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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 #2361 : Calculus Ii

$\int_{1}^{e}lnxdx$

Solve the integral using integration by parts.

Possible Answers:

$\pi$

e-1

Correct answer:

Explanation:

First we identify what are our u and dv are and then we derive u and take the integral of dv.

u=lnx

Derive.

$du=\frac{1}{x}dx$

Now the dv is a little tricky, because it looks like the only term there is lnx. We have to keep in mind that there is always a 1 attached to the front as a coefficient.

dv=1dx

Integrate.

v=x

Now we plug into our by parts equation.

$\int{udv}=uv-\int{vdu}$

$(lnx)x-\int_{1}^{e}x\frac{1}{x}dx$

Simplify the second part.

$xlnx-\int_{1}^{e}1dx$

Integrate the second part

$xlnx-x|_{1}^{e}$

Plug in the bounds.

elne-e-(1ln1-1)=e-e-(0-1)=-(-1)=1

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Example Question #243 : Definite Integrals

$\int_{0}^{1}56x(x+1)^6dx$

Solve the integral by using integration by parts.

Possible Answers:

768

770

769

767

Correct answer:

769

Explanation:

First things first. We must identify what our u and dv are. U is generally the simpler of the two and since 56x only has a coefficient we will go with that one. Dv will then be (x+1)^6 . We will need to take the derivative of u and integrate dv and then plug into the by parts equation.

u=56x

Derive.

du=56dx

and

dv=(x+1)^6dx

Integrate.

$v=\frac{(x+1)^7}{7}$

Plug into our by parts equation.

$\int{udv}=uv-\int{vdu}$

$56x\frac{(x+1)^7}{7}-\int_{0}^{1}\frac{(x+1)^7)}{7}56dx$

Integrate the second part.

$8x(x+1)^7-56\frac{(x+1)^8}{56}|_{0}^{1}$

Plug in the top value and the bottom value and subtract.

769

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Example Question #611 : Integrals

$\int_{0}^{2}x^2+3dx$

Possible Answers:

$\frac{1}{3}$

$\frac{28}{3}$

$\frac{2}{3}$

$\frac{8}{3}$

$\frac{26}{3}$

Correct answer:

$\frac{26}{3}$

Explanation:

First, you must integrate this expression. Evaluate each term separately. Remember to raise the exponent by 1 and also put that result on the denominator.

First term: $x^2=\frac{x^3}{3}$

Second term: 3=3x

Put those together to get $\frac{x^3}{3}+3x$ . Now evaluate at 2 and then 0. Subtract the results: $[\frac{(2)^3}{3}+3(2)]-0=\frac{26}{3}$ .

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Example Question #2361 : Calculus Ii

$\int_{0}^{3}3x^2+1dx$

Possible Answers:

Correct answer:

Explanation:

First, integrate the terms. Remember to raise the exponent by 1 and also put that result on the denominator:

$3(\frac{x^3}{3})+x=x^3+x$

Next, evaluate at 3 and then 0. Subtract the results:

[3^3+3]-0=30 .

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Example Question #2362 : Calculus Ii

Give the indefinite integral:

$\int \left (\frac{4}{3} x ^{5} - \frac{3}{4} x ^{3} + \frac{2}{7} x \right) \;\mathrm{d} x$

Possible Answers:

$\frac{2 }{9} x ^{6} - \frac{ 3}{16}x ^{4} + \frac{1}{7} x^{2} +C$

$\frac{4 }{15} x ^{6} - \frac{ 1}{4}x ^{4} + \frac{2}{7} x^{2} +C$

$\frac{2 }{9} x ^{4} - \frac{ 3}{16}x ^{2} +C$

$\frac{4 }{15} x ^{6} - \frac{ 1}{4}x ^{4} + \frac{2}{7} x +C$

$\frac{4 }{15} x ^{4} - \frac{ 1}{4}x ^{2} + C$

Correct answer:

$\frac{2 }{9} x ^{6} - \frac{ 3}{16}x ^{4} + \frac{1}{7} x^{2} +C$

Explanation:

$\int ax ^{n} \;\mathrm{d} x = a \cdot\frac{ x ^{n+ 1 }}{n+1}$ ,

$\int \left (\frac{4}{3} x ^{5} - \frac{3}{4} x ^{3} + \frac{2}{7} x \right) \;\mathrm{d} x$

$=\frac{4}{3}\cdot\frac{ x ^{5+1}}{5+1} - \frac{3}{4} \cdot\frac{ x ^{3+1}}{3+1} + \frac{2}{7} \cdot \frac{ x^{1+1} }{1+1}+C$

$=\frac{4}{3}\cdot\frac{ x ^{6}}{6} - \frac{3}{4} \cdot\frac{ x ^{4}}{4} + \frac{2}{7} \cdot \frac{ x^{2} }{2}+C$

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Example Question #612 : Integrals

Give the indefinite integral:

$\int \left ( 0.4 x ^{99} - 0.7 x ^{49} \right ) \;\mathrm{d} x$

Possible Answers:

$0.04 x ^{100} - 0.14 x ^{50} +C$

$0.004 x ^{100} - 0.007 x ^{50} +C$

$0.04 x ^{98} - 0.14 x ^{48} +C$

$0.004 x ^{100} - 0.014 x ^{50} +C$

$0.004 x ^{98} - 0.014 x ^{48} +C$

Correct answer:

$0.004 x ^{100} - 0.014 x ^{50} +C$

Explanation:

$\int ax ^{n} \;\mathrm{d} x= a \cdot\frac{ x ^{n+ 1 }}{n+1}$ ,

$\int \left ( 0.4 x ^{99} - 0.7 x ^{49} \right ) \;\mathrm{d} x$

$= 0.4 \cdot \frac{ x ^{99+1}}{99+1} - 0.7 \cdot \frac{ x ^{49+1} }{49+1} +C$

$= 0.4 \cdot \frac{ x ^{100}}{100} - 0.7 \cdot \frac{ x ^{50} }{50} +C$

$= \frac{0.4 }{100}x ^{100} - \frac{ 0.7 }{50} x ^{50} +C$

$=0.004 x ^{100} - 0.014 x ^{50} +C$

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Example Question #264 : Finding Integrals

Give the indefinite integral:

$\int \cot \frac{\pi }{6}x \; \mathrm{d} x$

Possible Answers:

$\frac{\pi } {6} \ln \left \left | \cos \frac{\pi }{6}x \right \right | + C$

$\frac{\pi } {6} \ln \left \left | \sin \frac{\pi }{6}x \right \right | + C$

$\frac{6} {\pi } \ln \left \left | \cos \frac{\pi }{6}x \right \right | + C$

$\frac{6} {\pi } \ln \left \left | \sin \frac{\pi }{6}x \cos \frac{\pi }{6}x \right \right | + C$

$\frac{6} {\pi } \ln \left \left | \sin \frac{\pi }{6}x \right \right | + C$

Correct answer:

$\frac{6} {\pi } \ln \left \left | \sin \frac{\pi }{6}x \right \right | + C$

Explanation:

Substitute $u = \frac{\pi }{6}x$ . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = \frac{\pi }{6}$ , $\mathrm{d} u = \frac{\pi }{6}\; \mathrm{d} x$ , and $\frac{6} {\pi } \; \mathrm{d} u = \mathrm{d} x$ .

The integral becomes

$\int \cot \frac{\pi }{6}x \; \mathrm{d}x$

$= \int \cot u \; \cdot \frac{6} {\pi } \; \mathrm{d}u$

$=\frac{6} {\pi } \int \cot u \; \mathrm{d}u$

$=\frac{6} {\pi } \left ( \ln | \sin u | + C \right )$

$=\frac{6} {\pi } \ln | \sin u | + \frac{6} {\pi } C$

$=\frac{6} {\pi } \ln | \sin u | + C$

$=\frac{6} {\pi } \ln \left \left | \sin \frac{\pi }{6}x \right \right | + C$

Note that the $\frac{6} {\pi }$ gets absorbed into the constant term in the second-to-last step.

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Example Question #613 : Integrals

Find the indefinite integral:

$\int (8x ^{3} - 6x ^{2} + 4x - 2) \;\mathrm{d} x$

Possible Answers:

$-2x ^{4} - 2x ^{3} - 2x^{2} - 2x+ C$

$\frac{8}{3} x ^{4} - 3x ^{3} + 4x^{2} - 2x+ C$

$2x ^{4}+2x ^{3} + 2x^{2} + 2x+ C$

$2x ^{4} - 2x ^{3} + 2x^{2} - 2x+ C$

$\frac{8}{3} x ^{4} + 3x ^{3} + 4x^{2} + 2x + C$

Correct answer:

$2x ^{4} - 2x ^{3} + 2x^{2} - 2x+ C$

Explanation:

Recall the rules of integration: $\int ax ^{n} \;\mathrm{d} x = a \cdot\frac{ x ^{n+ 1 }}{n+1}$ and $\int a \; dx = ax$

$\int (8x ^{3} - 6x ^{2} + 4x - 2) \;\mathrm{d} x$

$= 8\cdot \frac{ x ^{3+ 1} }{3+ 1 }- 6\cdot \frac{ x ^{2+ 1} }{2+1}+ 4\cdot \frac{ x^{1+ 1} }{1+1}- 2x+C$

$= 8\cdot \frac{ x ^{4} }{4 }- 6\cdot \frac{ x ^{3} }{3}+ 4\cdot \frac{ x^{2} }{2}- 2x + C$

$= 2x ^{4} - 2x ^{3} + 2x^{2} - 2x + C$

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Example Question #614 : Integrals

Give the indefinite integral:

$\int \sqrt{ e ^{x } }\; \mathrm{d} x$

Possible Answers:

$e^x \sqrt{e^{x}} + C$

$\sqrt{e^{x}} + C$

$\frac{\sqrt{e^{x}}}{2e^{x}} + C$

$2 \sqrt{e^{x}} + C$

$\frac{\sqrt{e^{x}}}{e^{x}} + C$

Correct answer:

$2 \sqrt{e^{x}} + C$

Explanation:

$\sqrt{e^{x}} =\left ( e^{x} \right ) ^{ \frac{1}{2}} = e^{\frac{1}{2}x}$ , so the integral can be rewritten as $\int e^{\frac{1}{2}x} \; \mathrm{d} x$ .

Substitute $u = \frac{1}{2}x$ , so $\frac{\mathrm{d} u}{\mathrm{d} x} = \frac{1}{2}$ and $\mathrm{d} x = 2 \cdot \mathrm{d} u$ .

The integral becomes

$\int e^{\frac{1}{2}x} \; \mathrm{d} x$

$= \int e^{u} \cdot \; 2 \; \mathrm{d} u$

$=2 \int e^{u}\; \mathrm{d} u$

$=2 \left ( e^{u} + C \right )$

$=2 \left (e^{\frac{1}{2}x} + C \right )$

$=2 \left (\sqrt{e^{x}} + C \right )$

$=2 \sqrt{e^{x}} + 2 \cdot C$

$=2 \sqrt{e^{x}} + C$

Note that the 2 gets absorbed into the constant term.

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Example Question #2365 : Calculus Ii

Give the indefinite integral:

$\int \sin 7x \; \mathrm{d} x$

Possible Answers:

$7 \cos 7x + C$

$-7 \cos 7x + C$

$- \cos 7x + C$

$- \frac{1}{7} \cos 7x + C$

$\frac{1}{7} \cos 7x + C$

Correct answer:

$- \frac{1}{7} \cos 7x + C$

Explanation:

Substitute u = 7x . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = 7$ and $\mathrm{d} u = 7\: \mathrm{d} x$ , or $\mathrm{d} x = \frac{1}{7} \mathrm{d} u$ .

The integral becomes

$\int \sin 7x \; \mathrm{d} x$

$= \int \sin u \cdot \frac{1}{7}\; \mathrm{d} u$

$= \frac{1}{7} \int \sin u \; \mathrm{d} u$

$= \frac{1}{7} (- \cos u + C)$

$= - \frac{1}{7} \cos 7x + \frac{1}{7}C = - \frac{1}{7} \cos 7x + C$

Note that the $\frac{1} {7 }$ gets absorbed into the constant term.

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #2361 : Calculus Ii

Example Question #243 : Definite Integrals

Example Question #611 : Integrals

Example Question #2361 : Calculus Ii

Example Question #2362 : Calculus Ii

Example Question #612 : Integrals

Example Question #264 : Finding Integrals

Example Question #613 : Integrals

Example Question #614 : Integrals

Example Question #2365 : Calculus Ii

All Calculus 2 Resources

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