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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 #194 : Indefinite Integrals

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{2^{(18x)}}{4})dx\end{align*}$

Possible Answers:

$\frac{2^{(18x)}}{(72ln(2))}+C$

$\frac{(9\cdot 2^{(18x)})}{(2ln(2))}+C$

$\frac{(9\cdot 2^{(18x)})}{(2ln(2))}$

$\frac{2^{(18x)}}{(72ln(2))}$

Correct answer:

$\frac{2^{(18x)}}{(72ln(2))}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\text{Note for this problem, the }a\text{ value is } 18\\&\int(\frac{2^{(18x)}}{4})dx=\frac{2^{(18x)}}{(72ln(2))}+C\\&\text{Now you'll note after integration we add a constant, C. This}\\&\text{is because we're working with an indefinite integral. Note how}\\&\text{if we take the derivative of this function, the constant would}\\&\text{go to zero.}\\&\end{align*}$

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Example Question #195 : Indefinite Integrals

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{38}{(45x)})dx\end{align*}$

Possible Answers:

$\frac{(38ln(x))}{45}+C$

$\frac{(760ln(x))}{9}$

$\frac{(38ln(x))}{45}$

$\frac{(760ln(x))}{9}+C$

Correct answer:

$\frac{(38ln(x))}{45}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int\frac{a}{u}=aln(u)=ln(u^{a})\\&\text{Note for this problem, the }a\text{ value is } 10\\&\int(\frac{38}{(45x)})dx=\frac{(38ln(x))}{45}+C\\&\text{Now you'll note after integration we add a constant, C. This}\\&\text{is because we're working with an indefinite integral. Note how}\\&\text{if we take the derivative of this function, the constant would}\\&\text{go to zero.}\\&\end{align*}$

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Example Question #196 : Indefinite Integrals

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(5\cdot 2^{(18x)})}{37})dx\end{align*}$

Possible Answers:

$\frac{(90\cdot 2^{(18x)})}{(37ln(2))}+C$

$\frac{(90\cdot 2^{(18x)})}{(37ln(2))}$

$\frac{(5\cdot 2^{(18x)})}{(666ln(2))}+C$

$\frac{(5\cdot 2^{(18x)})}{(666ln(2))}$

Correct answer:

$\frac{(5\cdot 2^{(18x)})}{(666ln(2))}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\text{Note for this problem, the }a\text{ value is } 18\\&\int(\frac{(5\cdot 2^{(18x)})}{37})dx=\frac{(5\cdot 2^{(18x)})}{(666ln(2))}+C\\&\text{Now you'll note after integration we add a constant, C. This}\\&\text{is because we're working with an indefinite integral. Note how}\\&\text{if we take the derivative of this function, the constant would}\\&\text{go to zero.}\\&\end{align*}$

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Example Question #197 : Indefinite Integrals

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(13\cdot 3^{(16x)})dx\end{align*}$

Possible Answers:

$\frac{(208\cdot 3^{(16x)})}{ln(3)}$

$\frac{(208\cdot 3^{(16x)})}{ln(3)}+C$

$\frac{(13\cdot 3^{(16x)})}{(16ln(3))}+C$

$\frac{(13\cdot 3^{(16x)})}{(16ln(3))}$

Correct answer:

$\frac{(13\cdot 3^{(16x)})}{(16ln(3))}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\text{Note for this problem, the }a\text{ value is } 16\\&\int(13\cdot 3^{(16x)})dx=\frac{(13\cdot 3^{(16x)})}{(16ln(3))}+C\\&\text{Now you'll note after integration we add a constant, C. This}\\&\text{is because we're working with an indefinite integral. Note how}\\&\text{if we take the derivative of this function, the constant would}\\&\text{go to zero.}\\&\end{align*}$

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Example Question #198 : Indefinite Integrals

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{e^{(9x)}}{3}+\frac{ (64x^{10})}{7})dx\end{align*}$

Possible Answers:

$3e^{(9x)} +\frac{ (6400x^{11})}{77}+C$

$\frac{e^{(9x)}}{27}+\frac{ (64x^{11})}{77}$

$3e^{(9x)} +\frac{ (6400x^{11})}{77}$

$\frac{e^{(9x)}}{27}+\frac{ (64x^{11})}{77}+C$

Correct answer:

$\frac{e^{(9x)}}{27}+\frac{ (64x^{11})}{77}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\text{The }a\text{ value for }\frac{e^{(9x)}}{3}\text{ is }9\\&\text{The }a\text{ value for }\frac{(64x^{10})}{7}\text{ is }10\\&\int(\frac{e^{(9x)}}{3}+\frac{ (64x^{10})}{7})dx=\frac{e^{(9x)}}{27}+\frac{ (64x^{11})}{77}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #199 : Indefinite Integrals

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{3^{(16x)}}{22}+\frac{ 1}{(11x^{11})})dx\end{align*}$

Possible Answers:

$\frac{3^{(16x)}}{(352ln(3))}-\frac{ 1}{(110x^{10})}+C$

$\frac{(8\cdot 3^{(16x)})}{(11ln(3))}-\frac{ 11}{(10x^{10})}+C$

$\frac{(8\cdot 3^{(16x)})}{(11ln(3))}-\frac{ 11}{(10x^{10})}$

$\frac{3^{(16x)}}{(352ln(3))}-\frac{ 1}{(110x^{10})}$

Correct answer:

$\frac{3^{(16x)}}{(352ln(3))}-\frac{ 1}{(110x^{10})}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\text{The }a\text{ value for }\frac{1}{(11x^{11})}\text{ is }-11\\&\text{The }a\text{ value for }\frac{3^{(16x)}}{22}\text{ is }16\\&\int(\frac{3^{(16x)}}{22}+\frac{ 1}{(11x^{11})})dx=\frac{3^{(16x)}}{(352ln(3))}-\frac{ 1}{(110x^{10})}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #200 : Indefinite Integrals

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{3^{(17x)}}{2}+\frac{ 3}{(7x)})dx\end{align*}$

Possible Answers:

$\frac{(3ln(x))}{7}+\frac{ 3^{(17x)}}{(34ln(3))}$

$\frac{(3ln(x))}{7}+\frac{ 3^{(17x)}}{(34ln(3))}+C$

$\frac{(27ln(x))}{343}+\frac{ (17\cdot 3^{(17x)})}{(2ln(3))}$

$\frac{(27ln(x))}{343}+\frac{ (17\cdot 3^{(17x)})}{(2ln(3))}+C$

Correct answer:

$\frac{(3ln(x))}{7}+\frac{ 3^{(17x)}}{(34ln(3))}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\int\frac{a}{u}=aln(u)=ln(u^{a})\\&\text{The }a\text{ value for }\frac{3^{(17x)}}{2}\text{ is }17\\&\text{The }a\text{ value for }\frac{3}{(7x)}\text{ is }\frac{3}{7}\\&\int(\frac{3^{(17x)}}{2}+\frac{ 3}{(7x)})dx=\frac{(3ln(x))}{7}+\frac{ 3^{(17x)}}{(34ln(3))}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #821 : Integrals

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(68cos(x + 9))}{7}+\frac{ 23}{(2x^{7})})dx\end{align*}$

Possible Answers:

$\frac{(68sin(x + 9))}{7}-\frac{ 23}{(12x^{6})}$

$\frac{(68sin(x + 9))}{7}-\frac{ 23}{(12x^{6})}+C$

$\frac{(5508sin(x + 9))}{7}-\frac{ 1127}{(12x^{6})}+C$

$\frac{(5508sin(x + 9))}{7}-\frac{ 1127}{(12x^{6})}$

Correct answer:

$\frac{(68sin(x + 9))}{7}-\frac{ 23}{(12x^{6})}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\text{In deriving the inside of}\frac{(68cos(x + 9))}{7}\text{, the lone constant goes to zero.}\\&\text{The }a\text{ value for }\frac{23}{(2x^{7})}\text{ is }-7\\&\int(\frac{(68cos(x + 9))}{7}+\frac{ 23}{(2x^{7})})dx=\frac{(68sin(x + 9))}{7}-\frac{ 23}{(12x^{6})}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #822 : Integrals

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{sin(6x)}{50}+\frac{ x^{8}}{29})dx\end{align*}$

Possible Answers:

$\frac{(64x^{9})}{261}-\frac{ (3cos(6x))}{25}$

$\frac{x^{9}}{261}-\frac{ cos(6x)}{300}+C$

$\frac{(64x^{9})}{261}-\frac{ (3cos(6x))}{25}+C$

$\frac{x^{9}}{261}-\frac{ cos(6x)}{300}$

Correct answer:

$\frac{x^{9}}{261}-\frac{ cos(6x)}{300}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\text{The }a\text{ value for }\frac{x^{8}}{29}\text{ is }8\\&\text{The }a\text{ value for }\frac{sin(6x)}{50}\text{ is }6\\&\int(\frac{sin(6x)}{50}+\frac{ x^{8}}{29})dx=\frac{x^{9}}{261}-\frac{ cos(6x)}{300}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #823 : Integrals

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(2cos(4x))}{7}+\frac{ (37e^{(15x)})}{8})dx\end{align*}$

Possible Answers:

$\frac{(37e^{(15x)})}{120}+\frac{ sin(4x)}{14}$

$\frac{(555e^{(15x)})}{8}+\frac{ (8sin(4x))}{7}$

$\frac{(37e^{(15x)})}{120}+\frac{ sin(4x)}{14}+C$

$\frac{(555e^{(15x)})}{8}+\frac{ (8sin(4x))}{7}+C$

Correct answer:

$\frac{(37e^{(15x)})}{120}+\frac{ sin(4x)}{14}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\text{The }a\text{ value for }\frac{(2cos(4x))}{7}\text{ is }4\\&\text{The }a\text{ value for }\frac{(37e^{(15x)})}{8}\text{ is }15\\&\int(\frac{(2cos(4x))}{7}+\frac{ (37e^{(15x)})}{8})dx=\frac{(37e^{(15x)})}{120}+\frac{ sin(4x)}{14}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #194 : Indefinite Integrals

Example Question #195 : Indefinite Integrals

Example Question #196 : Indefinite Integrals

Example Question #197 : Indefinite Integrals

Example Question #198 : Indefinite Integrals

Example Question #199 : Indefinite Integrals

Example Question #200 : Indefinite Integrals

Example Question #821 : Integrals

Example Question #822 : Integrals

Example Question #823 : Integrals

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