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Calculus 2 : Calculus II

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #81 : Introduction To Integrals

$\begin{align*}&\text{Using the method of midpoint Riemann sums approximate the integral:}\\&\int_{-5}^{6.4}(4\cdot 3^x)dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

2165.80

13644.55

251.84

746.83

Correct answer:

2165.80

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-5}^{6.4}(4\cdot 3^x)dx\\&\text{So the interval is}[-5,6.4]\text{ the subintervals have length}\frac{6.4-(-5)}{3}=3.8\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[-\frac{31}{10},\frac{7}{10},\frac{9}{2}]\\&\int_{-5}^{6.4}(4\cdot 3^x)dx=3.8[(\frac{(4\cdot 3^{(\frac{9}{10})})}{81})+(4\cdot 3^{(\frac{7}{10})})+(324\cdot 3^{(\frac{1}{2})})]\\&\int_{-5}^{6.4}(4\cdot 3^x)dx=2165.80\end{align*}$

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Example Question #81 : Integrals

$\begin{align*}&\text{Using }4\text{ intervals, and the method of midpoint Riemann sums approximate the integral:}\\&\int_{-4}^{8.8}(-10tan(6x))dx\end{align*}$

Possible Answers:

39.71

61.70

127.96

345.50

Correct answer:

345.50

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-4}^{8.8}(-10tan(6x))dx\\&\text{So the interval is}[-4,8.8]\text{ the subintervals have length}\frac{8.8-(-4)}{4}=3.2\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[-\frac{12}{5},\frac{4}{5},4,\frac{36}{5}]\\&\int_{-4}^{8.8}(-10tan(6x))dx=3.2[(10tan(\frac{72}{5}))+(-10tan(\frac{24}{5}))+(-10tan(24))+(-10tan(\frac{216}{5}))]\\&\int_{-4}^{8.8}(-10tan(6x))dx=345.50\end{align*}$

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Example Question #82 : Introduction To Integrals

$\begin{align*}&\text{Calculate the Riemann sums integral approximation of :}\\&\int_{-2}^{0.4}(-9e^{(6x)})dx\\&\text{Using midpoints over }4\text{ intervals.}\end{align*}$

Possible Answers:

-3.89

-99.14

-10.12

-1.61

Correct answer:

-10.12

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-2}^{0.4}(-9e^{(6x)})dx\\&\text{So the interval is}[-2,0.4]\text{ the subintervals have length}\frac{0.4-(-2)}{4}=0.6\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[-\frac{17}{10},-\frac{11}{10},-\frac{1}{2},\frac{1}{10}]\\&\int_{-2}^{0.4}(-9e^{(6x)})dx=0.6[(-9e^{(-\frac{51}{5})})+(-9e^{(-\frac{33}{5})})+(-9e^{(-3)})+(-9e^{(\frac{3}{5})})]\\&\int_{-2}^{0.4}(-9e^{(6x)})dx=-10.12\end{align*}$

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Example Question #83 : Introduction To Integrals

$\begin{align*}&\text{Using the method of midpoint Riemann sums approximate the integral:}\\&\int_{5}^{18.2}(-cos(3x))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

1.12

4.65

11.16

73.63

Correct answer:

11.16

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{5}^{18.2}(-cos(3x))dx\\&\text{So the interval is}[5,18.2]\text{ the subintervals have length}\frac{18.2-(5)}{3}=4.4\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{36}{5},\frac{58}{5},16]\\&\int_{5}^{18.2}(-cos(3x))dx=4.4[(-cos(\frac{108}{5}))+(-cos(\frac{174}{5}))+(-cos(48))]\\&\int_{5}^{18.2}(-cos(3x))dx=11.16\end{align*}$

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Example Question #84 : Introduction To Integrals

$\begin{align*}&\text{Using the method of midpoint Riemann sums approximate the integral:}\\&\int_{-2}^{2.4}(13sin(2x))dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

-51.73

-1.19

-20.22

-5.95

Correct answer:

-5.95

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-2}^{2.4}(13sin(2x))dx\\&\text{So the interval is}[-2,2.4]\text{ the subintervals have length}\frac{2.4-(-2)}{4}=1.1\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[-\frac{29}{20},-\frac{7}{20},\frac{3}{4},\frac{37}{20}]\\&\int_{-2}^{2.4}(13sin(2x))dx=1.1[(-13sin(\frac{29}{10}))+(-13sin(\frac{7}{10}))+(13sin(\frac{3}{2}))+(13sin(\frac{37}{10}))]\\&\int_{-2}^{2.4}(13sin(2x))dx=-5.95\end{align*}$

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Example Question #85 : Introduction To Integrals

$\begin{align*}&\text{Calculate the Riemann sums integral approximation of :}\\&\int_{-5}^{0.6}(10tan(6x))dx\\&\text{Using midpoints over }4\text{ intervals.}\end{align*}$

Possible Answers:

603.79

25.16

10.71

100.63

Correct answer:

100.63

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{-5}^{0.6}(10tan(6x))dx\\&\text{So the interval is}[-5,0.6]\text{ the subintervals have length}\frac{0.6-(-5)}{4}=1.4\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[-\frac{43}{10},-\frac{29}{10},-\frac{3}{2},-\frac{1}{10}]\\&\int_{-5}^{0.6}(10tan(6x))dx=1.4[(-10tan(\frac{129}{5}))+(-10tan(\frac{87}{5}))+(-10tan(9))+(-10tan(\frac{3}{5}))]\\&\int_{-5}^{0.6}(10tan(6x))dx=100.63\end{align*}$

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Example Question #86 : Introduction To Integrals

$\begin{align*}&\text{Using the method of midpoint Riemann sums approximate the integral:}\\&\int_{1}^{17.8}(-10cos(6x))dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

-165.68

-579.87

-977.50

-18.62

Correct answer:

-165.68

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{1}^{17.8}(-10cos(6x))dx\\&\text{So the interval is}[1,17.8]\text{ the subintervals have length}\frac{17.8-(1)}{4}=4.2\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{31}{10},\frac{73}{10},\frac{23}{2},\frac{157}{10}]\\&\int_{1}^{17.8}(-10cos(6x))dx=4.2[(-10cos(\frac{93}{5}))+(-10cos(\frac{219}{5}))+(-10cos(69))+(-10cos(\frac{471}{5}))]\\&\int_{1}^{17.8}(-10cos(6x))dx=-165.68\end{align*}$

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Example Question #87 : Introduction To Integrals

$\begin{align*}&\text{Using the method of midpoint Riemann sums approximate the integral:}\\&\int_{5}^{15.8}(-8ln(5x))dx\\&\text{Using }4\text{ intervals.}\end{align*}$

Possible Answers:

-34.79

-146.73

-1754.83

-337.47

Correct answer:

-337.47

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{5}^{15.8}(-8ln(5x))dx\\&\text{So the interval is}[5,15.8]\text{ the subintervals have length}\frac{15.8-(5)}{4}=2.7\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{127}{20},\frac{181}{20},\frac{47}{4},\frac{289}{20}]\\&\int_{5}^{15.8}(-8ln(5x))dx=2.7[(-8ln(\frac{127}{4}))+(-8ln(\frac{181}{4}))+(-8ln(\frac{235}{4}))+(-8ln(\frac{289}{4}))]\\&\int_{5}^{15.8}(-8ln(5x))dx=-337.47\end{align*}$

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Example Question #88 : Introduction To Integrals

$\begin{align*}&\text{Using }4\text{ intervals, and the method of midpoint Riemann sums approximate the integral:}\\&\int_{2}^{9.2}(-7x^{2})dx\end{align*}$

Possible Answers:

-1784.66

-356.93

-200.52

-892.33

Correct answer:

-1784.66

Explanation:

$\begin{align*}&\text{A Riemann sum integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx \frac{b-a}{n}(f(x_1)+f(x_2)+...+f(x_n))\\&\text{It is essentially a sum of n rectangles, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable heights :}f(x_i)\text{, which depend on the function value at }x_i\\&\text{We're asked to approximate}\int_{2}^{9.2}(-7x^{2})dx\\&\text{So the interval is}[2,9.2]\text{ the subintervals have length}\frac{9.2-(2)}{4}=1.8\\&\text{and since we are using the *mid*point of each interval, the x-values are:}\\&[\frac{29}{10},\frac{47}{10},\frac{13}{2},\frac{83}{10}]\\&\int_{2}^{9.2}(-7x^{2})dx=1.8[(-\frac{5887}{100})+(-\frac{15463}{100})+(-\frac{1183}{4})+(-\frac{48223}{100})]\\&\int_{2}^{9.2}(-7x^{2})dx=-1784.66\end{align*}$

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Example Question #89 : 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:

11.76

2.35

1.38

30.58

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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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #81 : Introduction To Integrals

Example Question #81 : Integrals

Example Question #82 : Introduction To Integrals

Example Question #83 : Introduction To Integrals

Example Question #84 : Introduction To Integrals

Example Question #85 : Introduction To Integrals

Example Question #86 : Introduction To Integrals

Example Question #87 : Introduction To Integrals

Example Question #88 : Introduction To Integrals

Example Question #89 : Introduction To Integrals

All Calculus 2 Resources

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