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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 #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 #229 : 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:

-12.68

-0.58

-1.98

-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 #230 : Definite Integrals

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

Possible Answers:

3.55

0.10

0.99

5.52

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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Example Question #231 : Definite Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{7}^{11.5}(\frac{(9x^{2})}{40}- 2cos(x + 1))dx\end{align*}$

Possible Answers:

90.45

15.33

832.16

198.99

Correct answer:

90.45

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being subtracted (rather than multiplied/divided),}\\&\text{we integrate separately and combine the results. This is known}\\&\text{as superposition.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int_{7}^{11.5}(\frac{(9x^{2})}{40}- 2cos(x + 1))dx=\frac{(3x^{3})}{40}- 2sin(x + 1)|_{7}^{11.5}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&=\frac{(3(11.5)^{3})}{40}- 2sin((11.5) + 1)-(\frac{(3(7)^{3})}{40}- 2sin((7) + 1))\\&=90.45\end{align*}$

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Example Question #232 : Definite Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{-10}^{-8}(\frac{34}{(5x^{3})}-\frac{ (4e^{(-x)})}{55})dx\end{align*}$

Possible Answers:

-266.37

-157.40

-1385.15

-407.40

Correct answer:

-1385.15

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being subtracted (rather than multiplied/divided),}\\&\text{we integrate separately and combine the results. This is known}\\&\text{as superposition.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\int_{-10}^{-8}(\frac{34}{(5x^{3})}-\frac{ (4e^{(-x)})}{55})dx=\frac{(4e^{(-x)})}{55}-\frac{ 17}{(5x^{2})}|_{-10}^{-8}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&=\frac{(4e^{(-(-8))})}{55}-\frac{ 17}{(5(-8)^{2})}-(\frac{(4e^{(-(-10))})}{55}-\frac{ 17}{(5(-10)^{2})})\\&=-1385.15\end{align*}$

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Example Question #231 : Definite Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{5}^{7.5}(\frac{(8\cdot 2^{(\frac{x}{3})})}{13}-\frac{ (8cos(3x))}{3})dx\end{align*}$

Possible Answers:

73.93

7.62

2.63

39.63

Correct answer:

7.62

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being subtracted (rather than multiplied/divided),}\\&\text{we integrate separately and combine the results. This is known}\end{align*}$ $\begin{align*}\\&\text{as superposition.}\\&\text{Utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int_{5}^{7.5}(\frac{(8\cdot 2^{(\frac{x}{3})})}{13}-\frac{ (8cos(3x))}{3})dx=\frac{(24\cdot 2^{(\frac{x}{3})})}{(13ln(2))}-\frac{ (8sin(3x))}{9}|_{5}^{7.5}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&=\frac{(24\cdot 2^{(\frac{(7.5)}{3})})}{(13ln(2))}-\frac{ (8sin(3(7.5)))}{9}-(\frac{(24\cdot 2^{(\frac{(5)}{3})})}{(13ln(2))}-\frac{ (8sin(3(5)))}{9})\\&=7.62\end{align*}$

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Example Question #232 : Definite Integrals

$\begin{align*}&\text{Evaluate the integral: }\\&\int_{8}^{9.5}(\frac{67}{x}-\frac{ 29}{(6\cdot 3^x)})dx\end{align*}$

Possible Answers:

5.23

1.19

11.51

57.57

Correct answer:

11.51

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being subtracted (rather than multiplied/divided),}\\&\text{we integrate separately and combine the results. This is known}\\&\text{as superposition.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\int_{8}^{9.5}(\frac{67}{x}-\frac{ 29}{(6\cdot 3^x)})dx=67ln(x) +\frac{ 29}{(6\cdot 3^xln(3))}|_{8}^{9.5}\\&\text{Subtracting the function value at the lower bound from that of the upper bound:}\\&=67ln((9.5)) +\frac{ 29}{(6\cdot 3^{(9.5)}ln(3))}-(67ln((8)) +\frac{ 29}{(6\cdot 3^{(8)}ln(3))})\\&=11.51\end{align*}$

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Example Question #233 : Definite Integrals

$\int_{a}^{b}f(x)dx=3a+4b.$ What is $\int_{a}^{b}(f(x)+3)=?$

Possible Answers:

3a+4b+2

7b+6a

7ab

Correct answer:

Explanation:

For this problem, we need to first simplify the integral by splitting in two integrals. $\int_{a}^{b}(f(x)+3)dx=\int_{a}^{b}f(x)dx+\int_{a}^{b}3dx.$

Now, the first integral was given in the problem statement. Therefore, we just need to add that to the evaluation of the second integral.

$\int_{a}^{b}f(x)dx+\int_{a}^{b}3dx=3a+4b+\int_{a}^{b}3dx$

$\int_{a}^{b}3dx=3x=3(b)-3(a).$

Add the two parts together:

3a+4b+3b-3a=7b.

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Example Question #234 : Definite Integrals

Evaluate this integral.

$\int_{0}^{\pi /2}cos^{3}(x)dx$

Possible Answers:

$\frac{1}{3}$

$-\frac{1}{3}$

$\frac{2}{3}$

$-\frac{2}{3}$

Correct answer:

$\frac{2}{3}$

Explanation:

$\int_{0}^{\pi /2}cos^{3}(x)dx=\int_{0}^{\pi /2}cos^{2}(x)*cos(x)dx =\int_{0}^{\pi /2}(1-sin^2(x))cos(x)dx$

Now let u=sin(x) thus du=cos(x)dx .

Also notice sin(0)=0 and $sin(\pi/2)=1$ .

So we have

$\int_{0}^{1}(1-u^2)du=u-\frac{1}{3}u^3$

evaluated between 0 and 1.

We have

$1-\frac{1}{3}*1^3=\frac{2}{3}$

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Example Question #231 : Definite Integrals

Determine the area bounded by f(x)=x and g(x)=x^2 between on $0\leq x\leq 1$ .

Possible Answers:

$\frac{1}{3}$

$\frac{2}{3}$

$\frac{1}{6}$

$\frac{5}{6}$

$\frac{1}{2}$

Correct answer:

$\frac{1}{6}$

Explanation:

First identify the higher and lower function.

The higher function: f(x)=x

The lower function: g(x)=x^2

Since $f(x)\geq g(x)$ between x=0 and x=1 we must integrate.

To integrate take the difference of the lower function from the higher function.

$\int_{0}^{1}(x-x^2)dx=\frac{1}{2}x^2-\frac{1}{3}x^3$

Evaluating between zero and one results in,

$\\=\left[\frac{1}{2}1^2-\frac{1}{3}1^3\right]-\left[\frac{1}{2}0^2-\frac{1}{3}0^3 \right ] \\\\=\frac{1}{2}-\frac{1}{3}$

$=\frac{1}{6}$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #228 : Definite Integrals

Example Question #229 : Definite Integrals

Example Question #230 : Definite Integrals

Example Question #231 : Definite Integrals

Example Question #232 : Definite Integrals

Example Question #231 : Definite Integrals

Example Question #232 : Definite Integrals

Example Question #233 : Definite Integrals

Example Question #234 : Definite Integrals

Example Question #231 : Definite Integrals

All Calculus 2 Resources

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