^2+\frac{4}{3}t=1.5\) gives \(t=6\pm\frac{3\sqrt{10}}{2}\), approximately \(1.26\) and \(10.74\). Thus, it charges between those times.

a) At noon: \(12\,\text{kWh}\) At 6:00 p.m.: \(19\,\text{kWh}\) b) \(6-\frac{3\sqrt{10}}{2}<t<6+\frac{3\sqrt{10}}{2}\), approximately \(1.26<t<10.74\) hours after 6:00 a.m.

- Use the sign of the rate to determine when the stored volume rises or falls. - A minimum can occur when the rate changes from negative to positive. - Estimate the signed area between the graph and the time axis. - Add the estimated net change to the initial volume.

1. The rate is negative from \(t=0\) to \(t=2\), positive from \(t=2\) to \(t=8\), and negative again from \(t=8\) to \(t=10\). Thus the volume first decreases, then increases, and then decreases. The positive accumulation from \(t=2\) to \(t=8\) is greater than the later loss from \(t=8\) to \(t=10\), so the global minimum occurs at \(t=2\). 2. Estimating the signed area under the graph from \(t=0\) to \(t=8\) gives a net change of about \(5\,\text{L}\). Therefore, the estimated amount after \(8\) hours is \(100\,\text{L}+5\,\text{L}\approx105\,\text{L}\).

a) \(t=2\) hours b) Approximately \(105\,\text{L}\)

- Locate the intersection points of the two graphs. - A constant flow rate produces a rectangular area. - Compare the area where \(f\) is above \(g\) with the areas where \(g\) is above \(f\).

1. a) The first rate is greater where its graph lies above \(g=5\). The intersection times are approximately \(t=2.4\) and \(t=7.6\), so \(f(t)>g(t)\) between them. 2. b) The second pipe carries \((5\,\text{L/min})(10\,\text{min})=50\,\text{L}\). 3. c) The area under \(f\) is slightly larger than the rectangular area under \(g\). A numerical check gives \(\int_0^{10}f(t)\,\text{d}t\approx50.30\,\text{L}\), so the first pipe carries slightly more water.

a) Approximately \(2.4<t<7.6\) minutes b) \(50\,\text{L}\) c) The first pipe, with approximately \(50.30\,\text{L}\), carries slightly more water.

- Maximize the rate function by analyzing its derivative. - A definite integral of the net flow rate gives the net change in volume. - Add the accumulated change to the initial volume. - Distinguish signed accumulation from accumulation of the absolute rate.

1. a) The derivative is \(f'(t)=0.06t^2-1.2t+4.5\). Solving \(f'(t)=0\) gives \(t=5\) and \(t=15\). Since \(f''(5)=-0.6<0\), the rate is greatest at \(t=5\,\text{h}\). Its value is \(f(5)=5\,\text{m}^3/\text{h}\). 2. b) An antiderivative is \(F(t)=0.005t^4-0.2t^3+2.25t^2-5t\). Therefore, \(\int_2^8f(t)\,\text{d}t=F(8)-F(2)=22.08-(-2.52)=24.6\,\text{m}^3\). 3. c) Since \(\int_0^{10}f(t)\,\text{d}t=25\,\text{m}^3\), the volume is \(100+25=125\,\text{m}^3\). 4. d) The integral gives the total magnitude of the basin's net volume change during the \(15\) hours, counting both increases and decreases as positive amounts.

a) At \(t=5\,\text{h}\); \(5\,\text{m}^3/\text{h}\) b) A net increase of \(24.6\,\text{m}^3\) c) \(125\,\text{m}^3\) d) The total magnitude of the net volume change, with both increases and decreases counted positively.

- Find each line equation from two points on the graph. - Accumulated volume is geometric area under the rate graph. - Split part c at \(t=2\), where the line formula changes. - Compare the area accumulated by \(t=2\) with half of the full triangular area.

1. a) On \([0,2)\), the line through \((0,0)\) and \((2,8)\) has slope \(4\), so \(r(t)=4t\). On \([2,5]\), the line through \((2,8)\) and \((5,0)\) has slope \(-\frac{8}{3}\), so \(r(t)=-\frac{8}{3}t+\frac{40}{3}\). Outside \([0,5]\), \(r(t)=0\). 2. b) The area from \(0\) to \(1\) is a triangle: \(\frac12(1)(4)=2\,\text{L}\). 3. c) The trapezoid area from \(1.5\) to \(2\) is \(\frac{6+8}{2}(0.5)=3.5\,\text{L}\). From \(2\) to \(3\), the trapezoid area is \(\frac{8+\frac{16}{3}}{2}(1)=\frac{20}{3}\,\text{L}\). Therefore, the total is \(\frac{7}{2}+\frac{20}{3}=\frac{61}{6}\,\text{L}\approx10.17\,\text{L}\). 4. d) The total area is a triangle with base \(5\) and height \(8\), so the total volume is \(20\,\text{L}\). By \(t=2\), the accumulated area is only \(\frac12(2)(8)=8\,\text{L}\), which is less than half of \(20\,\text{L}\). Therefore, the halfway time is greater than \(2\) hours.

a) \(r(t)=\begin{cases}4t & 0\le t<2\\-\frac{8}{3}t+\frac{40}{3} & 2\le t\le5\\0 & \text{otherwise}\end{cases}\) b) \(2\,\text{L}\) c) \(\frac{61}{6}\,\text{L}\approx10.17\,\text{L}\) d) Only \(8\,\text{L}\) of the total \(20\,\text{L}\) has accumulated by \(t=2\), so the halfway time must be greater than \(2\) hours.

- An interval probability is area under the density curve. - The cumulative distribution function gives \(P(X\le x)\). - Treat the mean as an improper integral and use the supplied antiderivative hint. - Evaluate \(f(0)\) and consider what happens as \(x\) becomes large.

1. a) The graph begins at \(f(0)=0.2\), decreases for \(x\ge 0\), and approaches the x-axis as \(x\to\infty\). For reference, \(f(5)\approx 0.0736\), \(f(10)\approx 0.0271\), and \(f(15)\approx 0.0100\). 2. b) \(P(X\le 5)=\int_0^5 0.2e^{-0.2x}\,dx=[-e^{-0.2x}]_0^5=1-e^{-1}\approx 0.6321\). 3. c) \(F(x)=0\) for \(x<0\). For \(x\ge 0\), \(F(x)=\int_0^x0.2e^{-0.2t}\,dt=1-e^{-0.2x}\). 4. d) An antiderivative of \(0.2xe^{-0.2x}\) is \(-e^{-0.2x}(x+5)\). Therefore, \(\mu=\left[-e^{-0.2x}(x+5)\right]_0^\infty=5\) minutes.

a) The graph starts at \((0, 0.2)\), decreases, and approaches the x-axis. b) \(P(X\le 5)=1-e^{-1}\approx 0.6321\), or about \(63.21\%\) c) \(F(x)=\begin{cases}0 & x<0\\1-e^{-0.2x} & x\ge 0\end{cases}\) d) \(\mu=5\) minutes

- A probability density must be nonnegative and integrate to \(1\). - For a continuous random variable, the mean is found using \(\int x f(x)\,dx\). - Convert percentages to decimals before using them as integration limits. - The cumulative distribution function is an antiderivative of the density with value \(0\) at the left endpoint.

1. a) The function is nonnegative on \([0, 1]\), and \(\int_0^1(6x-6x^2)\,dx=[3x^2-2x^3]_0^1=1\). Thus, it is a density. 2. b) \(E(X)=\int_0^1x(6x-6x^2)\,dx=[2x^3-1.5x^4]_0^1=0.5\). On average, one-half, or \(50\%\), of the daylight is obscured by clouds. 3. c) \(P(0.25\le X\le 0.75)=[3x^2-2x^3]_{0.25}^{0.75}=0.84375-0.15625=0.6875\). 4. d) For \(0\le x\le 1\), \(F(x)=\int_0^x(6t-6t^2)\,dt=3x^2-2x^3\).

a) \(f(x)\ge 0\) and \(\int_0^1f(x)\,dx=1\) b) \(\mu=0.5\). On average, \(50\%\) of the daylight is obscured. c) \(P(0.25\le X\le 0.75)=0.6875\), or \(68.75\%\) d) \(F(x)=3x^2-2x^3\) for \(0\le x\le 1\)

- Determine the volume represented by one grid square. - Look for symmetry in the rate graph when comparing areas. - Accumulated volume is a definite integral of the feed rate. - Expand the product before finding an antiderivative.

1. a) Each grid square has area \((0.2\,\text{min})(1\,\text{L/min})=0.2\,\text{L}\). Estimating the area under the curve from \(0\) to \(0.5\) gives about \(1.6\,\text{L}\). The exact volume is \(1.5625\,\text{L}\). 2. b) The graph is symmetric about \(t=1\), so the area over \([0.5,1]\) equals the area over \([1,1.5]\). The rate is higher on \([1,1.5]\) than on \([1.5,2]\), so the volume on \([0.5,1]\) is greater. 3. c) The volume is \(\int_{0.8}^{1.2}(15t-7.5t^2)\,\text{d}t=\left[7.5t^2-2.5t^3\right]_{0.8}^{1.2}=6.48-3.52=2.96\,\text{L}\).

a) Approximately \(1.6\,\text{L}\) b) More volume is added on \([0.5,1]\). c) \(2.96\,\text{L}\)

- A probability density must be nonnegative and integrate to \(1\). - Find each interval probability by integrating the corresponding density. - For a continuous random variable, \(E(X)=\int x f(x)\,dx\). - The graph's area concentration can help you anticipate which model has the smaller mean.

1. a) Both functions are nonnegative on \([0, 2]\). Also, \(\int_0^2\frac{3}{8}x^2\,dx=1\) and \(\int_0^2\frac{3}{8}(x-2)^2\,dx=1\), so both are valid densities. 2. b) For SprintBus, \(P(X>1)=1-\int_0^1\frac{3}{8}x^2\,dx=1-\frac{1}{8}=\frac{7}{8}=0.875\). For SteadyLine, \(P(X>1)=\int_1^2\frac{3}{8}(x-2)^2\,dx=\frac{1}{8}=0.125\). 3. c) For SprintBus, \(E(X)=\int_0^2x\frac{3}{8}x^2\,dx=1.5\) minutes. For SteadyLine, \(E(X)=\int_0^2x\frac{3}{8}(x-2)^2\,dx=0.5\) minute. SteadyLine has the smaller mean wait time.

a) Both functions are nonnegative and integrate to \(1\) on \([0, 2]\). b) SprintBus: \(0.875\), or \(87.5\%\); SteadyLine: \(0.125\), or \(12.5\%\) c) SprintBus: \(1.5\) minutes; SteadyLine: \(0.5\) minute. SteadyLine is preferable when minimizing expected wait time.

- Decide whether the requested time lies before or after the peak. - It is simpler to use the volume that remains after time \(k\). - Express the small right-hand region as a triangle.

1. Half of the total area lies to the left of \(t=2\). Since \(8.75>5\), the time \(k\) lies in \([2,4]\). 2. Use the remaining volume: after time \(k\), \(10-8.75=1.25\,\text{L}\) remains to be transferred. The region to the right of \(k\) is a triangle with base \(4-k\). 3. On the right side, the rate is \(q(t)=2.5(4-t)\), so its height at \(k\) is \(2.5(4-k)\). Set the triangular area equal to \(1.25\): \(\frac12(4-k)[2.5(4-k)]=1.25\). Thus, \(1.25(4-k)^2=1.25\), so \((4-k)^2=1\). Since \(k<4\), \(4-k=1\), giving \(k=3\,\text{min}\).

\(k=3\,\text{min}\)

- Use symmetry to compare the mass on either side of \(x=0\). - Mass is the integral of linear density over length. - Approximate the central area with trapezoids. - Determine whether the trapezoidal estimate is above or below the concave-down curve.

1. a) The density graph is symmetric about the y-axis, and the total mass is \(\int_{-2}^{2}d(x)\,\text{d}x=1\,\text{kg}\). Therefore, the part with \(x>0\) has mass \(0.5\,\text{kg}\). 2. b) On \([0,1.5]\), use a composite trapezoidal estimate with width \(0.5\). Since \(d\) is concave down, the trapezoidal sum is a lower bound. The values are \(d(0)=0.375\), \(d(0.5)=0.3515625\), \(d(1)=0.28125\), and \(d(1.5)=0.1640625\). 3. By symmetry, the lower estimate for the mass on \([-1.5,1.5]\) is \(2\cdot0.5\left[\frac12(0.375)+0.3515625+0.28125+\frac12(0.1640625)\right]=0.90234375\,\text{kg}\). Thus, the mass is greater than \(0.90\,\text{kg}\).

a) \(0.5\,\text{kg}\) b) A composite trapezoidal lower estimate is \(0.90234375\,\text{kg}\), so the central section has mass greater than \(0.90\,\text{kg}\).

- Accumulated volume is area under the feed-rate graph. - Use a rectangle in part a and a trapezoid in part b. - First locate the graph section containing \(k\). - Write an accumulation equation using the linear rate on \([4,6]\).

1. a) The rate is constant at \(1\,\text{L/min}\) from \(t=1\) to \(t=3\), so the volume is \((3-1)(1)=2\,\text{L}\). 2. b) From \(5\) to \(6\), the area is a trapezoid with heights \(r(5)=3\) and \(r(6)=5\): \(\frac{3+5}{2}(1)=4\,\text{L}\). 3. c) The accumulated volume through \(t=4\) is \(4\,\text{L}\), so \(k\in[4,6]\). On this interval, \(r(t)=2t-7\). Solve \(4+\int_4^k(2t-7)\,\text{d}t=7\). This simplifies to \(k^2-7k+9=0\). The solution in \([4,6]\) is \(k=\frac{7+\sqrt{13}}{2}\approx5.30\,\text{min}\).

a) \(2\,\text{L}\) b) \(4\,\text{L}\) c) \(k=\frac{7+\sqrt{13}}{2}\approx5.30\,\text{min}\)

- Accumulated rainfall is area under the intensity graph. - Read or calculate the graph’s heights at the requested times. - First find half of the total triangular area. - Determine which side of the peak contains the halfway time.

1. a) The area from \(0\) to \(2\) is a triangle with height \(r(2)=5\): \(\frac12(2)(5)=5\,\text{mm}\). 2. b) On \([4,10]\), the line has equation \(r(t)=\frac{5}{3}(10-t)\). Thus, \(r(6)=\frac{20}{3}\) and \(r(8)=\frac{10}{3}\). The trapezoid area is \(\frac12\left(\frac{20}{3}+\frac{10}{3}\right)(2)=10\,\text{mm}\). 3. c) The total triangular area is \(\frac12(10)(10)=50\,\text{mm}\), so half is \(25\,\text{mm}\). The area through \(t=4\) is \(\frac12(4)(10)=20\,\text{mm}\), so \(t_m\in[4,10]\). Solve \(\int_4^{t_m}\frac{5}{3}(10-t)\,\text{d}t=5\). This gives \(t_m^2-20t_m+70=0\), so the solution in the interval is \(t_m=10-\sqrt{30}\approx4.52\,\text{h}\).

a) \(5\,\text{mm}\) b) \(10\,\text{mm}\) c) \(t_m=10-\sqrt{30}\approx4.52\,\text{h}\)

- Interpret the sign of the rate before describing its contextual meaning. - A maximum of the water level can occur where its rate changes from positive to negative. - The integral of the rate gives the net change in height. - Add the accumulated change to the initial water level.

1. a) \(v(2)=0.8\cos\left(\frac{\pi}{3}\right)=0.4\,\text{m/h}\). At 10:00 a.m., the water level is rising at \(0.4\,\text{m/h}\); the positive sign indicates an increase. 2. b) The rate changes from positive to negative at \(t=3\), so the water level is highest at \(t=3\), or 11:00 a.m. 3. c) \(\int_0^3v(t)\,\text{d}t=\left[\frac{4.8}{\pi}\sin\left(\frac{\pi}{6}t\right)\right]_0^3=\frac{4.8}{\pi}\approx1.53\,\text{m}\). The water level rises by about \(1.53\,\text{m}\) from 8:00 a.m. to 11:00 a.m. 4. d) At 2:00 p.m., \(t=6\). Since \(\int_0^6v(t)\,\text{d}t=\frac{4.8}{\pi}(\sin\pi-\sin0)=0\), the water level is still \(3.50\,\text{m}\).

a) \(0.4\,\text{m/h}\); the water level is rising at that rate at 10:00 a.m. b) \(t=3\), or 11:00 a.m. c) \(\frac{4.8}{\pi}\,\text{m}\approx1.53\,\text{m}\), the increase in water level from 8:00 a.m. to 11:00 a.m. d) \(3.50\,\text{m}\)

- Use a definite integral to find the area under the parabola. - What value must an accumulation function have at its lower limit? - Use the sign of \(f\) to determine whether \(F\) increases or decreases. - Compare the shape of an integrand with the shape of its antiderivative.

1. a) The area is \(A=\int_{-4}^{4}(4-0.25x^2)\,dx\) \(=[4x-\frac{1}{12}x^3]_{-4}^{4}=\frac{64}{3}\approx21.33\). Therefore, the cross-sectional area is \(\frac{64}{3}\,\text{ft}^2\), or about \(21.33\,\text{ft}^2\). 2. b) Since \(F(0)=0\), the correct graph must pass through the origin. Only Candidate \(1\) does. 3. Also, \(f(x)>0\) for \(-4<x<4\), so \(F\) must increase throughout this interval. Candidate \(2\) is the original quadratic function rather than its accumulation function, and Candidate \(3\) has the wrong value at \(x=0\) and the wrong shape. Therefore, Candidate \(1\) is correct.

a) \(\frac{64}{3}\,\text{ft}^2\approx21.33\,\text{ft}^2\) b) Candidate \(1\) is the graph of \(F\); Candidate \(2\) is \(f\), and Candidate \(3\) fails \(F(0)=0\) and has the wrong shape.

- Signed area under a rate graph gives the change in the accumulated quantity. - A sign change of the rate at zero produces a local extremum of the accumulation function. - The accumulation function equals zero at its lower limit. - Positive net flow makes the volume change increase; negative net flow makes it decrease.

1. The signed-area balance consists of four triangles with areas \(-2\), \(2\), \(2\), and \(-2\) cubic meters. Their sum is \(0\), so the total volume change is \(0\,\text{m}^3\). 2. The accumulation function satisfies \(G(-2)=0\) and \(G^{\prime}(x)=g(x)\). 3. The rate \(g\) is zero at \(x=0\) and \(x=4\). At \(x=0\), it changes from negative to positive, so \(G\) has a local minimum. At \(x=4\), it changes from positive to negative, so \(G\) has a local maximum. 4. At \(x=2\), \(g(2)=2>0\), so \(G\) has no extremum there. The change in the slope of \(g\) changes the concavity of \(G\). 5. Graph 1 starts at \(G(-2)=0\), has the required minimum at \(0\) and maximum at \(4\), and ends at \(G(6)=0\).

a) \(0\,\text{m}^3\) b) Graph 1; it has \(G(-2)=G(6)=0\), a local minimum at \(x=0\), and a local maximum at \(x=4\).

- Total production is represented by area under a production-rate graph. - For Leon's statement, compare the full \(10\)-hour areas. - For Sarah's statement, use only the interval from \(0\) to \(4\). - A higher graph indicates a higher instantaneous production rate.

1. Leon: Total production is the area under each rate graph. Machine 1 produces \(\frac{60+10}{2}(10)=350\) parts. Machine 2 produces \(\frac{1}{2}(4)(40)+(6)(40)=80+240=320\) parts. Leon is incorrect. 2. Sarah: During the first \(4\) hours, Machine 1 produces \(\frac{60+40}{2}(4)=200\) parts, while Machine 2 produces \(\frac{1}{2}(4)(40)=80\) parts. Sarah is correct. 3. Jan: For \(t>4\), \(r_2(t)=40\), while \(r_1(t)<40\). Therefore, Machine 2 has the higher production rate after hour \(4\), so Jan is correct.

Leon: Incorrect; Machine 1 produces \(350\) parts, and Machine 2 produces \(320\) parts. Sarah: Correct; Machine 1 produces \(200\) parts, and Machine 2 produces \(80\) parts during the first \(4\) hours. Jan: Correct; for \(t>4\), Machine 2 has the higher hourly production rate.

- The volume increases where its rate of change is positive. - A definite integral of the rate represents accumulated net change. - Add the accumulated change to the initial volume. - Check the sign of the rate immediately before and after \(t=6\).

1. a) Factor \(q(t)=-0.5(t-2)(t-6)\). Since the parabola opens downward, \(q(t)>0\) on \((2,6)\), so the volume increases during that interval. 2. b) The integral represents the net change in water volume, in cubic meters, from \(t=2\) to \(t=6\). Its value is \(\frac{16}{3}\,\text{m}^3\). 3. c) \(V(6)=150+\int_0^6q(t)\,\text{d}t\). Since \(\int_0^6q(t)\,\text{d}t=\left[-\frac{1}{6}t^3+2t^2-6t\right]_0^6=0\), \(V(6)=150\,\text{m}^3\). 4. d) The rate changes from positive to negative at \(t=6\). Therefore, the volume increases immediately before \(t=6\) and decreases immediately afterward, giving a local maximum.

a) \(2<t<6\) b) The net change in water volume from hour \(2\) to hour \(6\), equal to \(\frac{16}{3}\,\text{m}^3\) c) \(150\,\text{m}^3\) d) The rate changes from positive to negative at \(t=6\).

- Build the volume function from the initial volume and accumulated rate. - A minimum occurs where the rate changes from negative to positive. - Average rate equals net change divided by elapsed time. - Factor out the known solution \(t=0\) when solving for the return time.

1. The volume is \(V(t)=40+\int_0^tr(x)\,\text{d}x=40-0.02\left(\frac{t^3}{3}-10t^2+64t\right)\). 2. \(V(4)=\frac{2824}{75}\approx37.65\,\text{m}^3\), and \(V(10)=\frac{608}{15}\approx40.53\,\text{m}^3\). 3. The rate is zero at \(t=4\) and \(t=16\). It changes from negative to positive at \(t=4\), so the minimum during the first \(10\) hours occurs at \(t=4\), with volume approximately \(37.65\,\text{m}^3\). 4. \(V(16)=\frac{3256}{75}\approx43.41\,\text{m}^3\). The average rate is \(\frac{V(16)-V(0)}{16}\approx0.213\,\text{m}^3/\text{h}\). 5. Set \(V(t)=40\). After factoring out the known solution \(t=0\), solve \(t^2-30t+192=0\). The first positive solution is \(t=15-\sqrt{33}\approx9.26\,\text{h}\).

a) After \(4\) hours: approximately \(37.65\,\text{m}^3\); after \(10\) hours: approximately \(40.53\,\text{m}^3\) b) At \(t=4\,\text{h}\); minimum volume approximately \(37.65\,\text{m}^3\) c) Approximately \(0.213\,\text{m}^3/\text{h}\) d) \(t=15-\sqrt{33}\approx9.26\,\text{h}\)

- Solve an exponential equation for part a. - Use the sign of the derivative to establish increasing behavior. - Integrate the flow rate and apply the initial condition. - After \(t=20\), use a linear accumulation model with constant rate.

1. Solve \(12-12e^{-0.4t}=6\): \(e^{-0.4t}=0.5\), so \(t=\frac{\ln2}{0.4}\approx1.73\,\text{min}\). 2. \(v'(t)=4.8e^{-0.4t}>0\) for all \(t\ge0\), so \(v\) is increasing. 3. \(v'(5)=4.8e^{-2}\approx0.65\,\text{L/min}^2\). 4. An antiderivative is \(12t+30e^{-0.4t}+C\). Using \(V(0)=40\) gives \(C=10\), so \(V(t)=12t+30e^{-0.4t}+10\). 5. \(V(10)=130+30e^{-4}\approx130.55\,\text{L}\). 6. \(V(20)=250+30e^{-8}\approx250.01\,\text{L}\). For \(t\ge20\), solve \(250.01+12(t-20)=500\), giving \(t\approx40.83\,\text{min}\).

a) Approximately \(1.73\,\text{min}\) b) \(v'(t)=4.8e^{-0.4t}>0\) c) Approximately \(0.65\,\text{L/min}^2\) d) \(V(t)=12t+30e^{-0.4t}+10\) e) Approximately \(130.55\,\text{L}\) f) Approximately \(40.83\,\text{min}\) after filling begins

- Differentiate to find the maximum power. - Use integration by parts for the product of \(t\) and an exponential. - Apply the initial energy to determine the constant. - Average power equals energy gained divided by time.

1. \(P'(t)=(40-20t)e^{-0.5t}\). The derivative changes from positive to negative at \(t=2\), so the maximum occurs there. \(P(2)=\frac{80}{e}\approx29.43\,\text{W}\). 2. Integration by parts gives \(\int40te^{-0.5t}\,\text{d}t=-80(t+2)e^{-0.5t}+C\). 3. Using \(E(0)=200\) gives \(C=360\), so \(E(t)=360-80(t+2)e^{-0.5t}\). 4. \(E(4)=360-480e^{-2}\approx295.04\,\text{Wh}\). 5. The average charging power is \(\frac{E(4)-E(0)}{4}\approx23.76\,\text{W}\). 6. Since \((t+2)e^{-0.5t}\to0\), \(E(t)\to360\,\text{Wh}\).

a) At \(t=2\,\text{h}\); maximum power \(\frac{80}{e}\,\text{W}\approx29.43\,\text{W}\) b) \(E(t)=360-80(t+2)e^{-0.5t}\) c) Approximately \(295.04\,\text{Wh}\) d) Approximately \(23.76\,\text{W}\) e) \(360\,\text{Wh}\)

- What must the integral of a density over its full support equal? - An interval probability is the integral of the density over that interval. - Recall the formulas for the mean and variance of a continuous random variable. - The density is zero outside \([0,2]\).

1. a) A density must integrate to \(1\): \(c\int_0^2(4-x^2)\,\text{d}x=c\left[4x-\frac{x^3}{3}\right]_0^2=c\frac{16}{3}=1\). Thus, \(c=\frac{3}{16}\). 2. b) \(P(X\le1.5)=\int_0^{1.5}\frac{3}{16}(4-x^2)\,\text{d}x=\frac{3}{16}\left[4x-\frac{x^3}{3}\right]_0^{1.5}=\frac{117}{128}\approx0.9141\). 3. c) \(E(X)=\int_0^2x f(x)\,\text{d}x=\frac{3}{16}\left[2x^2-\frac{x^4}{4}\right]_0^2=0.75\,\text{mm}\). 4. Also, \(E(X^2)=\frac{3}{16}\int_0^2(4x^2-x^4)\,\text{d}x=0.8\,\text{mm}^2\). Therefore, \(\operatorname{Var}(X)=0.8-0.75^2=0.2375\,\text{mm}^2\), and \(\sigma(X)=\sqrt{0.2375}\approx0.4873\,\text{mm}\).

a) \(c=\frac{3}{16}=0.1875\) b) \(P(X\le1.5)=\frac{117}{128}\approx0.9141\), or about \(91.41\%\) c) \(E(X)=0.75\,\text{mm}\); \(\sigma(X)\approx0.4873\,\text{mm}\)

- Integrate the flow rate to find accumulated volume. - Divide accumulated volume by the interval length to find an average rate. - “Decreasing most rapidly” refers to the minimum value of the derivative of the flow-rate function. - Use the second derivative to locate extrema of the first derivative.

1. An antiderivative is \(G(t)=(-500t^2-5000t-25{,}000)e^{-0.2t}+200t\). 2. a) \(\int_0^{12}g(t)\,\text{d}t=G(12)-G(0)\approx13{,}157.28\,\text{m}^3\). 3. b) The average flow rate is \(\frac{1}{12}\int_6^{18}g(t)\,\text{d}t\approx1401.54\,\text{m}^3/\text{h}\). 4. c) The flow rate decreases most rapidly where its derivative is least. Since \(g'(t)=(200t-20t^2)e^{-0.2t}\), analyze its critical points by solving \(g''(t)=(4t^2-80t+200)e^{-0.2t}=0\). 5. The solutions are \(t=10\pm5\sqrt{2}\). In the decreasing region after \(t=10\), \(g'(t)\) reaches its minimum at \(t=10+5\sqrt{2}\approx17.07\,\text{h}\).

a) Approximately \(13{,}157.28\,\text{m}^3\) b) Approximately \(1401.54\,\text{m}^3/\text{h}\) c) \(t=10+5\sqrt{2}\,\text{h}\approx17.07\,\text{h}\)

- Examine the sign of the production rate. - Use the trapezoid-area formula. - Use the second derivative to determine concavity. - Compare the approximation with the exact integral.

1. \(h(20)=0\), and \(h(t)=t^3(2-0.1t)>0\) for \(0<t<20\). Therefore, the accumulated production increases throughout the interval and is greatest at \(t=20\). 2. \(h(10)=1000\) and \(h(15)=1687.5\). The trapezoidal estimate is \(\frac{1000+1687.5}{2}(5)=6718.75\) units. 3. \(h''(t)=-1.2t^2+12t=-1.2t(t-10)<0\) on \((10,15]\), so the graph is concave down and the secant segment lies below it. 4. The exact production is \(\int_{10}^{15}(-0.1t^4+2t^3)\,\text{d}t=7125\) units. 5. The percent error is \(\frac{7125-6718.75}{7125}\cdot100\%\approx5.70\%\).

a) \(h(20)=0\), and the accumulated production is greatest at \(t=20\) b) \(6718.75\) units c) The graph is concave down, so the trapezoid underestimates; percent error approximately \(5.70\%\)

- What must the integral of a density over its full support equal? - Find an interval probability by integrating the density over that interval. - Use the first derivative to locate the maximum of \(f\). - First calculate the endpoints of the interval within one standard deviation of the mean.

1. a) A density must have total area \(1\): \(\int_0^4 kx(4-x)^2\,dx=1\). Since \(\int_0^4 x(4-x)^2\,dx=\frac{64}{3}\), \(k=\frac{3}{64}\). 2. b) An antiderivative is \(F(x)=\frac{3}{64}\left(8x^2-\frac{8}{3}x^3+\frac{1}{4}x^4\right)\). Thus, \(P(1\le X\le 3)=F(3)-F(1)=\frac{243}{256}-\frac{67}{256}=\frac{11}{16}=0.6875\). 3. c) The graph suggests a mode near \(1.3\). With \(f(x)=\frac{3}{64}(16x-8x^2+x^3)\), solve \(f'(x)=\frac{3}{64}(16-16x+3x^2)=0\). The critical values are \(x=4\) and \(x=\frac{4}{3}\). Comparing values at the critical points and endpoints shows the maximum occurs at \(x=\frac{4}{3}\approx 1.33\). 4. d) The interval within one standard deviation is \([1.6-0.8, 1.6+0.8]=[0.8, 2.4]\). Then \(P(0.8\le X\le 2.4)=F(2.4)-F(0.8)=0.8208-0.1808=0.64\).

a) \(k=\frac{3}{64}\) b) \(P(1\le X\le 3)=\frac{11}{16}=0.6875\), or \(68.75\%\) c) Estimate: about \(1.3\,\text{g}\); exact mode: \(\frac{4}{3}\,\text{g}\approx 1.33\,\text{g}\) d) \(P(0.8\le X\le 2.4)=0.64\), or \(64\%\)

- Interpret a zero rate in the context of an accumulated inventory. - To maximize the rate, analyze the derivative of the rate function on the stated interval. - The inventory function is an antiderivative of the rate function. - Use the initial inventory to determine the constant of integration.

1. a) Factor the rate: \(f(t)=t(t-3)(t-8)\). Thus, the rate is zero at \(t=0\), \(t=3\), and \(t=8\). At noon the net rate is zero; at 3:00 p.m. the rate changes from positive to negative, so inventory is greatest; at 8:00 p.m. the net loss stops. 2. b) Since \(f'(t)=3t^2-22t+24\), the critical points of \(f\) are \(t=\frac{4}{3}\) and \(t=6\). Only \(t=\frac{4}{3}\) lies in \([0,3]\). Also, \(f''\left(\frac{4}{3}\right)=-14<0\), so the rate has a maximum there. The maximum rate is \(f\left(\frac{4}{3}\right)=\frac{400}{27}\approx14.81\) units per hour. 3. c) Since \(I'(t)=f(t)\), \(I(t)=\frac{1}{4}t^4-\frac{11}{3}t^3+12t^2+C\). The condition \(I(0)=100\) gives \(C=100\). 4. d) \(I(3)=\frac{517}{4}=129.25\) units, and \(I(8)=\frac{44}{3}\approx14.67\) units. Inventory increases on \([0,3]\) and decreases on \([3,8]\), so its minimum occurs at an endpoint. Since both endpoint values are positive, the warehouse never runs out.

a) \(t=0\), \(t=3\), and \(t=8\); these correspond to noon, 3:00 p.m., and 8:00 p.m. b) At \(t=\frac{4}{3}\) hours, or 1:20 p.m.; the maximum rate is \(\frac{400}{27}\) units per hour \(\approx14.81\) units per hour. c) \(I(t)=\frac{1}{4}t^4-\frac{11}{3}t^3+12t^2+100\) d) \(I(3)=129.25\) units and \(I(8)=\frac{44}{3}\) units \(\approx14.67\) units; the warehouse never runs out.

- The stored volume increases when inflow exceeds outflow. - Find where the two rate graphs intersect. - The maximum occurs when the net rate changes from positive to negative. - The area between the rate graphs gives the net volume change.

1. The reservoir volume increases when \(i(t)>o(t)\). Solve \(-0.1t^2+2t+5=12\), giving \(t=10\pm\sqrt{30}\), or approximately \(4.52\) and \(15.48\) hours. 2. a) Since the inflow parabola lies above the outflow line between the intersection times, the volume increases on \((10-\sqrt{30},10+\sqrt{30})\). 3. b) The net rate changes from positive to negative at \(t=10+\sqrt{30}\), so the maximum volume occurs at \(t=10+\sqrt{30}\approx15.48\,\text{h}\). 4. c) The net volume added is \(\int_{10-\sqrt{30}}^{10+\sqrt{30}}(i(t)-o(t))\,\text{d}t=4\sqrt{30}\) thousand cubic meters. This is approximately \(21.91\times10^3\,\text{m}^3\), or \(21{,}909\,\text{m}^3\).

a) \(10-\sqrt{30}<t<10+\sqrt{30}\), approximately \(4.52<t<15.48\) hours b) \(t=10+\sqrt{30}\,\text{h}\approx15.48\,\text{h}\) c) \(4\sqrt{30}\times10^3\,\text{m}^3\approx21{,}909\,\text{m}^3\)

- Interpret area under a rate graph as accumulated volume. - Use triangle or trapezoid areas where possible. - For part c, write an accumulation equation after \(t=8\). - The pumping process begins only after the inflow ends.

1. The integral is the total volume entering during the \(20\)-minute storm. The graph forms a triangle, so \(\int_0^{20}r(t)\,\mathrm{d}t =\frac{1}{2}\cdot20\cdot40 =400\,\text{m}^3\). 2. During the first \(8\) minutes, the area is \(\frac{1}{2}\cdot8\cdot40=160\,\text{m}^3\). 3. After \(t=8\), the rate is \(r(t)=40-\frac{10}{3}(t-8)\). Set the accumulated volume equal to \(340\): \(160+\int_8^t\left(40-\frac{10}{3}(s-8)\right)\,\mathrm{d}s=340\). Let \(u=t-8\). Then \(40u-\frac{5}{3}u^2=180\). The solution in \(0\le u\le12\) is \(u=6\), so \(t=14\) minutes. 4. At \(t=20\), the basin contains \(400\,\text{m}^3\). Pumping at \(20\,\text{m}^3/\text{min}\) takes \(\frac{400}{20}=20\) minutes. Therefore, the basin is empty at \(t=40\) minutes.

a) The total inflow during the storm; \(400\,\text{m}^3\) b) \(160\,\text{m}^3\) c) \(t=14\) minutes d) \(t=40\) minutes

- Accumulate the failure rate from the beginning of the test to time \(t\). - Find the number on an interval by subtracting cumulative values. - First determine half of the total expected failures. - Pay attention to the negative sign when integrating a power of \((5-t)\).

1. a) For \(0\le t\le5\), \(H(t)=\int_0^t3(5-s)^2\,\text{d}s=125-(5-t)^3\). Also, \(H(t)=0\) for \(t<0\) and \(H(t)=125\) for \(t>5\). 2. b) The expected number is \(H(2)-H(1)=\left[125-3^3\right]-\left[125-4^3\right]=37\) sensors. 3. c) Half of the total \(125\) expected failures is \(62.5\). Solve \(125-(5-t)^3=62.5\). This gives \(t=5-\sqrt[3]{62.5}\approx1.031\) hours. Thus, half of the expected failures occur within about \(1\) hour \(2\) minutes.

a) \(H(t)=\begin{cases}0 & t<0\\125-(5-t)^3 & 0\le t\le5\\125 & t>5\end{cases}\) b) \(37\) sensors c) \(t=5-\sqrt[3]{62.5}\approx1.031\) hours, or about \(1\) hour \(2\) minutes

- Use the sign of the rate to determine whether the volume is increasing or decreasing. - The volume increases most rapidly when its rate function is greatest. - A definite integral of the rate gives the net change in volume. - For the graphical part, look for symmetry that makes two signed areas cancel.

1. a) Since \(f(t)=0.5t(t-2)(t-4)\), the rate is zero at \(t=0\), \(t=2\), and \(t=4\), corresponding to 10:00 a.m., noon, and 2:00 p.m. 2. b) \(f(1)=0.5-3+4=1.5\). At 11:00 a.m., the water volume is increasing at \(1.5\times10^3\,\text{m}^3/\text{h}\). 3. c) The volume increases most rapidly when \(f\) is greatest. Solve \(f'(t)=1.5t^2-6t+4=0\). The local maximum occurs at \(t=2-\frac{2\sqrt{3}}{3}\approx0.85\), about 10:51 a.m. 4. d) The rate is positive on \((0,2)\) and negative on \((2,4)\), so the volume increases until \(t=2\) and decreases afterward. 5. e) \(\int_0^4(0.5t^3-3t^2+4t)\,\text{d}t=\left[0.125t^4-t^3+2t^2\right]_0^4=32-64+32=0\). Therefore, the net change is zero. 6. f) The graph of \(g\) is point-symmetric about \((2,0)\). Thus, the positive signed area from \(t=0.5\) to \(t=2\) cancels the negative signed area from \(t=2\) to \(t=3.5\). The same volume is reached again at \(t=3.5\), or 1:30 p.m.

a) \(t=0\), \(t=2\), and \(t=4\) b) \(f(1)=1.5\); at 11:00 a.m., the volume is increasing at \(1.5\times10^3\,\text{m}^3/\text{h}\). c) \(t=2-\frac{2\sqrt{3}}{3}\approx0.85\), about 10:51 a.m. d) The rate changes from positive to negative at \(t=2\). e) The accumulated change is \(0\), so the final and initial volumes are equal. f) \(t=3.5\), or 1:30 p.m.