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Calculus 2 : Definite Integrals

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Example Questions

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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Example Question #228 : Definite Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{27}^{28.5}(\frac{(8sin(x + 1))}{11}-\frac{ (3\cdot 2^{(\frac{x}{3})})}{38})dx\end{align*}$

Possible Answers:

-12.15

-25.14

-72.92

-721.90

Correct answer:

-72.92

Explanation:

$\begin{align*}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\int_{27}^{28.5}(\frac{(8sin(x + 1))}{11}-\frac{ (3\cdot 2^{(\frac{x}{3})})}{38})dx=-\frac{ (8cos(x + 1))}{11}-\frac{ (9\cdot 2^{(\frac{x}{3})})}{(38ln(2))}|_{27}^{28.5}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&=-\frac{ (8cos((28.5) + 1))}{11}-\frac{ (9\cdot 2^{(\frac{(28.5)}{3})})}{(38ln(2))}-(-\frac{ (8cos((27) + 1))}{11}-\frac{ (9\cdot 2^{(\frac{(27)}{3})})}{(38ln(2))})\\&=-72.92\end{align*}$

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Example Question #221 : Definite Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-10}^{-8}(\frac{5}{(36x)}-\frac{ 3}{(71\cdot 2^{(\frac{x}{2})})})dx\end{align*}$

Possible Answers:

-1.98

-12.68

-0.58

-0.21

Correct answer:

-1.98

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being subtracted (rather than multiplied/divided),}\\&\text{we integrate separately and combine the results. This is known}\\&\text{as superposition.}\\&\text{Utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\int_{-10}^{-8}(\frac{5}{(36x)}-\frac{ 3}{(71\cdot 2^{(\frac{x}{2})})})dx=\frac{(5ln(x))}{36}+\frac{ 6}{(71\cdot 2^{(\frac{x}{2})}ln(2))}|_{-10}^{-8}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&=\frac{(5ln((-8)))}{36}+\frac{ 6}{(71\cdot 2^{(\frac{(-8)}{2})}ln(2))}-(\frac{(5ln((-10)))}{36}+\frac{ 6}{(71\cdot 2^{(\frac{(-10)}{2})}ln(2))})\\&=-1.98\end{align*}$

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Example Question #591 : Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{11}^{14}(\frac{5}{11}-\frac{ (4sin(x + 1))}{17})dx\end{align*}$

Possible Answers:

5.52

3.55

0.10

0.99

Correct answer:

0.99

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being subtracted (rather than multiplied/divided),}\\&\text{we integrate separately and combine the results. This is known}\\&\text{as superposition.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int_{11}^{14}(\frac{5}{11}-\frac{ (4sin(x + 1))}{17})dx=\frac{(5x)}{11}+\frac{ (4cos(x + 1))}{17}|_{11}^{14}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&=\frac{(5(14))}{11}+\frac{ (4cos((14) + 1))}{17}-(\frac{(5(11))}{11}+\frac{ (4cos((11) + 1))}{17})\\&=0.99\end{align*}$

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Calculus 2 : Definite Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

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

Example Question #228 : Definite Integrals

Example Question #221 : Definite Integrals

Example Question #591 : Integrals

All Calculus 2 Resources

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