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AP Calculus BC : AP Calculus BC

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Example Questions

Example Question #3 : Riemann Sum: Midpoint Evaluation

$\begin{align*}&\text{Calculate the Riemann sums integral approximation of :}\\&\int_{-4}^{14.4}(-20sin(13tan(2x)))dx\\&\text{Using midpoints over }4\text{ intervals.}\end{align*}$

Possible Answers:

-1119.16

-1534.84

-159.88

-69.51

Correct answer:

-159.88

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-4}^{14.4}(-20sin(13tan(2x)))dx\\&\text{So the interval is }[-4,14.4]\text{ the subintervals have length }\frac{14.4-(-4)}{4}=\frac{23}{5}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[-\frac{17}{10},\frac{29}{10},\frac{15}{2},\frac{121}{10}]\\&\int_{-4}^{14.4}(-20sin(13tan(2x)))dx=\frac{23}{5}[(20sin(13tan(\frac{17}{5})))+(-20sin(13tan(\frac{29}{5})))+(-20sin(13tan(15)))+(-20sin(13tan(\frac{121}{5})))]\\&\int_{-4}^{14.4}(-20sin(13tan(2x)))dx=-159.88\end{align*}$

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Example Question #4 : Riemann Sum: Midpoint Evaluation

$\begin{align*}&\text{Using the method of midpoint Riemann sums approximate the integral:}\\&\int_{5}^{21}(-16cos(sin(6x)))dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

-67.24

-1506.27

-188.28

-35.53

Correct answer:

-188.28

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{5}^{21}(-16cos(sin(6x)))dx\\&\text{So the interval is }[5,21]\text{ the subintervals have length }\frac{21-(5)}{4}=4\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[7,11,15,19]\\&\int_{5}^{21}(-16cos(sin(6x)))dx=4[(-16cos(sin(42)))+(-16cos(sin(66)))+(-16cos(sin(90)))+(-16cos(sin(114)))]\\&\int_{5}^{21}(-16cos(sin(6x)))dx=-188.28\end{align*}$

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Example Question #5 : Riemann Sum: Midpoint Evaluation

$\begin{align*}&\text{Using }3\text{ intervals, and the method of midpoint Riemann sums approximate the integral:}\\&\int_{2}^{14.6}(16cos(20e^{(3x)}))dx\end{align*}$

Possible Answers:

4.13

78.12

247.37

26.04

Correct answer:

26.04

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{2}^{14.6}(16cos(20e^{(3x)}))dx\\&\text{So the interval is }[2,14.6]\text{ the subintervals have length }\frac{14.6-(2)}{3}=\frac{21}{5}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{41}{10},\frac{83}{10},\frac{25}{2}]\\&\int_{2}^{14.6}(16cos(20e^{(3x)}))dx=\frac{21}{5}[(16cos(20e^{(\frac{123}{10})}))+(16cos(20e^{(\frac{249}{10})}))+(16cos(20e^{(\frac{75}{2})}))]\\&\int_{2}^{14.6}(16cos(20e^{(3x)}))dx=26.04\end{align*}$

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Example Question #6 : Riemann Sum: Midpoint Evaluation

$\begin{align*}&\text{Using }3\text{ intervals, and the method of midpoint Riemann sums approximate the integral:}\\&\int_{-3}^{9.9}(\frac{13}{3^{(5cos(3x))}})dx\end{align*}$

Possible Answers:

1155.72

2653.87

7165.44

70937.83

Correct answer:

7165.44

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-3}^{9.9}(\frac{13}{3^{(5cos(3x))}})dx\\&\text{So the interval is }[-3,9.9]\text{ the subintervals have length }\frac{9.9-(-3)}{3}=\frac{43}{10}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[-\frac{17}{20},\frac{69}{20},\frac{31}{4}]\\&\int_{-3}^{9.9}(\frac{13}{3^{(5cos(3x))}})dx=\frac{43}{10}[(\frac{13}{3^{(5cos(\frac{51}{20}))}})+(\frac{13}{3^{(5cos(\frac{207}{20}))}})+(\frac{13}{3^{(5cos(\frac{93}{4}))}})]\\&\int_{-3}^{9.9}(\frac{13}{3^{(5cos(3x))}})dx=7165.44\end{align*}$

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Example Question #7 : Riemann Sum: Midpoint Evaluation

$\begin{align*}&\text{Using the method of midpoint Riemann sums approximate the integral:}\\&\int_{1}^{9}(19cos(2x^{2}))dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

-36.91

-114.43

-4.45

-184.57

Correct answer:

-36.91

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{1}^{9}(19cos(2x^{2}))dx\\&\text{So the interval is }[1,9]\text{ the subintervals have length }\frac{9-(1)}{4}=2\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[2,4,6,8]\\&\int_{1}^{9}(19cos(2x^{2}))dx=2[(19cos(8))+(19cos(32))+(19cos(72))+(19cos(128))]\\&\int_{1}^{9}(19cos(2x^{2}))dx=-36.91\end{align*}$

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Example Question #8 : Riemann Sum: Midpoint Evaluation

$\begin{align*}&\text{Using }4\text{ intervals, and the method of midpoint Riemann sums approximate the integral:}\\&\int_{1}^{12.6}(-9sin(18e^{(3x)}))dx\end{align*}$

Possible Answers:

-391.30

-184.57

-73.83

-649.70

Correct answer:

-73.83

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{1}^{12.6}(-9sin(18e^{(3x)}))dx\\&\text{So the interval is }[1,12.6]\text{ the subintervals have length }\frac{12.6-(1)}{4}=\frac{29}{10}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{49}{20},\frac{107}{20},\frac{33}{4},\frac{223}{20}]\\&\int_{1}^{12.6}(-9sin(18e^{(3x)}))dx=\frac{29}{10}[(-9sin(18e^{(\frac{147}{20})}))+(-9sin(18e^{(\frac{321}{20})}))+(-9sin(18e^{(\frac{99}{4})}))+(-9sin(18e^{(\frac{669}{20})}))]\\&\int_{1}^{12.6}(-9sin(18e^{(3x)}))dx=-73.83\end{align*}$

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Example Question #9 : Riemann Sum: Midpoint Evaluation

$\begin{align*}&\text{Using the method of midpoint Riemann sums approximate the integral:}\\&\int_{0}^{10}(8cos(19e^{(x)}))dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

-125.94

-8.65

-4.50

-44.98

Correct answer:

-44.98

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{0}^{10}(8cos(19e^{(x)}))dx\\&\text{So the interval is }[0,10]\text{ the subintervals have length }\frac{10-(0)}{4}=\frac{5}{2}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{5}{4},\frac{15}{4},\frac{25}{4},\frac{35}{4}]\\&\int_{0}^{10}(8cos(19e^{(x)}))dx=\frac{5}{2}[(8cos(19e^{(\frac{5}{4})}))+(8cos(19e^{(\frac{15}{4})}))+(8cos(19e^{(\frac{25}{4})}))+(8cos(19e^{(\frac{35}{4})}))]\\&\int_{0}^{10}(8cos(19e^{(x)}))dx=-44.98\end{align*}$

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Example Question #11 : Riemann Sum: Midpoint Evaluation

$\begin{align*}&\text{Using the method of midpoint Riemann sums approximate the integral:}\\&\int_{4}^{9.4}(-15sin(x^{2}))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

2.88

0.51

0.99

0.35

Correct answer:

2.88

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{4}^{9.4}(-15sin(x^{2}))dx\\&\text{So the interval is }[4,9.4]\text{ the subintervals have length }\frac{9.4-(4)}{3}=\frac{9}{5}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{49}{10},\frac{67}{10},\frac{17}{2}]\\&\int_{4}^{9.4}(-15sin(x^{2}))dx=\frac{9}{5}[(-15sin(\frac{2401}{100}))+(-15sin(\frac{4489}{100}))+(-15sin(\frac{289}{4}))]\\&\int_{4}^{9.4}(-15sin(x^{2}))dx=2.88\end{align*}$

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Example Question #12 : Riemann Sum: Midpoint Evaluation

$\begin{align*}&\text{Using the method of midpoint Riemann sums approximate the integral:}\\&\int_{0}^{12.3}(-12sin(19e^{(4x)}))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

-70.62

-13.08

-183.60

-593.17

Correct answer:

-70.62

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{0}^{12.3}(-12sin(19e^{(4x)}))dx\\&\text{So the interval is }[0,12.3]\text{ the subintervals have length }\frac{12.3-(0)}{3}=\frac{41}{10}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{41}{20},\frac{123}{20},\frac{41}{4}]\\&\int_{0}^{12.3}(-12sin(19e^{(4x)}))dx=\frac{41}{10}[(-12sin(19e^{(\frac{41}{5})}))+(-12sin(19e^{(\frac{123}{5})}))+(-12sin(19e^{(41)}))]\\&\int_{0}^{12.3}(-12sin(19e^{(4x)}))dx=-70.62\end{align*}$

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Example Question #12 : Riemann Sum: Midpoint Evaluation

$\begin{align*}&\text{Using the method of midpoint Riemann sums approximate the integral:}\\&\int_{-2}^{4.6}(9tan(9tan(6x)))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

477.77

1194.43

51.93

2962.18

Correct answer:

477.77

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-2}^{4.6}(9tan(9tan(6x)))dx\\&\text{So the interval is }[-2,4.6]\text{ the subintervals have length }\frac{4.6-(-2)}{3}=\frac{11}{5}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[-\frac{9}{10},\frac{13}{10},\frac{7}{2}]\\&\int_{-2}^{4.6}(9tan(9tan(6x)))dx=\frac{11}{5}[(-9tan(9tan(\frac{27}{5})))+(9tan(9tan(\frac{39}{5})))+(9tan(9tan(21)))]\\&\int_{-2}^{4.6}(9tan(9tan(6x)))dx=477.77\end{align*}$

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AP Calculus BC : AP Calculus BC

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #3 : Riemann Sum: Midpoint Evaluation

Example Question #4 : Riemann Sum: Midpoint Evaluation

Example Question #5 : Riemann Sum: Midpoint Evaluation

Example Question #6 : Riemann Sum: Midpoint Evaluation

Example Question #7 : Riemann Sum: Midpoint Evaluation

Example Question #8 : Riemann Sum: Midpoint Evaluation

Example Question #9 : Riemann Sum: Midpoint Evaluation

Example Question #11 : Riemann Sum: Midpoint Evaluation

Example Question #12 : Riemann Sum: Midpoint Evaluation

Example Question #12 : Riemann Sum: Midpoint Evaluation

All AP Calculus BC Resources

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