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AP Calculus BC : Riemann Sum: Midpoint Evaluation

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Example Questions

Example Question #31 : Integrals

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

Possible Answers:

2.88

0.99

0.51

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 #32 : Integrals

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

Possible Answers:

-593.17

-70.62

-183.60

-13.08

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 #11 : 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

2962.18

51.93

1194.43

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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Example Question #151 : Ap Calculus Bc

$\begin{align*}&\text{Calculate the Riemann sums integral approximation of :}\\&\int_{-2}^{2.5}(-cos(11e^{(2x)}))dx\\&\text{Using midpoints over }3\text{ intervals.}\end{align*}$

Possible Answers:

-3.55

-21.98

-12.41

-0.40

Correct answer:

-3.55

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-2}^{2.5}(-cos(11e^{(2x)}))dx\\&\text{So the interval is }[-2,2.5]\text{ the subintervals have length }\frac{2.5-(-2)}{3}=\frac{3}{2}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[-\frac{5}{4},\frac{1}{4},\frac{7}{4}]\\&\int_{-2}^{2.5}(-cos(11e^{(2x)}))dx=\frac{3}{2}[(-cos(11e^{(-\frac{5}{2})}))+(-cos(11e^{(\frac{1}{2})}))+(-cos(11e^{(\frac{7}{2})}))]\\&\int_{-2}^{2.5}(-cos(11e^{(2x)}))dx=-3.55\end{align*}$

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Example Question #12 : Riemann Sum: Midpoint Evaluation

$\begin{align*}&\text{Using the method of midpoint Riemann sums approximate the integral:}\\&\int_{0}^{4.8}(-tan(14sin(3x)))dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

23.92

4.60

0.53

1.18

Correct answer:

4.60

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{0}^{4.8}(-tan(14sin(3x)))dx\\&\text{So the interval is }[0,4.8]\text{ the subintervals have length }\frac{4.8-(0)}{4}=\frac{6}{5}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{3}{5},\frac{9}{5},3,\frac{21}{5}]\\&\int_{0}^{4.8}(-tan(14sin(3x)))dx=\frac{6}{5}[(-tan(14sin(\frac{9}{5})))+(-tan(14sin(\frac{27}{5})))+(-tan(14sin(9)))+(-tan(14sin(\frac{63}{5})))]\\&\int_{0}^{4.8}(-tan(14sin(3x)))dx=4.60\end{align*}$

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AP Calculus BC : Riemann Sum: Midpoint Evaluation

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #31 : Integrals

Example Question #32 : Integrals

Example Question #11 : Riemann Sum: Midpoint Evaluation

Example Question #151 : Ap Calculus Bc

Example Question #12 : Riemann Sum: Midpoint Evaluation

All AP Calculus BC Resources

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