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Calculus 2 : Riemann Sums

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Example Question #1861 : Calculus Ii

$\begin{align*}&\text{Using the method of midpoint Riemann sums approximate the integral:}\\&\int_{-1}^{2}(-2sin(17cos(2x)))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

-1.25

-24.76

-16.51

-2.75

Correct answer:

-2.75

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-1}^{2}(-2sin(17cos(2x)))dx\\&\text{So the interval is }[-1,2]\text{ the subintervals have length }\frac{2-(-1)}{3}=1\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[-\frac{1}{2},\frac{1}{2},\frac{3}{2}]\\&\int_{-1}^{2}(-2sin(17cos(2x)))dx=1[(-2sin(17cos(1)))+(-2sin(17cos(1)))+(-2sin(17cos(3)))]\\&\int_{-1}^{2}(-2sin(17cos(2x)))dx=-2.75\end{align*}$

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Example Question #72 : Riemann Sums

$\begin{align*}&\text{Calculate the Riemann sums integral approximation of :}\\&\int_{5}^{16.6}(tan(12sin(6x)))dx\\&\text{Using midpoints over }4\text{ intervals.}\end{align*}$

Possible Answers:

4.04

12.92

124.00

78.79

Correct answer:

12.92

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{5}^{16.6}(tan(12sin(6x)))dx\\&\text{So the interval is }[5,16.6]\text{ the subintervals have length }\frac{16.6-(5)}{4}=\frac{29}{10}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{129}{20},\frac{187}{20},\frac{49}{4},\frac{303}{20}]\\&\int_{5}^{16.6}(tan(12sin(6x)))dx=\frac{29}{10}[(tan(12sin(\frac{387}{10})))+(tan(12sin(\frac{561}{10})))+(tan(12sin(\frac{147}{2})))+(tan(12sin(\frac{909}{10})))]\\&\int_{5}^{16.6}(tan(12sin(6x)))dx=12.92\end{align*}$

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Example Question #73 : Riemann Sums

$\begin{align*}&\text{Using the method of midpoint Riemann sums approximate the integral:}\\&\int_{-4}^{-3.6}(-13sin(10tan(2x)))dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

20.39

12.38

2.43

7.77

Correct answer:

2.43

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-4}^{-3.6}(-13sin(10tan(2x)))dx\\&\text{So the interval is }[-4,-3.6]\text{ the subintervals have length }\frac{-3.6-(-4)}{4}=\frac{1}{10}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[-\frac{79}{20},-\frac{77}{20},-\frac{15}{4},-\frac{73}{20}]\\&\int_{-4}^{-3.6}(-13sin(10tan(2x)))dx=\frac{1}{10}[(13sin(10tan(\frac{79}{10})))+(13sin(10tan(\frac{77}{10})))+(13sin(10tan(\frac{15}{2})))+(13sin(10tan(\frac{73}{10})))]\\&\int_{-4}^{-3.6}(-13sin(10tan(2x)))dx=2.43\end{align*}$

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Example Question #74 : Riemann Sums

$\begin{align*}&\text{Using the method of midpoint Riemann sums approximate the integral:}\\&\int_{5}^{23.8}(3ln(-16tan(2x)))dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

187.14

21.03

1048.00

729.86

Correct answer:

187.14

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{5}^{23.8}(3ln(-16tan(2x)))dx\\&\text{So the interval is }[5,23.8]\text{ the subintervals have length }\frac{23.8-(5)}{4}=\frac{47}{10}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{147}{20},\frac{241}{20},\frac{67}{4},\frac{429}{20}]\\&\int_{5}^{23.8}(3ln(-16tan(2x)))dx=\frac{47}{10}[(3ln(-16tan(\frac{147}{10})))+(3ln(-16tan(\frac{241}{10})))+(3ln(-16tan(\frac{67}{2})))+(3ln(-16tan(\frac{429}{10})))]\\&\int_{5}^{23.8}(3ln(-16tan(2x)))dx=187.14\end{align*}$

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Example Question #75 : Riemann Sums

$\begin{align*}&\text{Calculate the Riemann sums integral approximation of :}\\&\int_{4}^{15.2}(-7sin(9e^{(2x)}))dx\\&\text{Using midpoints over }4\text{ intervals.}\end{align*}$

Possible Answers:

-405.43

-6.65

-40.54

-18.43

Correct answer:

-40.54

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{4}^{15.2}(-7sin(9e^{(2x)}))dx\\&\text{So the interval is }[4,15.2]\text{ the subintervals have length }\frac{15.2-(4)}{4}=\frac{14}{5}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{27}{5},\frac{41}{5},11,\frac{69}{5}]\\&\int_{4}^{15.2}(-7sin(9e^{(2x)}))dx=\frac{14}{5}[(-7sin(9e^{(\frac{54}{5})}))+(-7sin(9e^{(\frac{82}{5})}))+(-7sin(9e^{(22)}))+(-7sin(9e^{(\frac{138}{5})}))]\\&\int_{4}^{15.2}(-7sin(9e^{(2x)}))dx=-40.54\end{align*}$

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Example Question #76 : Riemann Sums

$\begin{align*}&\text{Calculate the Riemann sums integral approximation of :}\\&\int_{-2}^{10.6}(-13cos(15e^{(x)}))dx\\&\text{Using midpoints over }3\text{ intervals.}\end{align*}$

Possible Answers:

-325.29

-49.29

-5.24

-14.08

Correct answer:

-49.29

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-2}^{10.6}(-13cos(15e^{(x)}))dx\\&\text{So the interval is }[-2,10.6]\text{ the subintervals have length }\frac{10.6-(-2)}{3}=\frac{21}{5}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{1}{10},\frac{43}{10},\frac{17}{2}]\\&\int_{-2}^{10.6}(-13cos(15e^{(x)}))dx=\frac{21}{5}[(-13cos(15e^{(\frac{1}{10})}))+(-13cos(15e^{(\frac{43}{10})}))+(-13cos(15e^{(\frac{17}{2})}))]\\&\int_{-2}^{10.6}(-13cos(15e^{(x)}))dx=-49.29\end{align*}$

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Example Question #77 : Riemann Sums

$\begin{align*}&\text{Using }3\text{ intervals, and the method of midpoint Riemann sums approximate the integral:}\\&\int_{-2}^{10}(-10sin(15sin(x)))dx\end{align*}$

Possible Answers:

-68.00

-8.39

-346.78

-18.89

Correct answer:

-68.00

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-2}^{10}(-10sin(15sin(x)))dx\\&\text{So the interval is }[-2,10]\text{ the subintervals have length }\frac{10-(-2)}{3}=4\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[0,4,8]\\&\int_{-2}^{10}(-10sin(15sin(x)))dx=4[(0)+(-10sin(15sin(4)))+(-10sin(15sin(8)))]\\&\int_{-2}^{10}(-10sin(15sin(x)))dx=-68.00\end{align*}$

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Example Question #78 : Riemann Sums

$\begin{align*}&\text{Using the method of midpoint Riemann sums approximate the integral:}\\&\int_{1}^{11.5}(5cos(8cos(x)))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

1.21

10.02

63.10

31.05

Correct answer:

10.02

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{1}^{11.5}(5cos(8cos(x)))dx\\&\text{So the interval is }[1,11.5]\text{ the subintervals have length }\frac{11.5-(1)}{3}=\frac{7}{2}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{11}{4},\frac{25}{4},\frac{39}{4}]\\&\int_{1}^{11.5}(5cos(8cos(x)))dx=\frac{7}{2}[(5cos(8cos(\frac{11}{4})))+(5cos(8cos(\frac{25}{4})))+(5cos(8cos(\frac{39}{4})))]\\&\int_{1}^{11.5}(5cos(8cos(x)))dx=10.02\end{align*}$

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Example Question #79 : Riemann Sums

$\begin{align*}&\text{Using the method of midpoint Riemann sums approximate the integral:}\\&\int_{5}^{7.4}(8cos(6e^{(3x)}))dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

-26.13

-2.21

-4.43

-40.74

Correct answer:

-4.43

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{5}^{7.4}(8cos(6e^{(3x)}))dx\\&\text{So the interval is }[5,7.4]\text{ the subintervals have length }\frac{7.4-(5)}{4}=\frac{3}{5}\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{53}{10},\frac{59}{10},\frac{13}{2},\frac{71}{10}]\\&\int_{5}^{7.4}(8cos(6e^{(3x)}))dx=\frac{3}{5}[(8cos(6e^{(\frac{159}{10})}))+(8cos(6e^{(\frac{177}{10})}))+(8cos(6e^{(\frac{39}{2})}))+(8cos(6e^{(\frac{213}{10})}))]\\&\int_{5}^{7.4}(8cos(6e^{(3x)}))dx=-4.43\end{align*}$

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Example Question #80 : Riemann Sums

$\begin{align*}&\text{Using the method of right Riemann sums approximate the integral:}\\&\int_{-4}^{10}(-14ln(2x^{2}))dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

-3683.50

-1812.52

-584.68

-4794.40

Correct answer:

-584.68

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-4}^{10}(-14ln(2x^{2}))dx\\&\text{So the interval is }[-4,10]\text{ the subintervals have length }\frac{10-(-4)}{4}=\frac{7}{2}\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[-\frac{1}{2},3,\frac{13}{2},10]\\&\int_{-4}^{10}(-14ln(2x^{2}))dx=\frac{7}{2}[(14ln(2))+(-14ln(18))+(-14ln(\frac{169}{2}))+(-14ln(200))]\\&\int_{-4}^{10}(-14ln(2x^{2}))dx=-584.68\end{align*}$

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Calculus 2 : Riemann Sums

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1861 : Calculus Ii

Example Question #72 : Riemann Sums

Example Question #73 : Riemann Sums

Example Question #74 : Riemann Sums

Example Question #75 : Riemann Sums

Example Question #76 : Riemann Sums

Example Question #77 : Riemann Sums

Example Question #78 : Riemann Sums

Example Question #79 : Riemann Sums

Example Question #80 : Riemann Sums

All Calculus 2 Resources

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