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Calculus 2 : Riemann Sums

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Example Questions

Example Question #90 : Introduction To Integrals

$\begin{align*}&\text{Using }4\text{ intervals, and the method of midpoint Riemann sums approximate the integral:}\\&\int_{3}^{10.2}(-14sin(6x))dx\end{align*}$

Possible Answers:

30.58

1.38

2.35

11.76

Correct answer:

11.76

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{3}^{10.2}(-14sin(6x))dx\\&\text{So the interval is}[3,10.2]\text{ the subintervals have length}\frac{10.2-(3)}{4}=1.8\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{39}{10},\frac{57}{10},\frac{15}{2},\frac{93}{10}]\\&\int_{3}^{10.2}(-14sin(6x))dx=1.8[(-14sin(\frac{117}{5}))+(-14sin(\frac{171}{5}))+(-14sin(45))+(-14sin(\frac{279}{5}))]\\&\int_{3}^{10.2}(-14sin(6x))dx=11.76\end{align*}$

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Example Question #91 : Integrals

$\begin{align*}&\text{Calculate the Riemann sums integral approximation of :}\\&\int_{4}^{11.5}(-5sin(4x))dx\\&\text{Using midpoints over }3\text{ intervals.}\end{align*}$

Possible Answers:

-9.59

-0.52

-30.82

-3.42

Correct answer:

-3.42

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{4}^{11.5}(-5sin(4x))dx\\&\text{So the interval is}[4,11.5]\text{ the subintervals have length}\frac{11.5-(4)}{3}=2.5\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{21}{4},\frac{31}{4},\frac{41}{4}]\\&\int_{4}^{11.5}(-5sin(4x))dx=2.5[(-5sin(21))+(-5sin(31))+(-5sin(41))]\\&\int_{4}^{11.5}(-5sin(4x))dx=-3.42\end{align*}$

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Example Question #92 : Integrals

$\begin{align*}&\text{Using the method of midpoint Riemann sums approximate the integral:}\\&\int_{3}^{13.2}(-19sin(4x))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

-108.85

-435.41

-696.65

-10.89

Correct answer:

-108.85

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{3}^{13.2}(-19sin(4x))dx\\&\text{So the interval is }[3,13.2]\text{ the subintervals have length }\frac{13.2-(3)}{3}=3.4\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{47}{10},\frac{81}{10},\frac{23}{2}]\\&\int_{3}^{13.2}(-19sin(4x))dx=3.4[(-19sin(\frac{94}{5}))+(-19sin(\frac{162}{5}))+(-19sin(46))]\\&\int_{3}^{13.2}(-19sin(4x))dx=-108.85\end{align*}$

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Example Question #93 : Integrals

$\begin{align*}&\text{Using }4\text{ intervals, and the method of right point Riemann sums approximate the integral:}\\&\int_{3}^{10.6}(16x^{2})dx\end{align*}$

Possible Answers:

66744.72

7852.32

1454.13

2181.20

Correct answer:

7852.32

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{3}^{10.6}(16x^{2})dx\\&\text{So the interval is }[3,10.6]\text{ the subintervals have length }\frac{10.6-(3)}{4}=1.9\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[\frac{49}{10},\frac{34}{5},\frac{87}{10},\frac{53}{5}]\\&\int_{3}^{10.6}(16x^{2})dx=1.9[(\frac{9604}{25})+(\frac{18496}{25})+(\frac{30276}{25})+(\frac{44944}{25})]\\&\int_{3}^{10.6}(16x^{2})dx=7852.32\end{align*}$

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Example Question #94 : Integrals

$\begin{align*}&\text{Using the method of right Riemann sums approximate the integral:}\\&\int_{-5}^{-1.4}(14x^{2})dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

389.09

3735.24

69.48

117.91

Correct answer:

389.09

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-5}^{-1.4}(14x^{2})dx\\&\text{So the interval is }[-5,-1.4]\text{ the subintervals have length }\frac{-1.4-(-5)}{3}=1.2\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[-\frac{19}{5},-\frac{13}{5},-\frac{7}{5}]\\&\int_{-5}^{-1.4}(14x^{2})dx=1.2[(\frac{5054}{25})+(\frac{2366}{25})+(\frac{686}{25})]\\&\int_{-5}^{-1.4}(14x^{2})dx=389.09\end{align*}$

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Example Question #95 : Integrals

$\begin{align*}&\text{Using }3\text{ intervals, and the method of right point Riemann sums approximate the integral:}\\&\int_{5}^{6.5}(19x^{2})dx\end{align*}$

Possible Answers:

3298.40

6802.95

1030.75

121.26

Correct answer:

1030.75

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{5}^{6.5}(19x^{2})dx\\&\text{So the interval is }[5,6.5]\text{ the subintervals have length }\frac{6.5-(5)}{3}=0.5\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[\frac{11}{2},6,\frac{13}{2}]\\&\int_{5}^{6.5}(19x^{2})dx=0.5[(\frac{2299}{4})+(684)+(\frac{3211}{4})]\\&\int_{5}^{6.5}(19x^{2})dx=1030.75\end{align*}$

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Example Question #96 : Integrals

$\begin{align*}&\text{Using }3\text{ intervals, and the method of right point Riemann sums approximate the integral:}\\&\int_{-1}^{13.7}(-15ln(6x))dx\end{align*}$

Possible Answers:

-86.46

-211.83

-847.34

-162.95

Correct answer:

-847.34

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-1}^{13.7}(-15ln(6x))dx\\&\text{So the interval is }[-1,13.7]\text{ the subintervals have length }\frac{13.7-(-1)}{3}=4.9\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[\frac{39}{10},\frac{44}{5},\frac{137}{10}]\\&\int_{-1}^{13.7}(-15ln(6x))dx=4.9[(-15ln(\frac{117}{5}))+(-15ln(\frac{264}{5}))+(-15ln(\frac{411}{5}))]\\&\int_{-1}^{13.7}(-15ln(6x))dx=-847.34\end{align*}$

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Example Question #97 : Integrals

$\begin{align*}&\text{Calculate the right-side Riemann sums integral approximation of :}\\&\int_{2}^{5.5}(7tan(2x))dx\\&\text{Using points over }5\text{ intervals.}\end{align*}$

Possible Answers:

-3593.81

-10332.19

-6064.55

-1123.06

Correct answer:

-1123.06

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{2}^{5.5}(7tan(2x))dx\\&\text{So the interval is }[2,5.5]\text{ the subintervals have length }\frac{5.5-(2)}{5}=0.7\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[\frac{27}{10},\frac{17}{5},\frac{41}{10},\frac{24}{5},\frac{11}{2}]\\&\int_{2}^{5.5}(7tan(2x))dx=0.7[(7tan(\frac{27}{5}))+(7tan(\frac{34}{5}))+(7tan(\frac{41}{5}))+(7tan(\frac{48}{5}))+(7tan(11))]\\&\int_{2}^{5.5}(7tan(2x))dx=-1123.06\end{align*}$

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Example Question #98 : Integrals

$\begin{align*}&\text{Using the method of right Riemann sums approximate the integral:}\\&\int_{-4}^{9.5}(3cos(5x))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

-1.06

-9.99

-4.00

-1.54

Correct answer:

-9.99

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-4}^{9.5}(3cos(5x))dx\\&\text{So the interval is }[-4,9.5]\text{ the subintervals have length }\frac{9.5-(-4)}{3}=4.5\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[\frac{1}{2},5,\frac{19}{2}]\\&\int_{-4}^{9.5}(3cos(5x))dx=4.5[(3cos(\frac{5}{2}))+(3cos(25))+(3cos(\frac{95}{2}))]\\&\int_{-4}^{9.5}(3cos(5x))dx=-9.99\end{align*}$

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Example Question #99 : Integrals

$\begin{align*}&\text{Calculate the right-side Riemann sums integral approximation of :}\\&\int_{-2}^{6.7}(-3sin(6x))dx\\&\text{Using points over }3\text{ intervals.}\end{align*}$

Possible Answers:

62.54

7.82

1.13

3.40

Correct answer:

7.82

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-2}^{6.7}(-3sin(6x))dx\\&\text{So the interval is }[-2,6.7]\text{ the subintervals have length }\frac{6.7-(-2)}{3}=2.9\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[\frac{9}{10},\frac{19}{5},\frac{67}{10}]\\&\int_{-2}^{6.7}(-3sin(6x))dx=2.9[(-3sin(\frac{27}{5}))+(-3sin(\frac{114}{5}))+(-3sin(\frac{201}{5}))]\\&\int_{-2}^{6.7}(-3sin(6x))dx=7.82\end{align*}$

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Calculus 2 : Riemann Sums

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #90 : Introduction To Integrals

Example Question #91 : Integrals

Example Question #92 : Integrals

Example Question #93 : Integrals

Example Question #94 : Integrals

Example Question #95 : Integrals

Example Question #96 : Integrals

Example Question #97 : Integrals

Example Question #98 : Integrals

Example Question #99 : Integrals

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