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Calculus 2 : Riemann Sums

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Example Question #61 : Riemann Sums

$\begin{align*}&\text{Calculate the Riemann sums integral approximation of :}\\&\int_{2}^{4.4}(7tan(16ln(6x)))dx\\&\text{Using left points over }4\text{ intervals.}\end{align*}$

Possible Answers:

-0.93

-8.37

-1.29

-20.09

Correct answer:

-8.37

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{2}^{4.4}(7tan(16ln(6x)))dx\\&\text{So the interval is }[2,4.4]\text{ the subintervals have length }\frac{4.4-(2)}{4}=\frac{3}{5}\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[2,\frac{13}{5},\frac{16}{5},\frac{19}{5}]\\&\int_{2}^{4.4}(7tan(16ln(6x)))dx=\frac{3}{5}[(7tan(16ln(12)))+(7tan(16ln(\frac{78}{5})))+(7tan(16ln(\frac{96}{5})))+(7tan(16ln(\frac{114}{5})))]\\&\int_{2}^{4.4}(7tan(16ln(6x)))dx=-8.37\end{align*}$

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Example Question #111 : Integrals

$\begin{align*}&\text{Using the method of left point Riemann sums approximate the integral:}\\&\int_{3}^{20.2}(160sin(6x)^{2})dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

283.62

1134.48

140.06

6806.89

Correct answer:

1134.48

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{3}^{20.2}(160sin(6x)^{2})dx\\&\text{So the interval is }[3,20.2]\text{ the subintervals have length }\frac{20.2-(3)}{4}=\frac{43}{10}\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[3,\frac{73}{10},\frac{58}{5},\frac{159}{10}]\\&\int_{3}^{20.2}(160sin(6x)^{2})dx=\frac{43}{10}[(160sin(18)^{2})+(160sin(\frac{219}{5})^{2})+(160sin(\frac{348}{5})^{2})+(160sin(\frac{477}{5})^{2})]\\&\int_{3}^{20.2}(160sin(6x)^{2})dx=1134.48\end{align*}$

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Example Question #112 : Integrals

$\begin{align*}&\text{Using the method of left point Riemann sums approximate the integral:}\\&\int_{-4}^{4.1}(11sin(17cos(x)))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

2.31

7.61

73.83

44.91

Correct answer:

7.61

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-4}^{4.1}(11sin(17cos(x)))dx\\&\text{So the interval is }[-4,4.1]\text{ the subintervals have length }\frac{4.1-(-4)}{3}=\frac{27}{10}\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[-4,-\frac{13}{10},\frac{7}{5}]\\&\int_{-4}^{4.1}(11sin(17cos(x)))dx=\frac{27}{10}[(11sin(17cos(4)))+(11sin(17cos(\frac{13}{10})))+(11sin(17cos(\frac{7}{5})))]\\&\int_{-4}^{4.1}(11sin(17cos(x)))dx=7.61\end{align*}$

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Example Question #113 : Integrals

$\begin{align*}&\text{Calculate the Riemann sums integral approximation of :}\\&\int_{0}^{3}(6cos(20e^{(x)}))dx\\&\text{Using left points over }3\text{ intervals.}\end{align*}$

Possible Answers:

-0.81

-6.95

-1.20

-2.24

Correct answer:

-6.95

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{0}^{3}(6cos(20e^{(x)}))dx\\&\text{So the interval is }[0,3]\text{ the subintervals have length }\frac{3-(0)}{3}=1\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[0,1,2]\\&\int_{0}^{3}(6cos(20e^{(x)}))dx=1[(6cos(20))+(6cos(20e^{(1)}))+(6cos(20e^{(2)}))]\\&\int_{0}^{3}(6cos(20e^{(x)}))dx=-6.95\end{align*}$

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Example Question #114 : Integrals

$\begin{align*}&\text{Using }4\text{ intervals, and the method of left point Riemann sums approximate the integral:}\\&\int_{-1}^{12.2}(5sin(9e^{(x)}))dx\end{align*}$

Possible Answers:

7.57

42.40

135.69

4.99

Correct answer:

42.40

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-1}^{12.2}(5sin(9e^{(x)}))dx\\&\text{So the interval is }[-1,12.2]\text{ the subintervals have length }\frac{12.2-(-1)}{4}=\frac{33}{10}\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[-1,\frac{23}{10},\frac{28}{5},\frac{89}{10}]\\&\int_{-1}^{12.2}(5sin(9e^{(x)}))dx=\frac{33}{10}[(5sin(9e^{(-1)}))+(5sin(9e^{(\frac{23}{10})}))+(5sin(9e^{(\frac{28}{5})}))+(5sin(9e^{(\frac{89}{10})}))]\\&\int_{-1}^{12.2}(5sin(9e^{(x)}))dx=42.40\end{align*}$

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Example Question #115 : Integrals

$\begin{align*}&\text{Using the method of left point Riemann sums approximate the integral:}\\&\int_{1}^{8.2}(-6cos(19x^{2}))dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

-211.88

-25.53

-51.06

-142.95

Correct answer:

-25.53

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{1}^{8.2}(-6cos(19x^{2}))dx\\&\text{So the interval is }[1,8.2]\text{ the subintervals have length }\frac{8.2-(1)}{4}=\frac{9}{5}\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[1,\frac{14}{5},\frac{23}{5},\frac{32}{5}]\\&\int_{1}^{8.2}(-6cos(19x^{2}))dx=\frac{9}{5}[(-6cos(19))+(-6cos(\frac{3724}{25}))+(-6cos(\frac{10051}{25}))+(-6cos(\frac{19456}{25}))]\\&\int_{1}^{8.2}(-6cos(19x^{2}))dx=-25.53\end{align*}$

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Example Question #116 : Integrals

$\begin{align*}&\text{Using the method of left point Riemann sums approximate the integral:}\\&\int_{2}^{8}(17cos(2sin(2x)))dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

4.18

36.79

7.36

10.22

Correct answer:

36.79

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{2}^{8}(17cos(2sin(2x)))dx\\&\text{So the interval is }[2,8]\text{ the subintervals have length }\frac{8-(2)}{4}=\frac{3}{2}\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[2,\frac{7}{2},5,\frac{13}{2}]\\&\int_{2}^{8}(17cos(2sin(2x)))dx=\frac{3}{2}[(17cos(2sin(4)))+(17cos(2sin(7)))+(17cos(2sin(10)))+(17cos(2sin(13)))]\\&\int_{2}^{8}(17cos(2sin(2x)))dx=36.79\end{align*}$

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Example Question #117 : Integrals

$\begin{align*}&\text{Using the method of midpoint Riemann sums approximate the integral:}\\&\int_{0}^{6.6}(-196cos(3x)^{2})dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

-2825.01

-974.14

-170.90

-115.97

Correct answer:

-974.14

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{0}^{6.6}(-196cos(3x)^{2})dx\\&\text{So the interval is }[0,6.6]\text{ the subintervals have length }\frac{6.6-(0)}{3}=\frac{11}{5}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{11}{10},\frac{33}{10},\frac{11}{2}]\\&\int_{0}^{6.6}(-196cos(3x)^{2})dx=\frac{11}{5}[(-196cos(\frac{33}{10})^{2})+(-196cos(\frac{99}{10})^{2})+(-196cos(\frac{33}{2})^{2})]\\&\int_{0}^{6.6}(-196cos(3x)^{2})dx=-974.14\end{align*}$

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Example Question #118 : Integrals

$\begin{align*}&\text{Using }4\text{ intervals, and the method of midpoint Riemann sums approximate the integral:}\\&\int_{3}^{8.6}(11cos(6e^{(4x)}))dx\end{align*}$

Possible Answers:

0.93

59.17

5.98

1.81

Correct answer:

5.98

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{3}^{8.6}(11cos(6e^{(4x)}))dx\\&\text{So the interval is }[3,8.6]\text{ the subintervals have length }\frac{8.6-(3)}{4}=\frac{7}{5}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{37}{10},\frac{51}{10},\frac{13}{2},\frac{79}{10}]\\&\int_{3}^{8.6}(11cos(6e^{(4x)}))dx=\frac{7}{5}[(11cos(6e^{(\frac{74}{5})}))+(11cos(6e^{(\frac{102}{5})}))+(11cos(6e^{(26)}))+(11cos(6e^{(\frac{158}{5})}))]\\&\int_{3}^{8.6}(11cos(6e^{(4x)}))dx=5.98\end{align*}$

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Example Question #119 : Integrals

$\begin{align*}&\text{Calculate the Riemann sums integral approximation of :}\\&\int_{-1}^{14.2}(-9cos(16e^{(2x)}))dx\\&\text{Using midpoints over }4\text{ intervals.}\end{align*}$

Possible Answers:

-5.18

-6.85

-167.71

-46.59

Correct answer:

-46.59

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-1}^{14.2}(-9cos(16e^{(2x)}))dx\\&\text{So the interval is }[-1,14.2]\text{ the subintervals have length }\frac{14.2-(-1)}{4}=\frac{19}{5}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{9}{10},\frac{47}{10},\frac{17}{2},\frac{123}{10}]\\&\int_{-1}^{14.2}(-9cos(16e^{(2x)}))dx=\frac{19}{5}[(-9cos(16e^{(\frac{9}{5})}))+(-9cos(16e^{(\frac{47}{5})}))+(-9cos(16e^{(17)}))+(-9cos(16e^{(\frac{123}{5})}))]\\&\int_{-1}^{14.2}(-9cos(16e^{(2x)}))dx=-46.59\end{align*}$

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Calculus 2 : Riemann Sums

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #61 : Riemann Sums

Example Question #111 : Integrals

Example Question #112 : Integrals

Example Question #113 : Integrals

Example Question #114 : Integrals

Example Question #115 : Integrals

Example Question #116 : Integrals

Example Question #117 : Integrals

Example Question #118 : Integrals

Example Question #119 : Integrals

All Calculus 2 Resources

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