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Calculus 2 : Integrals

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Example Questions

Example Question #2611 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(69x^{6})}{10}+\frac{ 41}{(6x^{13})})dx\end{align*}$

Possible Answers:

$\frac{(\frac{(69x^{19})}{70}-\frac{ 41}{72})}{x^{12}}$

$\frac{(1242x^{7})}{35}-\frac{ 6929}{(72x^{12})}$

$\frac{(1242x^{7})}{35}-\frac{ 6929}{(72x^{12})}+C$

$\frac{(\frac{(69x^{19})}{70}-\frac{ 41}{72})}{x^{12}}+C$

Correct answer:

$\frac{(\frac{(69x^{19})}{70}-\frac{ 41}{72})}{x^{12}}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\text{The }a\text{ value for }\frac{41}{(6x^{13})}\text{ is }-13\\&\text{The }a\text{ value for }\frac{(69x^{6})}{10}\text{ is }6\\&\int(\frac{(69x^{6})}{10}+\frac{ 41}{(6x^{13})})dx=\frac{(\frac{(69x^{19})}{70}-\frac{ 41}{72})}{x^{12}}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2612 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\int(\frac{(7x^{14})}{58}+\frac{ x^{20}}{71})dx\end{align*}$

Possible Answers:

$\frac{(7x^{15})}{870}+\frac{ x^{21}}{1491}$

$\frac{(7x^{15})}{870}+\frac{ x^{21}}{1491}+C$

$\frac{(686x^{15})}{435}+\frac{ (400x^{21})}{1491}$

$\frac{(686x^{15})}{435}+\frac{ (400x^{21})}{1491}+C$

Correct answer:

$\frac{(7x^{15})}{870}+\frac{ x^{21}}{1491}+C$

Explanation:

$\begin{align*}&\text{In performing an integral, consider how in a derivative we’d}\\&\text{use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside. Integrals, in}\\&\text{similar fashion, entail a division. Now, although we have two}\\&\text{functions, since they're being added, we integrate separately}\\&\text{and combine the results. This is known as superposition.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\text{The }a\text{ value for }\frac{x^{20}}{71}\text{ is }20\\&\text{The }a\text{ value for }\frac{(7x^{14})}{58}\text{ is }14\\&\int(\frac{(7x^{14})}{58}+\frac{ x^{20}}{71})dx=\frac{(7x^{15})}{870}+\frac{ x^{21}}{1491}+C\\&\text{Now you'll note after integration we add a constant, C, because}\\&\text{we're working with an indefinite integral.}\\&\end{align*}$

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Example Question #2613 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite double integral: }\\&\iint(4x^{17})dx\end{align*}$

Possible Answers:

$\frac{(2x^{19})}{171}$

$\frac{(2x^{19})}{171}+C_{1}x+C_{2}$

$\frac{(2x^{18})}{9}+C_{1}x$

$\frac{(2x^{19})}{171}+C_{1}+C_{2}$

Correct answer:

$\frac{(2x^{19})}{171}+C_{1}x+C_{2}$

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\text{Note for this problem, the }a\text{ value is } 17\\&\text{Taking the first integral we find:}\\&\frac{(2x^{18})}{9}+C_{1}\\&\text{Note after indefinite integration we add a constant, }C_1\\&\text{Integrating again, we'll add a new constant, while multiplying}\\&\text{our first one by }x:\\&\frac{(2x^{19})}{171}+C_{1}x+C_{2}\\&\end{align*}$

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Example Question #2614 : Calculus Ii

$\begin{align*}&\text{Calculate the integral: }\\&\iint(\frac{(3sin(11x))}{22})dx\end{align*}$

Possible Answers:

$-\frac{(3cos(11x))}{242}+C_{1}$

$-\frac{(3sin(11x))}{2662}+C_{1}x+C_{2}$

$-\frac{(3cos(11x))}{242}+C_{1}x$

$-\frac{(3sin(11x))}{2662}+11C_{1}+C_{2}$

Correct answer:

$-\frac{(3sin(11x))}{2662}+C_{1}x+C_{2}$

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\text{Note for this problem, the }a\text{ value is } 11\\&\text{Taking the first integral we find:}\\&-\frac{(3cos(11x))}{242}+C_{1}\\&\text{Note that after indefinite integration we add a constant, }C_1\\&\text{Integrating again, we'll add a new constant, while multiplying}\\&\text{our first one by }x:\\&-\frac{(3sin(11x))}{2662}+C_{1}x+C_{2}\\&\end{align*}$

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Example Question #2615 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\iint(\frac{(5\cdot 2^{(17x)})}{26})dx\end{align*}$

Possible Answers:

$\frac{(5\cdot 2^{(17x)})}{(7514ln(2)^{2})}+17C_{1}+C_{2}$

$\frac{(5\cdot 2^{(17x)})}{(7514ln(2)^{2})}+C_{1}x+C_{2}$

$\frac{(5\cdot 2^{(17x)})}{(442ln(2))}+C_{1}x$

$\frac{(5\cdot 2^{(17x)})}{(442ln(2))}+C_{1}$

Correct answer:

$\frac{(5\cdot 2^{(17x)})}{(7514ln(2)^{2})}+C_{1}x+C_{2}$

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\text{Note for this problem, the }a\text{ value is } 17\\&\text{Taking the first integral we find:}\\&\frac{(5\cdot 2^{(17x)})}{(442ln(2))}+C_{1}\\&\text{Note that after indefinite integration we add a constant, }C_1\\&\text{Integrating again, we'll add a new constant, while multiplying}\\&\text{our first one by }x:\\&\frac{(5\cdot 2^{(17x)})}{(7514ln(2)^{2})}+C_{1}x+C_{2}\\&\end{align*}$

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Example Question #2616 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite double integral: }\\&\iint(\frac{(14\cdot 2^{(\frac{x}{3})})}{5})dx\end{align*}$

Possible Answers:

$\frac{(42\cdot 2^{(\frac{x}{3})})}{(5ln(2))}+C_{1}$

$\frac{(126\cdot 2^{(\frac{x}{3})})}{(5ln(2)^{2})}+\frac{1}{3}C_{1}+C_{2}$

$\frac{(42\cdot 2^{(\frac{x}{3})})}{(5ln(2))}+C_{1}x$

$\frac{(126\cdot 2^{(\frac{x}{3})})}{(5ln(2)^{2})}+C_{1}x+C_{2}$

Correct answer:

$\frac{(126\cdot 2^{(\frac{x}{3})})}{(5ln(2)^{2})}+C_{1}x+C_{2}$

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\text{Note for this problem, the }a\text{ value is } \frac{1}{3}\\&\text{Taking the first integral we find:}\\&\frac{(42\cdot 2^{(\frac{x}{3})})}{(5ln(2))}+C_{1}\\&\text{Note that after indefinite integration we add a constant, }C_1\\&\text{Integrating again, we'll add a new constant, while multiplying}\\&\text{our first one by }x:\\&\frac{(126\cdot 2^{(\frac{x}{3})})}{(5ln(2)^{2})}+C_{1}x+C_{2}\\&\end{align*}$

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Example Question #2617 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\iint(\frac{(17sin(x + 10))}{4})dx\end{align*}$

Possible Answers:

$-\frac{(17sin(x + 10))}{4}+C_{1}x+C_{2}$

$-\frac{(17cos(x + 10))}{4}+C_{1}x$

$-\frac{(17sin(x + 10))}{4}+10C_{1}+C_{2}$

$-\frac{(17cos(x + 10))}{4}+C_{1}$

Correct answer:

$-\frac{(17sin(x + 10))}{4}+C_{1}x+C_{2}$

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\text{Note in deriving the inside, the lone constant goes to zero.}\\&\text{Taking the first integral we find:}\\&-\frac{(17cos(x + 10))}{4}+C_{1}\\&\text{Note that after indefinite integration we add a constant, }C_1\\&\text{Integrating again, we'll add a new constant, while multiplying}\\&\text{our first one by }x:\\&-\frac{(17sin(x + 10))}{4}+C_{1}x+C_{2}\\&\end{align*}$

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Example Question #2618 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite double integral: }\\&\iint(\frac{(3cos(x + 5))}{79})dx\end{align*}$

Possible Answers:

$\frac{(3sin(x + 5))}{79}+C_{1}x$

$\frac{(3sin(x + 5))}{79}+C_{1}$

$-\frac{(3cos(x + 5))}{79}+5C_{1}+C_{2}$

$-\frac{(3cos(x + 5))}{79}+C_{1}x+C_{2}$

Correct answer:

$-\frac{(3cos(x + 5))}{79}+C_{1}x+C_{2}$

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[cos(ax)]=\frac{sin(ax)}{a}\\&\int[sin(ax)]=-\frac{cos(ax)}{a}\\&\text{Note in deriving the inside, the lone constant goes to zero.}\\&\text{Taking the first integral we find:}\\&\frac{(3sin(x + 5))}{79}+C_{1}\\&\text{Note that after indefinite integration we add a constant, }C_1\\&\text{Integrating again, we'll add a new constant, while multiplying}\\&\text{our first one by }x:\\&-\frac{(3cos(x + 5))}{79}+C_{1}x+C_{2}\\&\end{align*}$

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Example Question #2619 : Calculus Ii

$\begin{align*}&\text{Evaluate the indefinite integral: }\\&\iint(\frac{(7x^{16})}{3})dx\end{align*}$

Possible Answers:

$\frac{(7x^{17})}{51}+C_{1}x$

$\frac{(7x^{18})}{918}+16C_{1}+C_{2}$

$\frac{(7x^{18})}{918}+C_{1}x+C_{2}$

$\frac{(7x^{17})}{51}+C_{1}$

Correct answer:

$\frac{(7x^{18})}{918}+C_{1}x+C_{2}$

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int x^{a}=\frac{x^{a+1}}{a+1}\\&\text{Note for this problem, the }a\text{ value is } 16\\&\text{Taking the first integral we find:}\\&\frac{(7x^{17})}{51}+C_{1}\\&\text{Note that after indefinite integration we add a constant, }C_1\\&\text{Integrating again, we'll add a new constant, while multiplying}\\&\text{our first one by }x:\\&\frac{(7x^{18})}{918}+C_{1}x+C_{2}\\&\end{align*}$

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Example Question #2620 : Calculus Ii

$\begin{align*}&\text{Calculate the integral: }\\&\iint(\frac{(34\cdot 2^{(20x)})}{7})dx\end{align*}$

Possible Answers:

$\frac{(17\cdot 2^{(20x)})}{(1400ln(2)^{2})}+20C_{1}+C_{2}$

$\frac{(17\cdot 2^{(20x)})}{(1400ln(2)^{2})}+C_{1}x+C_{2}$

$\frac{(17\cdot 2^{(20x)})}{(70ln(2))}+C_{1}x$

$\frac{(17\cdot 2^{(20x)})}{(70ln(2))}+C_{1}$

Correct answer:

$\frac{(17\cdot 2^{(20x)})}{(1400ln(2)^{2})}+C_{1}x+C_{2}$

Explanation:

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\\&\text{can prove useful. After all, an integral is essentially the}\\&\text{opposite operation of a derivative. Consider how in a derivative}\\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\\&\text{of a function by the derivative of the outside? Integrals, in}\\&\text{similar fashion, entail a division.}\\&\text{Utilizing integral rules:}\\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\\&\text{Note for this problem, the }a\text{ value is } 20\\&\text{Taking the first integral we find:}\\&\frac{(17\cdot 2^{(20x)})}{(70ln(2))}+C_{1}\\&\text{Note that after indefinite integration we add a constant, }C_1\\&\text{Integrating again, we'll add a new constant, while multiplying}\\&\text{our first one by }x:\\&\frac{(17\cdot 2^{(20x)})}{(1400ln(2)^{2})}+C_{1}x+C_{2}\\&\end{align*}$

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Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #2611 : Calculus Ii

Example Question #2612 : Calculus Ii

Example Question #2613 : Calculus Ii

Example Question #2614 : Calculus Ii

Example Question #2615 : Calculus Ii

Example Question #2616 : Calculus Ii

Example Question #2617 : Calculus Ii

Example Question #2618 : Calculus Ii

Example Question #2619 : Calculus Ii

Example Question #2620 : Calculus Ii

All Calculus 2 Resources

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