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AP Calculus BC : Riemann Sum: Right Evaluation

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Example Questions

Example Question #2 : Riemann Sums

Given a function $y=\frac{5}{x}$ , find the Right Riemann Sum of the function on the interval [0,4] divided into four sub-intervals.

Possible Answers:

$\frac{123}{12}$

$\frac{122}{12}$

$\frac{125}{12}$

$\frac{124}{12}$

$\frac{126}{12}$

Correct answer:

$\frac{125}{12}$

Explanation:

In order to find the Riemann Sum of a given function, we need to approximate the area under the line or curve resulting from the function using rectangles spaced along equal sub-intervals of a given interval. Since we have an interval [0,4] divided into sub-intervals, we'll be using rectangles with vertices at x=0,1,2,3,4 .

To approximate the area under the curve, we need to find the areas of each rectangle in the sub-intervals. We already know the width or base of each rectangle is b=1 because the rectangles are spaced unit apart. Since we're looking for the Right Riemann Sum of $y=\frac{5}{x}$ , we want to find the heights of each rectangle by taking the values of each rightmost function value on each sub-interval, as follows:

$f(4)=\frac{5}{4}$

$f(3)=\frac{5}{3}$

$f(2)=\frac{5}{2}$

$f(1)=\frac{5}{1}=5$

Putting it all together, the Right Riemann Sum is

$A=bh=1(\frac{5}{4}+\frac{5}{3}+\frac{5}{2}+5)=\frac{15}{12}+\frac{20}{12}+\frac{30}{12}+\frac{60}{12}=\frac{125}{12}$ .

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Example Question #1 : Riemann Sum: Right Evaluation

$\begin{align*}&\text{Using the method of right Riemann sums approximate the integral:}\\&\int_{1}^{15.8}(-15sin(14e^{(4x)}))dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

3.35

5.20

20.79

191.30

Correct answer:

20.79

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{1}^{15.8}(-15sin(14e^{(4x)}))dx\\&\text{So the interval is }[1,15.8]\text{ the subintervals have length }\frac{15.8-(1)}{4}=\frac{37}{10}\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[\frac{47}{10},\frac{42}{5},\frac{121}{10},\frac{79}{5}]\\&\int_{1}^{15.8}(-15sin(14e^{(4x)}))dx=\frac{37}{10}[(-15sin(14e^{(4(\frac{47}{10}))}))+(-15sin(14e^{(4(\frac{42}{5}))}))+(-15sin(14e^{(4(\frac{121}{10}))}))+(-15sin(14e^{(4(\frac{79}{5}))}))]\\&\int_{1}^{15.8}(-15sin(14e^{(4x)}))dx=20.79\end{align*}$

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Example Question #1 : Riemann Sum: Right Evaluation

$\begin{align*}&\text{Using the method of right Riemann sums approximate the integral:}\\&\int_{4}^{14}(-8sin(12e^{(x)}))dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

-17.23

-51.69

-89.59

-143.01

Correct answer:

-17.23

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{4}^{14}(-8sin(12e^{(x)}))dx\\&\text{So the interval is }[4,14]\text{ the subintervals have length }\frac{14-(4)}{4}=\frac{5}{2}\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[\frac{13}{2},9,\frac{23}{2},14]\\&\int_{4}^{14}(-8sin(12e^{(x)}))dx=\frac{5}{2}[(-8sin(12e^{((\frac{13}{2}))}))+(-8sin(12e^{((9))}))+(-8sin(12e^{((\frac{23}{2}))}))+(-8sin(12e^{((14))}))]\\&\int_{4}^{14}(-8sin(12e^{(x)}))dx=-17.23\end{align*}$

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Example Question #4 : Riemann Sum: Right Evaluation

$\begin{align*}&\text{Using the method of right Riemann sums approximate the integral:}\\&\int_{0}^{6}(5x - 1000sin(3x)^{2})dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

-3390.69

-1412.79

-21361.33

-403.65

Correct answer:

-3390.69

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{0}^{6}(5x - 1000sin(3x)^{2})dx\\&\text{So the interval is }[0,6]\text{ the subintervals have length }\frac{6-(0)}{4}=\frac{3}{2}\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[\frac{3}{2},3,\frac{9}{2},6]\\&\int_{0}^{6}(5x - 1000sin(3x)^{2})dx=\frac{3}{2}[(5(\frac{3}{2}) - 1000sin(3(\frac{3}{2}))^{2})+(5(3) - 1000sin(3(3))^{2})+(5(\frac{9}{2}) - 1000sin(3(\frac{9}{2}))^{2})+(5(6) - 1000sin(3(6))^{2})]\\&\int_{0}^{6}(5x - 1000sin(3x)^{2})dx=-3390.69\end{align*}$

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Example Question #5 : Riemann Sum: Right Evaluation

$\begin{align*}&\text{Using the method of right Riemann sums approximate the integral:}\\&\int_{-4}^{-1.6}(15cos(12sin(6x)) + 5x^{3})dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

-52.18

-1732.45

-208.73

-31.63

Correct answer:

-208.73

Explanation:

$\begin{align*}\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[-\frac{16}{5},-\frac{12}{5},-\frac{8}{5}]\\&\int_{-4}^{-1.6}(15cos(12sin(6x)) + 5x^{3})dx\approx\frac{4}{5}[(15cos(12sin(6(-\frac{16}{5}))) + 5(-\frac{16}{5})^{3})+(15cos(12sin(6(-\frac{12}{5}))) + 5(-\frac{12}{5})^{3})+(15cos(12sin(6(-\frac{8}{5}))) + 5(-\frac{8}{5})^{3})]\\&\int_{-4}^{-1.6}(15cos(12sin(6x)) + 5x^{3})dx\approx-208.73\end{align*}$

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Example Question #2 : Riemann Sum: Right Evaluation

$\begin{align*}&\text{Using the method of right Riemann sums approximate the integral:}\\&\int_{0}^{13.6}(2x + 4sin(3tan(5x)))dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

214.27

1735.56

1092.76

89.28

Correct answer:

214.27

Explanation:

$\begin{align*}\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[\frac{17}{5},\frac{34}{5},\frac{51}{5},\frac{68}{5}]\\&\int_{0}^{13.6}(2x + 4sin(3tan(5x)))dx\approx\frac{17}{5}[(2(\frac{17}{5}) + 4sin(3tan(5(\frac{17}{5}))))+(2(\frac{34}{5}) + 4sin(3tan(5(\frac{34}{5}))))+(2(\frac{51}{5}) + 4sin(3tan(5(\frac{51}{5}))))+(2(\frac{68}{5}) + 4sin(3tan(5(\frac{68}{5}))))]\\&\int_{0}^{13.6}(2x + 4sin(3tan(5x)))dx\approx214.27\end{align*}$

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Example Question #1 : Riemann Sum: Right Evaluation

$\begin{align*}&\text{Using }3\text{ intervals, and the method of right point Riemann sums approximate the integral:}\\&\int_{-4}^{7.1}(4x - 15sin(13x^{2}))dx\end{align*}$

Possible Answers:

647.06

74.37

21.25

505.75

Correct answer:

74.37

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-4}^{7.1}(4x - 15sin(13x^{2}))dx\\&\text{So the interval is }[-4,7.1]\text{ the subintervals have length }\frac{7.1-(-4)}{3}=\frac{37}{10}\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[-\frac{3}{10},\frac{17}{5},\frac{71}{10}]\\&\int_{-4}^{7.1}(4x - 15sin(13x^{2}))dx\approx\frac{37}{10}[(4(-\frac{3}{10}) - 15sin(13(-\frac{3}{10})^{2}))+(4(\frac{17}{5}) - 15sin(13(\frac{17}{5})^{2}))+(4(\frac{71}{10}) - 15sin(13(\frac{71}{10})^{2}))]\\&\int_{-4}^{7.1}(4x - 15sin(13x^{2}))dx\approx74.37\end{align*}$

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Example Question #7 : Riemann Sum: Right Evaluation

$\begin{align*}&\text{Calculate the right-side Riemann sums integral approximation of :}\\&\int_{-1}^{15}(3x + 5sin(2e^{(2x)}))dx\\&\text{Using points over }4\text{ intervals.}\end{align*}$

Possible Answers:

431.64

3927.89

1553.89

2417.16

Correct answer:

431.64

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-1}^{15}(3x + 5sin(2e^{(2x)}))dx\\&\text{So the interval is }[-1,15]\text{ the subintervals have length }\frac{15-(-1)}{4}=4\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[3,7,11,15]\\&\int_{-1}^{15}(3x + 5sin(2e^{(2x)}))dx\approx4[(3(3) + 5sin(2e^{(2(3))}))+(3(7) + 5sin(2e^{(2(7))}))+(3(11) + 5sin(2e^{(2(11))}))+(3(15) + 5sin(2e^{(2(15))}))]\\&\int_{-1}^{15}(3x + 5sin(2e^{(2x)}))dx\approx431.64\end{align*}$

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Example Question #8 : Riemann Sum: Right Evaluation

$\begin{align*}&\text{Calculate the right-side Riemann sums integral approximation of :}\\&\int_{3}^{4.8}(4x + 16cos(13e^{(3x)}))dx\\&\text{Using points over }3\text{ intervals.}\end{align*}$

Possible Answers:

3.88

34.93

136.22

6.35

Correct answer:

34.93

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{3}^{4.8}(4x + 16cos(13e^{(3x)}))dx\\&\text{So the interval is }[3,4.8]\text{ the subintervals have length }\frac{4.8-(3)}{3}=\frac{3}{5}\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[\frac{18}{5},\frac{21}{5},\frac{24}{5}]\\&\int_{3}^{4.8}(4x + 16cos(13e^{(3x)}))dx\approx\frac{3}{5}[(4(\frac{18}{5}) + 16cos(13e^{(3(\frac{18}{5}))}))+(4(\frac{21}{5}) + 16cos(13e^{(3(\frac{21}{5}))}))+(4(\frac{24}{5}) + 16cos(13e^{(3(\frac{24}{5}))}))]\\&\int_{3}^{4.8}(4x + 16cos(13e^{(3x)}))dx\approx34.93\end{align*}$

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Example Question #1 : Riemann Sum: Right Evaluation

$\begin{align*}&\text{Calculate the right-side Riemann sums integral approximation of :}\\&\int_{-5}^{15}(4x + 18sin(tan(5x)))dx\\&\text{Using points over }4\text{ intervals.}\end{align*}$

Possible Answers:

57.29

527.09

2635.46

1265.02

Correct answer:

527.09

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-5}^{15}(4x + 18sin(tan(5x)))dx\\&\text{So the interval is }[-5,15]\text{ the subintervals have length }\frac{15-(-5)}{4}=5\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[0,5,10,15]\\&\int_{-5}^{15}(4x + 18sin(tan(5x)))dx\approx5[(4(0) + 18sin(tan(5(0))))+(4(5) + 18sin(tan(5(5))))+(4(10) + 18sin(tan(5(10))))+(4(15) + 18sin(tan(5(15))))]\\&\int_{-5}^{15}(4x + 18sin(tan(5x)))dx\approx527.09\end{align*}$

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AP Calculus BC : Riemann Sum: Right Evaluation

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

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Example Question #2 : Riemann Sums

Example Question #1 : Riemann Sum: Right Evaluation

Example Question #1 : Riemann Sum: Right Evaluation

Example Question #4 : Riemann Sum: Right Evaluation

Example Question #5 : Riemann Sum: Right Evaluation

Example Question #2 : Riemann Sum: Right Evaluation

Example Question #1 : Riemann Sum: Right Evaluation

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Example Question #1 : Riemann Sum: Right Evaluation

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