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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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

Example Questions

Example Question #2591 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(11e^{(13x)})}{40}-\frac{ (45cos(x + 2))}{11})dx\end{align*}$

Possible Answers:

$\frac{(143e^{(13x)})}{40}-\frac{ (180sin(x + 2))}{11}+C$

$\frac{(11e^{(13x)})}{520}-\frac{ (45sin(x + 2))}{11}+C$

$\frac{(11e^{(13x)})}{520}-\frac{ (45sin(x + 2))}{11}$

$\frac{(143e^{(13x)})}{40}-\frac{ (180sin(x + 2))}{11}$

Correct answer:

$\frac{(11e^{(13x)})}{520}-\frac{ (45sin(x + 2))}{11}+C$

Explanation:

$\begin{align*}\\&\text{as superposition.}\\&\text{Utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\text{The }a\text{ value for }\frac{(11e^{(13x)})}{40}\text{ is }13\\&\text{In deriving the inside of}\frac{(45cos(x + 2))}{11}\text{, the lone constant goes to zero.}\\&\int(\frac{(11e^{(13x)})}{40}-\frac{ (45cos(x + 2))}{11})dx=\frac{(11e^{(13x)})}{520}-\frac{ (45sin(x + 2))}{11}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2592 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(29\cdot 2^{(\frac{x}{7})})}{10}-\frac{ (59x^{16})}{3})dx\end{align*}$

Possible Answers:

$\frac{(203\cdot 2^{(\frac{x}{7})})}{(10ln(2))}-\frac{ (59x^{17})}{51}+C$

$\frac{(29\cdot 2^{(\frac{x}{7})})}{(70ln(2))}-\frac{ (15104x^{17})}{51}+C$

$\frac{(203\cdot 2^{(\frac{x}{7})})}{(10ln(2))}-\frac{ (59x^{17})}{51}$

$\frac{(29\cdot 2^{(\frac{x}{7})})}{(70ln(2))}-\frac{ (15104x^{17})}{51}$

Correct answer:

$\frac{(203\cdot 2^{(\frac{x}{7})})}{(10ln(2))}-\frac{ (59x^{17})}{51}+C$

Explanation:

$\begin{align*}\\&\text{as superposition.}\\&\text{Utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\text{The }a\text{ value for }\frac{(29\cdot 2^{(\frac{x}{7})})}{10}\text{ is }\frac{1}{7}\\&\text{The }a\text{ value for }\frac{(59x^{16})}{3}\text{ is }16\\&\int(\frac{(29\cdot 2^{(\frac{x}{7})})}{10}-\frac{ (59x^{16})}{3})dx=\frac{(203\cdot 2^{(\frac{x}{7})})}{(10ln(2))}-\frac{ (59x^{17})}{51}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2593 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(2x^{10})}{17}- 8cos(13x))dx\end{align*}$

Possible Answers:

$\frac{(200x^{11})}{187}- 104sin(13x)$

$\frac{(2x^{11})}{187}-\frac{ (8sin(13x))}{13}$

$\frac{(200x^{11})}{187}- 104sin(13x)+C$

$\frac{(2x^{11})}{187}-\frac{ (8sin(13x))}{13}+C$

Correct answer:

$\frac{(2x^{11})}{187}-\frac{ (8sin(13x))}{13}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being subtracted (rather than multiplied/divided),}\\&\text{we integrate separately and combine the results. This is known}\\&\text{as superposition.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\text{The }a\text{ value for }\frac{(2x^{10})}{17}\text{ is }10\\&\text{The }a\text{ value for }8cos(13x)\text{ is }13\\&\int(\frac{(2x^{10})}{17}- 8cos(13x))dx=\frac{(2x^{11})}{187}-\frac{ (8sin(13x))}{13}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2594 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{5}{(132x)}-\frac{ (9x^{5})}{2})dx\end{align*}$

Possible Answers:

$\frac{(5ln(x))}{132}-\frac{ (3x^{6})}{4}$

$\frac{(125ln(x))}{2299968}-\frac{ (75x^{6})}{4}+C$

$\frac{(125ln(x))}{2299968}-\frac{ (75x^{6})}{4}$

$\frac{(5ln(x))}{132}-\frac{ (3x^{6})}{4}+C$

Correct answer:

$\frac{(5ln(x))}{132}-\frac{ (3x^{6})}{4}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being subtracted (rather than multiplied/divided),}\\&\text{we integrate separately and combine the results. This is known}\\&\text{as superposition.}\\&\text{Utilizing integral rules:}\\&\int\frac{a}{u}=aln(u)=ln(u^{a})\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\text{The }a\text{ value for }\frac{5}{(132x)}\text{ is }\frac{5}{132}\\&\text{The }a\text{ value for }\frac{(9x^{5})}{2}\text{ is }5\\&\int(\frac{5}{(132x)}-\frac{ (9x^{5})}{2})dx=\frac{(5ln(x))}{132}-\frac{ (3x^{6})}{4}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2595 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{cos(x + 20)}{2}-\frac{ (23x^{18})}{6})dx\end{align*}$

Possible Answers:

$200sin(x + 20) -\frac{ (1242x^{19})}{19}+C$

$\frac{sin(x + 20)}{2}-\frac{ (23x^{19})}{114}+C$

$200sin(x + 20) -\frac{ (1242x^{19})}{19}$

$\frac{sin(x + 20)}{2}-\frac{ (23x^{19})}{114}$

Correct answer:

$\frac{sin(x + 20)}{2}-\frac{ (23x^{19})}{114}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being subtracted (rather than multiplied/divided),}\\&\text{we integrate separately and combine the results. This is known}\\&\text{as superposition.}\\&\text{Utilizing integral rules:}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\text{In deriving the inside of}\frac{cos(x + 20)}{2}\text{, the lone constant goes to zero.}\\&\text{The }a\text{ value for }\frac{(23x^{18})}{6}\text{ is }18\\&\int(\frac{cos(x + 20)}{2}-\frac{ (23x^{18})}{6})dx=\frac{sin(x + 20)}{2}-\frac{ (23x^{19})}{114}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2596 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(31x^{6})}{3}-\frac{ 2}{x})dx\end{align*}$

Possible Answers:

$\frac{(31x^{7})}{21}- 2ln(x)+C$

$\frac{(372x^{7})}{7}- 8ln(x)+C$

$\frac{(31x^{7})}{21}- 2ln(x)$

$\frac{(372x^{7})}{7}- 8ln(x)$

Correct answer:

$\frac{(31x^{7})}{21}- 2ln(x)+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being subtracted (rather than multiplied/divided),}\\&\text{we integrate separately and combine the results. This is known}\\&\text{as superposition.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int\frac{a}{u}=aln(u)=ln(u^{a})\\&\text{The }a\text{ value for }\frac{(31x^{6})}{3}\text{ is }6\\&\text{The }a\text{ value for }\frac{2}{x}\text{ is }2\\&\int(\frac{(31x^{6})}{3}-\frac{ 2}{x})dx=\frac{(31x^{7})}{21}- 2ln(x)+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2597 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{4}{(39x^{15})}- 80cos(5x))dx\end{align*}$

Possible Answers:

$- 16sin(5x) -\frac{ 2}{(273x^{14})}+C$

$- 400sin(5x) -\frac{ 150}{(91x^{14})}+C$

$- 16sin(5x) -\frac{ 2}{(273x^{14})}$

$- 400sin(5x) -\frac{ 150}{(91x^{14})}$

Correct answer:

$- 16sin(5x) -\frac{ 2}{(273x^{14})}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being subtracted (rather than multiplied/divided),}\\&\text{we integrate separately and combine the results. This is known}\\&\text{as superposition.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\text{The }a\text{ value for }\frac{4}{(39x^{15})}\text{ is }-15\\&\text{The }a\text{ value for }80cos(5x)\text{ is }5\\&\int(\frac{4}{(39x^{15})}- 80cos(5x))dx=- 16sin(5x) -\frac{ 2}{(273x^{14})}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2598 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(76cos(x + 6))}{3}-\frac{ 25}{(102x)})dx\end{align*}$

Possible Answers:

$\frac{(76sin(x + 6))}{3}-\frac{ (25ln(x))}{102}+C$

$912sin(x + 6) -\frac{ (15625ln(x))}{1061208}+C$

$\frac{(76sin(x + 6))}{3}-\frac{ (25ln(x))}{102}$

$912sin(x + 6) -\frac{ (15625ln(x))}{1061208}$

Correct answer:

$\frac{(76sin(x + 6))}{3}-\frac{ (25ln(x))}{102}+C$

Explanation:

$\begin{align*}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int\frac{a}{u}=aln(u)=ln(u^{a})\\&\text{In deriving the inside of}\frac{(76cos(x + 6))}{3}\text{, the lone constant goes to zero.}\\&\text{The }a\text{ value for }\frac{25}{(102x)}\text{ is }\frac{25}{102}\\&\int(\frac{(76cos(x + 6))}{3}-\frac{ 25}{(102x)})dx=\frac{(76sin(x + 6))}{3}-\frac{ (25ln(x))}{102}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2599 : Calculus Ii

Find the solution to the Integral:

$\int \frac{ln^7(x)}{x}dx$

Possible Answers:

8ln(x)+c

8ln^8(x)+c

7ln^6(x)+c

$\frac{ln^8(x)}{8}+c$

$\frac{}1{}8ln^2(x)+c$

Correct answer:

$\frac{ln^8(x)}{8}+c$

Explanation:

To Evaluate this Integral we must use substitution:

Let

u=ln(x) and $du=\frac{1}{x}dx$

Therefore:

$\int \frac{ln^7(x)}{x}dx=\int u^7du$

$\int u^7du=\frac{1}{8}u^8+c=\frac{1}{8}ln^8(x)+c$

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Example Question #2600 : Calculus Ii

Evaluate the integrand:

$\int \frac{dx}{9x+7}dx$

Possible Answers:

$\frac{1}{9}ln(9x+7)+c$

$\frac{1}{9}ln{x}+7x+c$

ln(9x-7)+c

$ln(9x^{10}+7x)+c$

ln(9x+7)+c

Correct answer:

$\frac{1}{9}ln(9x+7)+c$

Explanation:

You have to use substitution to solve the integrand:

Let u=9x+7 and du=9dx and

$dx=\frac{du}{9}$

Therefore:

$\int \frac{dx}{9x+7}=\frac{1}{9}\int \frac{du}{u}=\frac{1}{9}ln(9x+7)+c$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #2591 : Calculus Ii

Example Question #2592 : Calculus Ii

Example Question #2593 : Calculus Ii

Example Question #2594 : Calculus Ii

Example Question #2595 : Calculus Ii

Example Question #2596 : Calculus Ii

Example Question #2597 : Calculus Ii

Example Question #2598 : Calculus Ii

Example Question #2599 : Calculus Ii

Example Question #2600 : Calculus Ii

All Calculus 2 Resources

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