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Calculus 2 : Integrals

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Example Questions

Example Question #2601 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(7sin(10x))}{4}+\frac{ (19x^{2})}{5})dx\end{align*}$

Possible Answers:

$\frac{(76x^{3})}{15}-\frac{ (35cos(10x))}{2}$

$\frac{(19x^{3})}{15}-\frac{ (7cos(10x))}{40}+C$

$\frac{(19x^{3})}{15}-\frac{ (7cos(10x))}{40}$

$\frac{(76x^{3})}{15}-\frac{ (35cos(10x))}{2}+C$

Correct answer:

$\frac{(19x^{3})}{15}-\frac{ (7cos(10x))}{40}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\text{The }a\text{ value for }\frac{(19x^{2})}{5}\text{ is }2\\&\text{The }a\text{ value for }\frac{(7sin(10x))}{4}\text{ is }10\\&\int(\frac{(7sin(10x))}{4}+\frac{ (19x^{2})}{5})dx=\frac{(19x^{3})}{15}-\frac{ (7cos(10x))}{40}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2602 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(3\cdot 3^{(\frac{x}{3})} + 11x^{15})dx\end{align*}$

Possible Answers:

$\frac{3^{(\frac{x}{3})}}{ln(3)}+\frac{ (2475x^{16})}{16}$

$\frac{(9\cdot 3^{(\frac{x}{3})})}{ln(3)}+\frac{ (11x^{16})}{16}$

$\frac{3^{(\frac{x}{3})}}{ln(3)}+\frac{ (2475x^{16})}{16}+C$

$\frac{(9\cdot 3^{(\frac{x}{3})})}{ln(3)}+\frac{ (11x^{16})}{16}+C$

Correct answer:

$\frac{(9\cdot 3^{(\frac{x}{3})})}{ln(3)}+\frac{ (11x^{16})}{16}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\text{The }a\text{ value for }11x^{15}\text{ is }15\\&\text{The }a\text{ value for }3\cdot 3^{(\frac{x}{3})}\text{ is }\frac{1}{3}\\&\int(3\cdot 3^{(\frac{x}{3})} + 11x^{15})dx=\frac{(9\cdot 3^{(\frac{x}{3})})}{ln(3)}+\frac{ (11x^{16})}{16}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2603 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(41sin(9x))}{10}+\frac{ (2\cdot 3^{(7x)})}{21})dx\end{align*}$

Possible Answers:

$\frac{(2\cdot 3^{(7x)})}{(147ln(3))}-\frac{ (41cos(9x))}{90}$

$\frac{(2\cdot 3^{(7x)})}{(147ln(3))}-\frac{ (41cos(9x))}{90}+C$

$\frac{(2\cdot 3^{(7x)})}{(3ln(3))}-\frac{ (369cos(9x))}{10}+C$

$\frac{(2\cdot 3^{(7x)})}{(3ln(3))}-\frac{ (369cos(9x))}{10}$

Correct answer:

$\frac{(2\cdot 3^{(7x)})}{(147ln(3))}-\frac{ (41cos(9x))}{90}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\text{The }a\text{ value for }\frac{(2\cdot 3^{(7x)})}{21}\text{ is }7\\&\text{The }a\text{ value for }\frac{(41sin(9x))}{10}\text{ is }9\\&\int(\frac{(41sin(9x))}{10}+\frac{ (2\cdot 3^{(7x)})}{21})dx=\frac{(2\cdot 3^{(7x)})}{(147ln(3))}-\frac{ (41cos(9x))}{90}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2604 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{7}{(1170x)}+\frac{ (45x^{7})}{8})dx\end{align*}$

Possible Answers:

$\frac{(187ln(x))}{1170}+\frac{ (45x^{8})}{8}$

$\frac{(7ln(x))}{1170}+\frac{ (45x^{8})}{64}+C$

$\frac{(187ln(x))}{1170}+\frac{ (45x^{8})}{8}+C$

$\frac{(7ln(x))}{1170}+\frac{ (45x^{8})}{64}$

Correct answer:

$\frac{(7ln(x))}{1170}+\frac{ (45x^{8})}{64}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int\frac{a}{u}=aln(u)=ln(u^{a})\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\text{The }a\text{ value for }\frac{7}{(1170x)}\text{ is }\frac{7}{1170}\\&\text{The }a\text{ value for }\frac{(45x^{7})}{8}\text{ is }7\\&\int(\frac{7}{(1170x)}+\frac{ (45x^{7})}{8})dx=\frac{(7ln(x))}{1170}+\frac{ (45x^{8})}{64}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2605 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(3cos(x + 19))}{14}+\frac{ (38x^{6})}{3})dx\end{align*}$

Possible Answers:

$\frac{(1083sin(x + 19))}{14}+\frac{ (456x^{7})}{7}+C$

$\frac{(3sin(x + 19))}{14}+\frac{ (38x^{7})}{21}+C$

$\frac{(1083sin(x + 19))}{14}+\frac{ (456x^{7})}{7}$

$\frac{(3sin(x + 19))}{14}+\frac{ (38x^{7})}{21}$

Correct answer:

$\frac{(3sin(x + 19))}{14}+\frac{ (38x^{7})}{21}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\text{The }a\text{ value for }\frac{(38x^{6})}{3}\text{ is }6\\&\text{In deriving the inside of}\frac{(3cos(x + 19))}{14}\text{, the lone constant goes to zero.}\\&\int(\frac{(3cos(x + 19))}{14}+\frac{ (38x^{6})}{3})dx=\frac{(3sin(x + 19))}{14}+\frac{ (38x^{7})}{21}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2606 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{7}{(6x)}+\frac{ 4}{(9x^{6})})dx\end{align*}$

Possible Answers:

$\frac{(343ln(x))}{216}-\frac{ 16}{(5x^{5})}$

$\frac{(7ln(x))}{6}-\frac{ 4}{(45x^{5})}+C$

$\frac{(343ln(x))}{216}-\frac{ 16}{(5x^{5})}+C$

$\frac{(7ln(x))}{6}-\frac{ 4}{(45x^{5})}$

Correct answer:

$\frac{(7ln(x))}{6}-\frac{ 4}{(45x^{5})}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int\frac{a}{u}=aln(u)=ln(u^{a})\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\text{The }a\text{ value for }\frac{7}{(6x)}\text{ is }\frac{7}{6}\\&\text{The }a\text{ value for }\frac{4}{(9x^{6})}\text{ is }-6\\&\int(\frac{7}{(6x)}+\frac{ 4}{(9x^{6})})dx=\frac{(7ln(x))}{6}-\frac{ 4}{(45x^{5})}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2607 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(13x^{6})}{3}+\frac{ (9x^{7})}{49})dx\end{align*}$

Possible Answers:

$\frac{(156x^{7})}{7}+\frac{ (9x^{8})}{8}+C$

$\frac{(x^{7}\cdot (27x + 728))}{1176}$

$\frac{(156x^{7})}{7}+\frac{ (9x^{8})}{8}$

$\frac{(x^{7}\cdot (27x + 728))}{1176}+C$

Correct answer:

$\frac{(x^{7}\cdot (27x + 728))}{1176}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\text{The }a\text{ value for }\frac{(9x^{7})}{49}\text{ is }7\\&\text{The }a\text{ value for }\frac{(13x^{6})}{3}\text{ is }6\\&\int(\frac{(13x^{6})}{3}+\frac{ (9x^{7})}{49})dx=\frac{(x^{7}\cdot (27x + 728))}{1176}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #231 : Indefinite Integrals

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{sin(14x)}{14}+\frac{ (32x^{18})}{5})dx\end{align*}$

Possible Answers:

$\frac{(32x^{19})}{95}-\frac{ cos(14x)}{196}+C$

$\frac{(32x^{19})}{95}-\frac{ cos(14x)}{196}$

$\frac{(10368x^{19})}{95}- cos(14x)$

$\frac{(10368x^{19})}{95}- cos(14x)+C$

Correct answer:

$\frac{(32x^{19})}{95}-\frac{ cos(14x)}{196}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\text{The }a\text{ value for }\frac{sin(14x)}{14}\text{ is }14\\&\text{The }a\text{ value for }\frac{(32x^{18})}{5}\text{ is }18\\&\int(\frac{sin(14x)}{14}+\frac{ (32x^{18})}{5})dx=\frac{(32x^{19})}{95}-\frac{ cos(14x)}{196}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #232 : Indefinite Integrals

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(5sin(x + 8))}{37}+\frac{ (55x^{14})}{2})dx\end{align*}$

Possible Answers:

$\frac{(1078x^{15})}{3}-\frac{ (320cos(x + 8))}{37}+C$

$\frac{(11x^{15})}{6}-\frac{ (5cos(x + 8))}{37}+C$

$\frac{(11x^{15})}{6}-\frac{ (5cos(x + 8))}{37}$

$\frac{(1078x^{15})}{3}-\frac{ (320cos(x + 8))}{37}$

Correct answer:

$\frac{(11x^{15})}{6}-\frac{ (5cos(x + 8))}{37}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\text{In deriving the inside of}\frac{(5sin(x + 8))}{37}\text{, the lone constant goes to zero.}\\&\text{The }a\text{ value for }\frac{(55x^{14})}{2}\text{ is }14\\&\int(\frac{(5sin(x + 8))}{37}+\frac{ (55x^{14})}{2})dx=\frac{(11x^{15})}{6}-\frac{ (5cos(x + 8))}{37}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #233 : Indefinite Integrals

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(4e^{(18x)} +\frac{ x^{16}}{3})dx\end{align*}$

Possible Answers:

$\frac{(2e^{(18x)})}{9}+\frac{ x^{17}}{51}$

$\frac{(2e^{(18x)})}{9}+\frac{ x^{17}}{51}+C$

$72e^{(18x)} +\frac{ (256x^{17})}{51}$

$72e^{(18x)} +\frac{ (256x^{17})}{51}+C$

Correct answer:

$\frac{(2e^{(18x)})}{9}+\frac{ x^{17}}{51}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int[e^{ax}]=\frac{e^{ax}}{a}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\text{The }a\text{ value for }4e^{(18x)}\text{ is }18\\&\text{The }a\text{ value for }\frac{x^{16}}{3}\text{ is }16\\&\int(4e^{(18x)} +\frac{ x^{16}}{3})dx=\frac{(2e^{(18x)})}{9}+\frac{ x^{17}}{51}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #2601 : Calculus Ii

Example Question #2602 : Calculus Ii

Example Question #2603 : Calculus Ii

Example Question #2604 : Calculus Ii

Example Question #2605 : Calculus Ii

Example Question #2606 : Calculus Ii

Example Question #2607 : Calculus Ii

Example Question #231 : Indefinite Integrals

Example Question #232 : Indefinite Integrals

Example Question #233 : Indefinite Integrals

All Calculus 2 Resources

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