Free math worksheets

- Compare the total interval length with the width of one subinterval. - The number of equal pieces must cover the interval exactly. - For a right-endpoint sum, identify where the last subinterval ends.

1. The interval length is \(11-2=9\). 2. The number of subintervals is \(9\div0.75=12\). 3. The final right endpoint is the interval's right bound, \(11\).

a) \(12\) subintervals b) \(11\)

- Identify which table entries correspond to the left edges of the four subintervals. - Include the common width as a factor in every rectangle contribution. - Keep negative function values as negative signed areas.

1. The subinterval width is \(\Delta x=1\). 2. The left endpoints are \(0, 1, 2, 3\), giving the signed contributions \(3\cdot1+5\cdot1+4\cdot1+(-1)\cdot1\). 3. The approximation is \(11\).

The left-endpoint approximation is \(11\).

- Compute each subinterval width separately from consecutive x-values. - Pair each width with the function value at that subinterval’s right endpoint. - Add the signed rectangle contributions.

1. The subintervals have widths \(1\), \(2\), and \(3\). 2. The right-endpoint values are \(g(2)=5\), \(g(4)=1\), and \(g(7)=-2\). 3. The Riemann sum is \(1\cdot5+2\cdot1+3\cdot(-2)=1\).

The right-endpoint approximation is \(1\).

- The listed times are centers of equal time intervals rather than endpoints. - Use the same time width with each measured rate. - Check that multiplying the rate unit by the time unit gives the requested quantity.

1. Each subinterval has width \(2\) minutes. 2. The midpoint sum is \(2\cdot(18+24+21+15)=156\). 3. Multiplying packages per minute by minutes gives packages.

Approximately \(156\) packages are processed.

- Treat each adjacent pair of measurements as one trapezoid. - Use the actual width of each time interval. - Average the two endpoint speeds on each subinterval before multiplying by its duration.

1. On \([0, 1]\), the trapezoid contributes \(1\cdot\frac{2+6}{2}=4\). 2. On \([1, 2.5]\), the contribution is \(1.5\cdot\frac{6+10}{2}=12\). 3. On \([2.5, 4]\), the contribution is \(1.5\cdot\frac{10+4}{2}=10.5\). 4. The approximate displacement is \(4+12+10.5=26.5\) feet.

The trapezoidal approximation is \(26.5\,\text{ft}\).

- Identify which table value is not used by a left-endpoint sum. - Include the subinterval width outside the sum of heights. - Set the resulting expression equal to the stated approximation.

1. The width is \(\Delta x=2\), and the left endpoints are \(0, 2, 4, 6\). 2. The sum is \(2[3+a+(-1)+5]=20\). 3. This simplifies to \(2(a+7)=20\), so \(a=3\).

\(a=3\)

- Find each width from consecutive partition values. - Use the specified sample point from the same subinterval as its width. - Add one height-times-width term for every subinterval.

1. The subinterval widths are \(1\), \(2\), and \(3\). 2. Pair each width with the function value at its selected sample point. 3. The sum is \(f(0.4)+2f(2.5)+3f(5)\).

\(f(0.4)+2f(2.5)+3f(5)\)

- Match the first, second, and third sample points to the first, second, and third subintervals. - Endpoints may belong to a subinterval, so do not reject a set merely for using them. - Check each proposed point against the interval assigned to it.

1. The sample point for each term must lie in its corresponding subinterval: \([-2, 0]\), \([0, 3]\), and \([3, 5]\). 2. Set A is valid: \(-1\), \(2\), and \(4\) lie in the required intervals. 3. Set B is valid because left endpoints are allowed. 4. Set C is invalid because \(4\notin[0, 3]\), so its second sample point is outside the second subinterval.

Sets A and B are valid. Set C is invalid because its second sample point, \(4\), is not in \([0, 3]\).

- Read the graph height at each integer partition point. - Use all but the final height for the left sum and all but the initial height for the right sum. - The graph’s direction gives a quick check on which sum should be larger.

1. The left heights are \(4, 3, 3, 1\), so \(L_4=(4+3+3+1)(1)=11\). 2. The right heights are \(3, 3, 1, 0\), so \(R_4=(3+3+1+0)(1)=7\). 3. Therefore, \(L_4>R_4\), consistent with the graph's overall nonincreasing behavior.

a) \(L_4=11\) b) \(R_4=7\) c) \(L_4\) is larger.

- Interpret the factor outside the sum as a time width. - List the input times generated by the index. - Combine the rate unit with the width unit to identify the accumulated quantity.

1. The width is \(5\) minutes, and the sample times are \(5, 10, 15, 20, 25, 30\), the right endpoints of six intervals covering \([0, 30]\). 2. Rate multiplied by time approximates a number of vehicles. 3. The sum approximates the total number of vehicles entering during the first \(30\) minutes, in vehicles.

a) Six \(5\)-minute subintervals on \([0, 30]\), using right endpoints b) The number of vehicles entering during the first \(30\) minutes, measured in vehicles

- Check whether every term has both a height and a width. - Find the width of ten equal subintervals spanning a unit interval. - Preserve the existing sample points if they already match the stated endpoint choice.

1. The sample points \(\frac{i}{10}\) are correct right endpoints for ten equal subintervals of \([0, 1]\). 2. Each rectangle must also include the width \(\Delta x=\frac{1}{10}\). 3. The corrected approximation is \(\frac{1}{10}\sum_{i=1}^{10}f\left(\frac{i}{10}\right)\).

The width factor is missing. The corrected sum is \(\frac{1}{10}\sum_{i=1}^{10}f\left(\frac{i}{10}\right)\).

- Test the proposed formula at the first index. - The first left endpoint must equal the interval’s left bound. - Shift the index so the first increment is zero rather than one full width.

1. The proposed formula gives \(x_1=2+\frac{6}{n}\), which is the first right endpoint, not the first left endpoint. 2. Left endpoints begin at \(2\) when \(i=1\). 3. The correct formula is \(x_{i-1}=2+\frac{6(i-1)}{n}\), for \(i=1,\ldots,n\).

The proposal uses right endpoints. The left endpoint for subinterval \(i\) is \(2+\frac{6(i-1)}{n}\).

- Find the common width from the full interval and number of entries. - Add the function values with their signs intact. - Distinguish a definite integral from a request for total geometric area.

1. The interval length is \(5\), so each of five subintervals has width \(1\). 2. The sum is \((-4-2+1-3+5)\cdot1=-3\). 3. A Riemann sum for a definite integral approximates signed accumulation, so values below the x-axis contribute negatively.

The right-endpoint sum is \(-3\). Negative heights remain negative because the sum approximates signed, not total, area.

- Determine the common subinterval width from the midpoint locations. - Multiply the sum of sampled heights by that width. - Dividing an integral by the interval length changes the quantity being estimated. - Distinguish total accumulation from average value.

1. a) The four subintervals have width \(1\), so the midpoint approximation is \((2+5+4+1)\cdot1=12\). 2. b) The student may have divided \(12\) by the interval length \(4\), obtaining \(3\). That calculation estimates the average value of \(f\), not the integral itself.

a) \(12\) b) The value \(3\) comes from \(12\div4\), which estimates the average value rather than the integral.

- Do not assume midpoint sums require equal-width subintervals. - Match each midpoint value to the width of its own interval. - Keep the negative midpoint contribution signed.

1. The subinterval widths are \(2\), \(3\), and \(5\). 2. Multiply each midpoint value by its own width: \(2\cdot4+3\cdot(-1)+5\cdot3\). 3. The sum is \(8-3+15=20\).

The midpoint Riemann sum is \(20\).

- Create the four time subintervals before selecting measurements. - Right endpoints include the final timestamp but exclude the initial one. - Left endpoints reverse which boundary timestamp is omitted.

1. The right endpoints are \(4, 8, 12, 16\), so use \(r(4), r(8), r(12), r(16)\). 2. The width is \(4\) minutes, giving \(4\cdot[r(4)+r(8)+r(12)+r(16)]\). 3. A left-endpoint estimate uses \(r(0), r(4), r(8), r(12)\), omitting the final reading instead of the initial reading.

a) \(r(4), r(8), r(12), r(16)\) b) \(4\cdot[r(4)+r(8)+r(12)+r(16)]\) c) Use \(r(0), r(4), r(8), r(12)\).

- Write one generic height-times-width term for a constant function. - Factor the common function value out of the sum. - Use what the sum of all subinterval widths must equal.

1. A general Riemann sum is \(\sum 7\Delta x_i=7\sum\Delta x_i\). 2. The subinterval widths add to the full interval length: \(\sum\Delta x_i=5-(-3)=8\). 3. Every Riemann sum therefore equals \(7\cdot8=56\), regardless of the partition or sample points.

Every Riemann sum equals \(56\).

- Use the outside factor to identify the subinterval width. - Test the first and last index values to see where the sample points fall. - Combine the starting value with the total interval length.

1. The width is \(\frac{3}{n}\), so the interval length is \(3\) and begins at \(2\); the interval is \([2, 5]\). 2. At \(i=0\), the first sample point is \(2\), so the sum uses left endpoints. 3. The limiting integral is \(\int_2^5f(x)\,\text{d}x\).

a) \([2, 5]\) b) Left endpoints c) \(\int_2^5f(x)\,\text{d}x\)

- Generate the four sample points from the index before evaluating the function. - Keep exact fractional values rather than converting to decimals. - Apply the common width after adding the sampled heights.

1. The right endpoints are \(\frac{1}{2}, 1, \frac{3}{2}, 2\). 2. The function values are \(-\frac{1}{4}, 0, \frac{3}{4}, 2\). 3. Multiply their sum by \(\frac{1}{2}\): \(R_4=\frac{1}{2}\left(-\frac{1}{4}+0+\frac{3}{4}+2\right)=\frac{5}{4}\).

\(R_4=\frac{5}{4}\)

- Compare each rectangle height with the graph value at the left and right edges of its base. - Read the common width and the three rectangle heights from the figure. - Use the direction of the graph to judge whether the rectangles lie mostly below or above it.

1. The rectangle heights are \(f(0)=1\), \(f(1)=2\), and \(f(2)=3\), so the rectangles use left endpoints. 2. With width \(1\), the sum is \((1+2+3)\cdot1=6\). 3. Since \(f\) is increasing, the left-endpoint sum is an underestimate.

a) Left endpoints b) \(6\) c) Underestimate

- Check both the width factor and the location of every sample point. - Sample points for an integral over \([0, 1]\) must stay in that interval. - Determine whether the index-generated inputs align with a partition of the fixed interval.

1. In A, \(\Delta x=\frac{1}{n}\) and the sample points \(\frac{i}{n}\) are right endpoints in \([0, 1]\), so A is a Riemann sum for the integral. 2. In B, the sample points are \(1, 2, \ldots, n\), which do not remain inside \([0, 1]\). 3. Therefore, B is not a Riemann sum for the stated integral.

Expression A is the Riemann sum. Expression B is not, because its sample points are not in the subintervals of \([0, 1]\).

- First determine the width of each rectangle. - For an upper sum of an increasing function, decide whether to use the left or right endpoint of each subinterval. - Express each sample point in terms of the summation index. - Use the formula for the sum of the first \(n\) squares to simplify the larger sum.

1. For \(n=4\), the subinterval width is \(\Delta x=\frac{4-0}{4}=1\). Because \(f\) is increasing on \([0, 4]\), each upper rectangle uses the right endpoint \(x_k=k\). Thus, \(U_4=\sum_{k=1}^{4} f(x_k)\Delta x=\sum_{k=1}^{4}\frac{1}{2}k^2\cdot 1=\frac{1}{2}(1^2+2^2+3^2+4^2)=15\). 2. For \(n=8\), \(\Delta x=\frac{4}{8}=0.5\), and the right endpoints are \(x_k=0.5k\). Therefore, \(U_8=\sum_{k=1}^{8}f(0.5k)\cdot 0.5=\sum_{k=1}^{8}\frac{1}{16}k^2\). Using \(\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}\), \(U_8=\frac{1}{16}\cdot\frac{8\cdot 9\cdot 17}{6}=\frac{51}{4}=12.75\).

1) \(U_4=\sum_{k=1}^{4}\frac{1}{2}k^2=15\) 2) \(U_8=\sum_{k=1}^{8}\frac{1}{16}k^2=12.75\)

- Determine whether the function is increasing or decreasing on the interval. This tells you which endpoint gives each rectangle's minimum height. - Write the subinterval width in terms of the number of rectangles. - Express the \(x\)-coordinate of the \(k\)th sample point. - Substitute that sample point into the function to get the height of the \(k\)th rectangle.

1. The subinterval width is \(\Delta x=\frac{2-0}{5}=0.4\). Because \(f\) is decreasing on \([0, 2]\), the minimum on each subinterval occurs at its right endpoint, \(x_k=0.4k\). Therefore, \(L_5=\sum_{k=1}^{5}\left(4-(0.4k)^2\right)\cdot 0.4\). The function values at \(0.4\), \(0.8\), \(1.2\), \(1.6\), and \(2\) are \(3.84\), \(3.36\), \(2.56\), \(1.44\), and \(0\). Their sum is \(11.2\), so \(L_5=11.2\cdot 0.4=4.48\). 2. For \(n\) subintervals, \(\Delta x=\frac{2}{n}\), and the right endpoints are \(x_k=\frac{2k}{n}\). Thus, \(L_n=\sum_{k=1}^{n}f\left(\frac{2k}{n}\right)\frac{2}{n}=\sum_{k=1}^{n}\left(4-\left(\frac{2k}{n}\right)^2\right)\frac{2}{n}\).

1) \(L_5=4.48\) 2) \(L_n=\sum_{k=1}^{n}\left(4-\frac{4k^2}{n^2}\right)\frac{2}{n}\)

- Combine partial grid squares to estimate whole square units. - For a midpoint Riemann sum, use the midpoint of each equal subinterval as the sample point. - For the exact value, find an antiderivative and apply the Fundamental Theorem of Calculus.

1. For a), counting full and partial grid squares gives a reasonable estimate of about \(14.5\) to \(15\) square units. 2. For b), \(\Delta x=2\), and the midpoints are \(x=1\) and \(x=3\). Since \(f(1)=2.75\) and \(f(3)=4.75\), \(M_2=2(2.75+4.75)=15\). 3. For c), an antiderivative is \(F(x)=-\frac{1}{12}x^3+x^2+x\). Therefore, \(\int_{0}^{4}f(x)\,dx=F(4)-F(0)=-\frac{16}{3}+20=\frac{44}{3}\approx14.67\). 4. Both estimates are close to the exact value; the midpoint estimate is about \(0.33\) square unit too large.

a) About \(14.5\) to \(15\) square units b) \(15\) square units c) \(\frac{44}{3}\) square units, or approximately \(14.67\) square units; both estimates are close, and the midpoint estimate is \(\frac{1}{3}\) square unit too large.

- Locate where the graph crosses the x-axis before judging signed area. - Areas below the x-axis contribute negatively, and areas above it contribute positively. - Estimate areas by combining grid squares or simple geometric regions. - Compare magnitudes only after accounting for the signs of the two integrals.

1. a) The graph crosses the x-axis at approximately \(x=0.3\) and \(x=6.5\). On the displayed interval \([-1, 10]\), \(f\) is negative from \(x=-1\) to about \(x=0.3\), positive from about \(x=0.3\) to \(x=6.5\), and negative from about \(x=6.5\) to \(x=10\). 2. b) Estimating signed areas from the grid gives \(\int_{-1}^{1}f(x)\,\text{d}x\approx-3.4\) and \(\int_{1}^{6}f(x)\,\text{d}x\approx6.4\). 3. c) Since \(|6.4|>|-3.4|\), the second integral has the greater absolute value. The broad region above the x-axis on \([1, 6]\) has more area than the net region below the axis on \([-1, 1]\).

a) The x-intercepts are approximately \(x=0.3\) and \(x=6.5\). On \([-1, 10]\), \(f\) is negative for approximately \(-1\le x<0.3\), positive for approximately \(0.3<x<6.5\), and negative for approximately \(6.5<x\le10\). b) \(\int_{-1}^{1}f(x)\,\text{d}x\approx-3.4\); \(\int_{1}^{6}f(x)\,\text{d}x\approx6.4\) c) \(\int_{1}^{6}f(x)\,\text{d}x\) has the greater absolute value.

- Compare a concave-up curve with the secant segment joining two points on it. - Think about how the graph bends relative to a rectangle centered in a subinterval. - Use the same concavity behavior on all three subintervals.

1. For a concave-up graph, secant segments lie above the graph, so the trapezoids overestimate the integral. 2. Midpoint rectangles use heights below the average graph height on each symmetric subinterval, so the midpoint sum underestimates. 3. Thus \(M_3<\int_0^6f(x)\,\text{d}x<T_3\).

\(M_3\) is an underestimate and \(T_3\) is an overestimate: \(M_3<\int_0^6f(x)\,\text{d}x<T_3\).

- Find the total interval length before dividing it into equal parts. - A right endpoint begins at the left bound and advances by a whole-number multiple of the width. - Substitute that endpoint expression into the integrand and include one width factor.

1. The interval length is \(3-(-2)=5\), so \(\Delta x=\frac{5}{n}\). 2. The right endpoint is \(x_i=-2+\frac{5i}{n}\). 3. The required limit is \(\lim_{n\to\infty}\sum_{i=1}^n\left[\left(-2+\frac{5i}{n}\right)^2+1\right]\frac{5}{n}\).

\(\Delta x=\frac{5}{n}\), \(x_i=-2+\frac{5i}{n}\), and \(\lim_{n\to\infty}\sum_{i=1}^n\left[\left(-2+\frac{5i}{n}\right)^2+1\right]\frac{5}{n}\).

- Express the common width from the interval length and number of subintervals. - Locate each midpoint halfway between two consecutive partition points. - Use the midpoint as the input of the function in every term.

1. The width is \(\Delta x=\frac{5-1}{n}=\frac{4}{n}\). 2. The midpoint of subinterval \(i\) is \(1+\left(i-\frac{1}{2}\right)\frac{4}{n}\). 3. Therefore, \(M_n=\frac{4}{n}\sum_{i=1}^n\ln\left(1+\frac{4i-2}{n}\right)\).

\(M_n=\frac{4}{n}\sum_{i=1}^n\ln\left(1+\frac{4i-2}{n}\right)\)

- Use the endpoint heights and actual width of each subinterval. - Allow a trapezoid to contribute signed area when an endpoint is below the x-axis. - Compare each trapezoid’s top edge with the graph segment over the same interval.

1. The trapezoids contribute \(1\cdot\frac{1+3}{2}=2\), \(2\cdot\frac{3+(-1)}{2}=2\), and \(2\cdot\frac{-1+2}{2}=1\). 2. The total is \(2+2+1=5\). 3. On each subinterval, the graph itself is a line segment, so each trapezoid exactly matches the signed region under the graph.

a) \(5\) b) It is exact because \(g\) is linear on every subinterval of the partition.

- List the four right endpoints before substituting into the function. - Keep the common width as a factor of the entire sum. - Use the stated numerical value to form an equation for the parameter.

1. The width is \(\Delta x=\frac{1}{2}\), and the right endpoints are \(\frac{1}{2}, 1, \frac{3}{2}, 2\). 2. The sum is \(\frac{1}{2}\left[a\left(\frac{1}{2}+1+\frac{3}{2}+2\right)+4\right]=\frac{5a}{2}+2\). 3. Set \(\frac{5a}{2}+2=5\). Then \(\frac{5a}{2}=3\), so \(a=\frac{6}{5}\).

\(a=\frac{6}{5}\)

- First separate all left sums from all right sums using monotonicity. - Then compare coarse and refined partitions within the same endpoint choice. - Place the exact integral between the underestimates and overestimates.

1. For an increasing function, every left sum is an underestimate and every right sum is an overestimate. 2. Refining the partition raises a left sum because more rectangles use larger left-endpoint values. 3. Refining the partition lowers a right sum because more rectangles use smaller right-endpoint values. 4. Therefore, \(L_4<L_8<\int_0^4f(x)\,\text{d}x<R_8<R_4\).

\(L_4<L_8<\int_0^4f(x)\,\text{d}x<R_8<R_4\)

- Use monotonicity to decide which endpoint sum is below and which is above the integral. - The average of two bounds is the midpoint of the possible-value interval. - Half the width of the interval gives the best guaranteed upper bound on the error.

1. Since \(f\) is increasing, \(7.2<\int_1^7f(x)\,\text{d}x<9.1\). 2. The midpoint of the bound interval is \(\frac{7.2+9.1}{2}=8.15\). 3. Every possible integral value is less than \(0.95\) from \(8.15\), so \(0.95\) is the best guaranteed upper bound on the absolute error.

a) \((7.2, 9.1)\) b) \(0.95\)

- Use signed velocity values when approximating change in position. - Use magnitudes of velocity when approximating path length. - Multiply each sum of sampled rates by the common time width.

1. The displacement approximation is \(1.5\cdot[3+(-2)+(-4)+1]=-3\) meters. 2. For total distance, use speed values: \(1.5\cdot(3+2+4+1)=15\) meters. 3. The different signs affect displacement but not total distance.

Approximate displacement: \(-3\,\text{m}\) Approximate total distance: \(15\,\text{m}\)

- Use one trapezoid for each pair of consecutive measurements. - Use each interval's actual duration. - Average the endpoint temperatures on each interval before multiplying by its width. - Track the product of the vertical and horizontal units.

1. a) The trapezoidal approximation is \(1\cdot\frac{64+70}{2}+2\cdot\frac{70+76}{2}+3\cdot\frac{76+68}{2}=429\). 2. b) Multiplying degrees Fahrenheit by hours gives degree-hours. The integral accumulates temperature over time, so \(429\) is not a single temperature reading. A temperature estimate would require an additional interpretation, such as dividing by the interval length to find an average.

a) \(429\) b) The units are degree-hours. The value is an accumulated temperature-time quantity, not a temperature.

- Relate the function value at the midpoint of a line segment to the endpoint values. - Write each one-subinterval approximation symbolically. - Interpret the region under a linear graph as a geometric figure.

1. For a linear function, \(f\left(\frac{a+b}{2}\right)=\frac{f(a)+f(b)}{2}\). 2. The midpoint rule gives \((b-a)f\left(\frac{a+b}{2}\right)\), and the trapezoidal rule gives \((b-a)\frac{f(a)+f(b)}{2}\); these are equal. 3. Writing \(f(x)=mx+c\), direct integration gives \(\int_a^bf(x)\,\text{d}x=(b-a)\frac{f(a)+f(b)}{2}\), so the common expression is exact.

a) Both methods give \((b-a)\frac{f(a)+f(b)}{2}\). b) For \(f(x)=mx+c\), direct integration gives \(\int_a^bf(x)\,\text{d}x=(b-a)\frac{f(a)+f(b)}{2}\), so this expression is exact.

- Use the defined value at \(x=2\) when an endpoint sum samples that point. - Find the exact integral from the two horizontal regions and their signs. - Compare the width of a rectangle with the width of a single point. - Distinguish the behavior of a particular finite sum from the value of the definite integral.

1. a) With width \(1\), \(L_4=f(0)+f(1)+f(2)+f(3)=2+2+5-1=8\). Also, \(R_4=f(1)+f(2)+f(3)+f(4)=2+5-1-1=5\). 2. b) The rectangle above the x-axis has signed area \(2\cdot2=4\), and the rectangle below the x-axis has signed area \(-1\cdot2=-2\). Thus, \(\int_0^4f(x)\,\text{d}x=4-2=2\). 3. c) Both endpoint sums happen to use \(x=2\) as a sample point, so the height \(5\) enters one rectangle in each sum. A value at a single x-coordinate has no width, so changing only that isolated value does not change the signed area represented by the definite integral.

a) \(L_4=8\) and \(R_4=5\) b) \(2\) c) The endpoint sums sample \(f(2)\), but a change at one isolated x-value contributes no area to the definite integral.

- Pair one function height with the width of the same subinterval. - Substitute the selected sample point into \(f(x)=1/x\). - The partition covers the same interval as the limiting definite integral. - A sample point must be selected from the subinterval whose width it multiplies.

1. a) The function height on subinterval \(i\) is \(f(c_i)=\frac{1}{c_i}\), and the width is \(x_i-x_{i-1}\). Thus, the sum is \(\sum_{i=1}^n\frac{x_i-x_{i-1}}{c_i}\). 2. b) As the partition is refined, the sums approach \(\int_1^4\frac{1}{x}\,\text{d}x\). 3. c) Each sample point must lie in its own subinterval: \(c_i\in[x_{i-1}, x_i]\).

a) \(\sum_{i=1}^n\frac{x_i-x_{i-1}}{c_i}\) b) \(\int_1^4\frac{1}{x}\,\text{d}x\) c) \(c_i\in[x_{i-1}, x_i]\) for every \(i\)

- Pair subintervals that are mirror images across the y-axis. - Choose sample points that preserve that same symmetry. - Use the defining sign relationship of an odd function to compare paired terms.

1. Pair each subinterval on the left of \(0\) with its reflected subinterval on the right. 2. Choose sample points in reflected pairs, \(c\) and \(-c\). 3. The paired widths are equal and \(f(-c)=-f(c)\), so each pair contributes \(f(c)\Delta x+f(-c)\Delta x=0\). 4. All four pairs cancel, making the entire Riemann sum \(0\).

Choose sample points symmetrically: whenever \(c\) is selected in a right-side subinterval, select \(-c\) in the reflected left-side subinterval. Equal-width contributions cancel because \(f\) is odd.