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Calculus 2 : Integrals

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Example Questions

Example Question #52 : Riemann Sums

$\begin{align*}&\text{Using }4\text{ intervals, and the method of right point Riemann sums approximate the integral:}\\&\int_{-5}^{3.4}(-16tan(6x))dx\end{align*}$

Possible Answers:

-505.32

-27549.82

-1332.20

-2930.83

Correct answer:

-2930.83

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-5}^{3.4}(-16tan(6x))dx\\&\text{So the interval is }[-5,3.4]\text{ the subintervals have length }\frac{3.4-(-5)}{4}=2.1\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[-\frac{29}{10},-\frac{4}{5},\frac{13}{10},\frac{17}{5}]\\&\int_{-5}^{3.4}(-16tan(6x))dx=2.1[(16tan(\frac{87}{5}))+(16tan(\frac{24}{5}))+(-16tan(\frac{39}{5}))+(-16tan(\frac{102}{5}))]\\&\int_{-5}^{3.4}(-16tan(6x))dx=-2930.83\end{align*}$

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Example Question #101 : Introduction To Integrals

$\begin{align*}&\text{Using the method of right Riemann sums approximate the integral:}\\&\int_{3}^{4.2}(2\cdot 5^x)dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

10189.48

180.09

591.72

1242.62

Correct answer:

1242.62

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{3}^{4.2}(2\cdot 5^x)dx\\&\text{So the interval is }[3,4.2]\text{ the subintervals have length }\frac{4.2-(3)}{3}=0.4\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[\frac{17}{5},\frac{19}{5},\frac{21}{5}]\\&\int_{3}^{4.2}(2\cdot 5^x)dx=0.4[(250\cdot 5^{(\frac{2}{5})})+(250\cdot 5^{(\frac{4}{5})})+(1250\cdot 5^{(\frac{1}{5})})]\\&\int_{3}^{4.2}(2\cdot 5^x)dx=1242.62\end{align*}$

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Example Question #54 : Riemann Sums

$\begin{align*}&\text{Using }4\text{ intervals, and the method of right point Riemann sums approximate the integral:}\\&\int_{4}^{17.2}(14cos(3x))dx\end{align*}$

Possible Answers:

-114.89

-87.85

-22.53

-2.48

Correct answer:

-22.53

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{4}^{17.2}(14cos(3x))dx\\&\text{So the interval is }[4,17.2]\text{ the subintervals have length }\frac{17.2-(4)}{4}=3.3\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[\frac{73}{10},\frac{53}{5},\frac{139}{10},\frac{86}{5}]\\&\int_{4}^{17.2}(14cos(3x))dx=3.3[(14cos(\frac{219}{10}))+(14cos(\frac{159}{5}))+(14cos(\frac{417}{10}))+(14cos(\frac{258}{5}))]\\&\int_{4}^{17.2}(14cos(3x))dx=-22.53\end{align*}$

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Example Question #1851 : Calculus Ii

$\begin{align*}&\text{Using the method of right Riemann sums approximate the integral:}\\&\int_{1}^{16.5}(-3cos(x))dx\\&\text{Using }5\text{ intervals.}\end{align*}$

Possible Answers:

5.93

49.80

1.85

0.91

Correct answer:

5.93

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{1}^{16.5}(-3cos(x))dx\\&\text{So the interval is }[1,16.5]\text{ the subintervals have length }\frac{16.5-(1)}{5}=3.1\\&\text{and since we are using the *right*point of each interval, the x-values are:}\\&[\frac{41}{10},\frac{36}{5},\frac{103}{10},\frac{67}{5},\frac{33}{2}]\\&\int_{1}^{16.5}(-3cos(x))dx=3.1[(-3cos(\frac{41}{10}))+(-3cos(\frac{36}{5}))+(-3cos(\frac{103}{10}))+(-3cos(\frac{67}{5}))+(-3cos(\frac{33}{2}))]\\&\int_{1}^{16.5}(-3cos(x))dx=5.93\end{align*}$

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Example Question #101 : Introduction To Integrals

$\begin{align*}&\text{Using the method of left point Riemann sums approximate the integral:}\\&\int_{5}^{21}(-16cos(sin(6x)))dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

-33.98

-64.31

-1440.60

-180.08

Correct answer:

-180.08

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{5}^{21}(-16cos(sin(6x)))dx\\&\text{So the interval is }[5,21]\text{ the subintervals have length }\frac{21-(5)}{4}=4\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[5,9,13,17]\\&\int_{5}^{21}(-16cos(sin(6x)))dx=4[(-16cos(sin(30)))+(-16cos(sin(54)))+(-16cos(sin(78)))+(-16cos(sin(102)))]\\&\int_{5}^{21}(-16cos(sin(6x)))dx=-180.08\end{align*}$

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Example Question #51 : Riemann Sums

$\begin{align*}&\text{Using the method of left point Riemann sums approximate the integral:}\\&\int_{-4}^{4.1}(10cos(6sin(5x)))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

1.39

3.07

0.81

7.36

Correct answer:

7.36

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-4}^{4.1}(10cos(6sin(5x)))dx\\&\text{So the interval is }[-4,4.1]\text{ the subintervals have length }\frac{4.1-(-4)}{3}=\frac{27}{10}\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[-4,-\frac{13}{10},\frac{7}{5}]\\&\int_{-4}^{4.1}(10cos(6sin(5x)))dx=\frac{27}{10}[(10cos(6sin(20)))+(10cos(6sin(\frac{13}{2})))+(10cos(6sin(7)))]\\&\int_{-4}^{4.1}(10cos(6sin(5x)))dx=7.36\end{align*}$

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Example Question #51 : Riemann Sums

$\begin{align*}&\text{Using the method of left point Riemann sums approximate the integral:}\\&\int_{5}^{6.8}(-2tan(16x^{2}))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

0.59

4.88

0.24

0.10

Correct answer:

0.59

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{5}^{6.8}(-2tan(16x^{2}))dx\\&\text{So the interval is }[5,6.8]\text{ the subintervals have length }\frac{6.8-(5)}{3}=\frac{3}{5}\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[5,\frac{28}{5},\frac{31}{5}]\\&\int_{5}^{6.8}(-2tan(16x^{2}))dx=\frac{3}{5}[(-2tan(400))+(-2tan(\frac{12544}{25}))+(-2tan(\frac{15376}{25}))]\\&\int_{5}^{6.8}(-2tan(16x^{2}))dx=0.59\end{align*}$

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Example Question #51 : Riemann Sums

$\begin{align*}&\text{Using }3\text{ intervals, and the method of left point Riemann sums approximate the integral:}\\&\int_{5}^{18.8}(-3ln(10ln(6x)))dx\end{align*}$

Possible Answers:

-425.94

-18.55

-882.30

-152.12

Correct answer:

-152.12

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{5}^{18.8}(-3ln(10ln(6x)))dx\\&\text{So the interval is }[5,18.8]\text{ the subintervals have length }\frac{18.8-(5)}{3}=\frac{23}{5}\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[5,\frac{48}{5},\frac{71}{5}]\\&\int_{5}^{18.8}(-3ln(10ln(6x)))dx=\frac{23}{5}[(-3ln(10ln(30)))+(-3ln(10ln(\frac{288}{5})))+(-3ln(10ln(\frac{426}{5})))]\\&\int_{5}^{18.8}(-3ln(10ln(6x)))dx=-152.12\end{align*}$

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Example Question #102 : Integrals

$\begin{align*}&\text{Using the method of left point Riemann sums approximate the integral:}\\&\int_{0}^{4.8}(-tan(14sin(3x)))dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

1.50

13.02

3.34

67.70

Correct answer:

13.02

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{0}^{4.8}(-tan(14sin(3x)))dx\\&\text{So the interval is }[0,4.8]\text{ the subintervals have length }\frac{4.8-(0)}{4}=\frac{6}{5}\\&\text{and since we are using the *left*point of each interval, the x-values are:}\\&[0,\frac{6}{5},\frac{12}{5},\frac{18}{5}]\\&\int_{0}^{4.8}(-tan(14sin(3x)))dx=\frac{6}{5}[(0)+(-tan(14sin(\frac{18}{5})))+(-tan(14sin(\frac{36}{5})))+(-tan(14sin(\frac{54}{5})))]\\&\int_{0}^{4.8}(-tan(14sin(3x)))dx=13.02\end{align*}$

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Example Question #103 : Introduction To Integrals

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

Possible Answers:

-0.93

-1.29

-20.09

-8.37

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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Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #52 : Riemann Sums

Example Question #101 : Introduction To Integrals

Example Question #54 : Riemann Sums

Example Question #1851 : Calculus Ii

Example Question #101 : Introduction To Integrals

Example Question #51 : Riemann Sums

Example Question #51 : Riemann Sums

Example Question #51 : Riemann Sums

Example Question #102 : Integrals

Example Question #103 : Introduction To Integrals

All Calculus 2 Resources

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