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Calculus 2 : Riemann Sums

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Example Question #81 : Riemann Sums

$\begin{align*}&\text{Using the method of right Riemann sums approximate the integral:}\\&\int_{-2}^{1}(12cos(5e^{(6x)}))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

64.93

246.20

3.92

27.05

Correct answer:

27.05

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-2}^{1}(12cos(5e^{(6x)}))dx\\&\text{So the interval is }[-2,1]\text{ the subintervals have length }\frac{1-(-2)}{3}=1\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[-1,0,1]\\&\int_{-2}^{1}(12cos(5e^{(6x)}))dx=1[(12cos(5e^{(-6)}))+(12cos(5))+(12cos(5e^{(6)}))]\\&\int_{-2}^{1}(12cos(5e^{(6x)}))dx=27.05\end{align*}$

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Example Question #82 : Riemann Sums

$\begin{align*}&\text{Calculate the right-side Riemann sums integral approximation of :}\\&\int_{4}^{14.2}(14sin(15ln(6x)))dx\\&\text{Using points over }3\text{ intervals.}\end{align*}$

Possible Answers:

-3.11

-26.74

-7.43

-5.14

Correct answer:

-26.74

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{4}^{14.2}(14sin(15ln(6x)))dx\\&\text{So the interval is }[4,14.2]\text{ the subintervals have length }\frac{14.2-(4)}{3}=\frac{17}{5}\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[\frac{37}{5},\frac{54}{5},\frac{71}{5}]\\&\int_{4}^{14.2}(14sin(15ln(6x)))dx=\frac{17}{5}[(14sin(15ln(6(\frac{37}{5}))))+(14sin(15ln(6(\frac{54}{5}))))+(14sin(15ln(6(\frac{71}{5}))))]\\&\int_{4}^{14.2}(14sin(15ln(6x)))dx=-26.74\end{align*}$

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Example Question #83 : Riemann Sums

$\begin{align*}&\text{Using the method of right Riemann sums approximate the integral:}\\&\int_{-3}^{0}(4tan(19x^{2}))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

0.93

19.78

31.86

3.35

Correct answer:

3.35

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-3}^{0}(4tan(19x^{2}))dx\\&\text{So the interval is }[-3,0]\text{ the subintervals have length }\frac{0-(-3)}{3}=1\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[-2,-1,0]\\&\int_{-3}^{0}(4tan(19x^{2}))dx=1[(4tan(19(-2)^{2}))+(4tan(19(-1)^{2}))+(4tan(19(0)^{2}))]\\&\int_{-3}^{0}(4tan(19x^{2}))dx=3.35\end{align*}$

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Example Question #84 : Riemann Sums

$\begin{align*}&\text{Using }3\text{ intervals, and the method of right point Riemann sums approximate the integral:}\\&\int_{-3}^{11.4}(4sin(5tan(x)))dx\end{align*}$

Possible Answers:

150.03

23.81

221.47

83.35

Correct answer:

23.81

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-3}^{11.4}(4sin(5tan(x)))dx\\&\text{So the interval is }[-3,11.4]\text{ the subintervals have length }\frac{11.4-(-3)}{3}=\frac{24}{5}\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[\frac{9}{5},\frac{33}{5},\frac{57}{5}]\\&\int_{-3}^{11.4}(4sin(5tan(x)))dx=\frac{24}{5}[(4sin(5tan((\frac{9}{5}))))+(4sin(5tan((\frac{33}{5}))))+(4sin(5tan((\frac{57}{5}))))]\\&\int_{-3}^{11.4}(4sin(5tan(x)))dx=23.81\end{align*}$

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Example Question #85 : Riemann Sums

$\begin{align*}&\text{Calculate the right-side Riemann sums integral approximation of :}\\&\int_{-5}^{9.1}(-13tan(16sin(6x)))dx\\&\text{Using points over }3\text{ intervals.}\end{align*}$

Possible Answers:

-46.76

-5.08

-112.23

-233.81

Correct answer:

-46.76

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-5}^{9.1}(-13tan(16sin(6x)))dx\\&\text{So the interval is }[-5,9.1]\text{ the subintervals have length }\frac{9.1-(-5)}{3}=\frac{47}{10}\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[-\frac{3}{10},\frac{22}{5},\frac{91}{10}]\\&\int_{-5}^{9.1}(-13tan(16sin(6x)))dx=\frac{47}{10}[(-13tan(16sin(6(-\frac{3}{10}))))+(-13tan(16sin(6(\frac{22}{5}))))+(-13tan(16sin(6(\frac{91}{10}))))]\\&\int_{-5}^{9.1}(-13tan(16sin(6x)))dx=-46.76\end{align*}$

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Example Question #86 : Riemann Sums

$\begin{align*}&\text{Calculate the right-side Riemann sums integral approximation of :}\\&\int_{5}^{20}(-5sin(20sin(3x)))dx\\&\text{Using points over }3\text{ intervals.}\end{align*}$

Possible Answers:

4.13

6.77

11.55

39.26

Correct answer:

39.26

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{5}^{20}(-5sin(20sin(3x)))dx\\&\text{So the interval is }[5,20]\text{ the subintervals have length }\frac{20-(5)}{3}=5\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[10,15,20]\\&\int_{5}^{20}(-5sin(20sin(3x)))dx=5[(-5sin(20sin(3(10))))+(-5sin(20sin(3(15))))+(-5sin(20sin(3(20))))]\\&\int_{5}^{20}(-5sin(20sin(3x)))dx=39.26\end{align*}$

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Example Question #87 : Riemann Sums

$\begin{align*}&\text{Using }4\text{ intervals, and the method of right point Riemann sums approximate the integral:}\\&\int_{4}^{4.4}(15cos(16tan(3x)))dx\end{align*}$

Possible Answers:

1.17

0.49

24.54

2.92

Correct answer:

2.92

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{4}^{4.4}(15cos(16tan(3x)))dx\\&\text{So the interval is }[4,4.4]\text{ the subintervals have length }\frac{4.4-(4)}{4}=\frac{1}{10}\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[\frac{41}{10},\frac{21}{5},\frac{43}{10},\frac{22}{5}]\\&\int_{4}^{4.4}(15cos(16tan(3x)))dx=\frac{1}{10}[(15cos(16tan(3(\frac{41}{10}))))+(15cos(16tan(3(\frac{21}{5}))))+(15cos(16tan(3(\frac{43}{10}))))+(15cos(16tan(3(\frac{22}{5}))))]\\&\int_{4}^{4.4}(15cos(16tan(3x)))dx=2.92\end{align*}$

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Example Question #88 : Riemann Sums

$\begin{align*}&\text{Calculate the right-side Riemann sums integral approximation of :}\\&\int_{-4}^{14.8}(2tan(2e^{(2x)}))dx\\&\text{Using points over }4\text{ intervals.}\end{align*}$

Possible Answers:

-29.51

-512.27

-795.86

-91.48

Correct answer:

-91.48

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-4}^{14.8}(2tan(2e^{(2x)}))dx\\&\text{So the interval is }[-4,14.8]\text{ the subintervals have length }\frac{14.8-(-4)}{4}=\frac{47}{10}\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[\frac{7}{10},\frac{27}{5},\frac{101}{10},\frac{74}{5}]\\&\int_{-4}^{14.8}(2tan(2e^{(2x)}))dx=\frac{47}{10}[(2tan(2e^{(2(\frac{7}{10}))}))+(2tan(2e^{(2(\frac{27}{5}))}))+(2tan(2e^{(2(\frac{101}{10}))}))+(2tan(2e^{(2(\frac{74}{5}))}))]\\&\int_{-4}^{14.8}(2tan(2e^{(2x)}))dx=-91.48\end{align*}$

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Example Question #131 : Introduction To Integrals

$\begin{align*}&\text{Calculate the right-side Riemann sums integral approximation of :}\\&\int_{-4}^{8.8}(-17tan(2e^{(3x)}))dx\\&\text{Using points over }4\text{ intervals.}\end{align*}$

Possible Answers:

139.92

36.82

6.35

3.72

Correct answer:

36.82

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-4}^{8.8}(-17tan(2e^{(3x)}))dx\\&\text{So the interval is }[-4,8.8]\text{ the subintervals have length }\frac{8.8-(-4)}{4}=\frac{16}{5}\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[-\frac{4}{5},\frac{12}{5},\frac{28}{5},\frac{44}{5}]\\&\int_{-4}^{8.8}(-17tan(2e^{(3x)}))dx=\frac{16}{5}[(-17tan(2e^{(3(-\frac{4}{5}))}))+(-17tan(2e^{(3(\frac{12}{5}))}))+(-17tan(2e^{(3(\frac{28}{5}))}))+(-17tan(2e^{(3(\frac{44}{5}))}))]\\&\int_{-4}^{8.8}(-17tan(2e^{(3x)}))dx=36.82\end{align*}$

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Example Question #90 : Riemann Sums

$\begin{align*}&\text{Using }3\text{ intervals, and the method of right point Riemann sums approximate the integral:}\\&\int_{-4}^{10.1}(17cos(13cos(3x)))dx\end{align*}$

Possible Answers:

57.94

1670.47

1095.69

179.62

Correct answer:

179.62

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-4}^{10.1}(17cos(13cos(3x)))dx\\&\text{So the interval is }[-4,10.1]\text{ the subintervals have length }\frac{10.1-(-4)}{3}=\frac{47}{10}\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[\frac{7}{10},\frac{27}{5},\frac{101}{10}]\\&\int_{-4}^{10.1}(17cos(13cos(3x)))dx=\frac{47}{10}[(17cos(13cos(3(\frac{7}{10}))))+(17cos(13cos(3(\frac{27}{5}))))+(17cos(13cos(3(\frac{101}{10}))))]\\&\int_{-4}^{10.1}(17cos(13cos(3x)))dx=179.62\end{align*}$

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Calculus 2 : Riemann Sums

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #81 : Riemann Sums

Example Question #82 : Riemann Sums

Example Question #83 : Riemann Sums

Example Question #84 : Riemann Sums

Example Question #85 : Riemann Sums

Example Question #86 : Riemann Sums

Example Question #87 : Riemann Sums

Example Question #88 : Riemann Sums

Example Question #131 : Introduction To Integrals

Example Question #90 : Riemann Sums

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