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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 #212 : Definite Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{6}^{9.5}(\frac{8}{x}+\frac{ (7x^{2})}{4})dx\end{align*}$

Possible Answers:

134.93

377.81

53.97

41.98

Correct answer:

377.81

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int_{6}^{9.5}(\frac{8}{x}+\frac{ (7x^{2})}{4})dx=8ln(x) +\frac{ (7x^{3})}{12}|_{6}^{9.5}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&8ln((9.5)) +\frac{ (7(9.5)^{3})}{12}-(8ln((6)) +\frac{ (7(6)^{3})}{12})\\&377.81\end{align*}$

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Example Question #213 : Definite Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{6}^{9}(\frac{sin(x + 1)}{2}+\frac{ 1}{(8x^{2})})dx\end{align*}$

Possible Answers:

0.80

8.03

0.15

0.22

Correct answer:

0.80

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int_{6}^{9}(\frac{sin(x + 1)}{2}+\frac{ 1}{(8x^{2})})dx=-\frac{ cos(x + 1)}{2}-\frac{ 1}{(8x)}|_{6}^{9}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&=-\frac{ cos((9) + 1)}{2}-\frac{ 1}{(8(9))}-(-\frac{ cos((6) + 1)}{2}-\frac{ 1}{(8(6))})\\&=0.80\end{align*}$

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Example Question #2331 : Calculus Ii

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{20}^{22.5}(\frac{(29sin(3x))}{12}+\frac{ 1}{(6\cdot 2^{(\frac{x}{2})})})dx\end{align*}$

Possible Answers:

-0.25

-0.73

-0.13

-6.87

Correct answer:

-0.73

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\int_{20}^{22.5}(\frac{(29sin(3x))}{12}+\frac{ 1}{(6\cdot 2^{(\frac{x}{2})})})dx=-\frac{ (29cos(3x))}{36}-\frac{ 1}{(3\cdot 2^{(\frac{x}{2})}ln(2))}|_{20}^{22.5}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&=-\frac{ (29cos(3(22.5)))}{36}-\frac{ 1}{(3\cdot 2^{(\frac{(22.5)}{2})}ln(2))}-(-\frac{ (29cos(3(20)))}{36}-\frac{ 1}{(3\cdot 2^{(\frac{(20)}{2})}ln(2))})\\&=-0.73\end{align*}$

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Example Question #231 : Finding Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-28}^{-24.5}(\frac{(54cos(3x))}{7}+\frac{ 5}{(2x)})dx\end{align*}$

Possible Answers:

36.28

0.63

15.15

3.99

Correct answer:

3.99

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int_{-28}^{-24.5}(\frac{(54cos(3x))}{7}+\frac{ 5}{(2x)})dx=\frac{(18sin(3x))}{7}+\frac{ (5ln(x))}{2}|_{-28}^{-24.5}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&=\frac{(18sin(3(-24.5)))}{7}+\frac{ (5ln((-24.5)))}{2}-(\frac{(18sin(3(-28)))}{7}+\frac{ (5ln((-28)))}{2})\\&=3.99\end{align*}$

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Example Question #232 : Finding Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-14}^{-12}(13sin(x + 1) +\frac{ 4}{(21x^{2})})dx\end{align*}$

Possible Answers:

11.74

36.40

2.26

115.07

Correct answer:

11.74

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int_{-14}^{-12}(13sin(x + 1) +\frac{ 4}{(21x^{2})})dx=- 13cos(x + 1) -\frac{ 4}{(21x)}|_{-14}^{-12}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&=- 13cos((-12) + 1) -\frac{ 4}{(21(-12))}-(- 13cos((-14) + 1) -\frac{ 4}{(21(-14))})\\&=11.74\end{align*}$

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Example Question #233 : Finding Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{6}^{8}(\frac{(29cos(x + 1))}{3}-\frac{ 8}{(7x)})dx\end{align*}$

Possible Answers:

-15.64

-2.70

-10.51

-0.29

Correct answer:

-2.70

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being subtracted (rather than multiplied/divided),}\\&\text{we integrate separately and combine the results. This is known}\\&\text{as superposition.}\\&\text{Utilizing integral rules:}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int_{6}^{8}(\frac{(29cos(x + 1))}{3}-\frac{ 8}{(7x)})dx=\frac{(29sin(x + 1))}{3}-\frac{ (8ln(x))}{7}|_{6}^{8}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&=\frac{(29sin((8) + 1))}{3}-\frac{ (8ln((8)))}{7}-(\frac{(29sin((6) + 1))}{3}-\frac{ (8ln((6)))}{7})\\&=-2.70\end{align*}$

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Example Question #221 : Definite Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{12}^{13.5}(\frac{45}{(4x)}-\frac{ 61}{(9\cdot 3^{(\frac{x}{2})})})dx\end{align*}$

Possible Answers:

0.22

1.32

2.89

0.13

Correct answer:

1.32

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being subtracted (rather than multiplied/divided),}\\&\text{we integrate separately and combine the results. This is known}\\&\text{as superposition.}\\&\text{Utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\int_{12}^{13.5}(\frac{45}{(4x)}-\frac{ 61}{(9\cdot 3^{(\frac{x}{2})})})dx=\frac{(45ln(x))}{4}+\frac{ 122}{(9\cdot 3^{(\frac{x}{2})}ln(3))}|_{12}^{13.5}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&=\frac{(45ln((13.5)))}{4}+\frac{ 122}{(9\cdot 3^{(\frac{(13.5)}{2})}ln(3))}-(\frac{(45ln((12)))}{4}+\frac{ 122}{(9\cdot 3^{(\frac{(12)}{2})}ln(3))})\\&=1.32\end{align*}$

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Example Question #235 : Finding Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-13}^{-11}(21cos(3x) -\frac{ 61}{10})dx\end{align*}$

Possible Answers:

-44.83

-120.79

-12.45

-1.92

Correct answer:

-12.45

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being subtracted (rather than multiplied/divided),}\\&\text{we integrate separately and combine the results. This is known}\\&\text{as superposition.}\\&\text{Utilizing integral rules:}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int_{-13}^{-11}(21cos(3x) -\frac{ 61}{10})dx=7sin(3x) -\frac{ (61x)}{10}|_{-13}^{-11}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&=7sin(3(-11)) -\frac{ (61(-11))}{10}-(7sin(3(-13)) -\frac{ (61(-13))}{10})\\&=-12.45\end{align*}$

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Example Question #236 : Finding Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{12}^{13.5}(\frac{(3\cdot 2^{(\frac{x}{3})})}{19}-\frac{ 11}{(2x^{2})})dx\end{align*}$

Possible Answers:

4.48

0.65

16.57

0.48

Correct answer:

4.48

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being subtracted (rather than multiplied/divided),}\\&\text{we integrate separately and combine the results. This is known}\\&\text{as superposition.}\\&\text{Utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int_{12}^{13.5}(\frac{(3\cdot 2^{(\frac{x}{3})})}{19}-\frac{ 11}{(2x^{2})})dx=\frac{(9\cdot 2^{(\frac{x}{3})})}{(19ln(2))}+\frac{ 11}{(2x)}|_{12}^{13.5}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&=\frac{(9\cdot 2^{(\frac{(13.5)}{3})})}{(19ln(2))}+\frac{ 11}{(2(13.5))}-(\frac{(9\cdot 2^{(\frac{(12)}{3})})}{(19ln(2))}+\frac{ 11}{(2(12))})\\&=4.48\end{align*}$

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Example Question #222 : Definite Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-10}^{-8}(\frac{74}{(7\cdot 3^{(\frac{x}{2})})}-\frac{ e^{(x)}}{2})dx\end{align*}$

Possible Answers:

3117.70

19641.51

29306.37

1247.08

Correct answer:

3117.70

Explanation:

$\begin{align*}\\&\text{as superposition.}\\&\text{Utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\int_{-10}^{-8}(\frac{74}{(7\cdot 3^{(\frac{x}{2})})}-\frac{ e^{(x)}}{2})dx=-\frac{ e^{(x)}}{2}-\frac{ 148}{(7\cdot 3^{(\frac{x}{2})}ln(3))}|_{-10}^{-8}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&=-\frac{ e^{((-8))}}{2}-\frac{ 148}{(7\cdot 3^{(\frac{(-8)}{2})}ln(3))}-(-\frac{ e^{((-10))}}{2}-\frac{ 148}{(7\cdot 3^{(\frac{(-10)}{2})}ln(3))})\\&=3117.70\end{align*}$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #212 : Definite Integrals

Example Question #213 : Definite Integrals

Example Question #2331 : Calculus Ii

Example Question #231 : Finding Integrals

Example Question #232 : Finding Integrals

Example Question #233 : Finding Integrals

Example Question #221 : Definite Integrals

Example Question #235 : Finding Integrals

Example Question #236 : Finding Integrals

Example Question #222 : Definite Integrals

All Calculus 2 Resources

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