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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 #442 : Finding Integrals

$\int \frac{5x^2+20x-10}{5}dx$

Possible Answers:

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

x^3+2x^2-2x+C

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

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

$\frac{x^3}{3}+2x^2-2x$

Correct answer:

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

Explanation:

First, chop up the fraction into three simplified terms:
$\int x^2+4x-2dx$

Now integrate.

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

Now add a C because it is an indefinite integral:

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

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Example Question #2542 : Calculus Ii

Calculate the following integral: $\int cos^{2}x+x\sqrt{x^{2}+1}dx$

Possible Answers:

$\frac{1}{2}x+\frac{1}{4}sin(2x)+(\sqrt{x^{2}+1})^{3}+C$

$\frac{1}{4}sin(2x)+\frac{1}{3}(\sqrt{x^{2}+1})^{3}+C$

$\frac{1}{2}x+\frac{1}{4}sin(x)+\frac{1}{3}(\sqrt{x^{2}+1})^{3}+C$

$\frac{1}{2}x+\frac{1}{4}sin(2x)+\frac{1}{3}(\sqrt{x^{2}+1})^{3}+C$

Correct answer:

$\frac{1}{2}x+\frac{1}{4}sin(2x)+\frac{1}{3}(\sqrt{x^{2}+1})^{3}+C$

Explanation:

Separate integral into two separate integrals:

$\int cos^{2}x+x\sqrt{x^{2}+1}dx=\int cos^{2}xdx+\int x\sqrt{x^{2}+1}dx$

Solve $\int cos^{2}xdx$ first.

Use power-reducing identity to simplify integral: $\int \frac{1+cos(2x)}{2}dx$ .

Factor $\frac{1}{2}$ out of the integral: $\int \frac{1}{2}+\frac{cosx}{2}dx$

Separate into two integrals: $\frac{1}{2}\int 1dx+\frac{1}{2}\int cos (2x)dx=\frac{1}{2}x+\frac{1}{2}\int cos (2x)dx$ .

Use the following substitution for the second integral: u=2x du=2 $\frac{1}{2}du=dx$

Plug in substitution and solve: $\frac{1}{4}\int cosudu=\frac{1}{4}sinu+C=\frac{1}{4}sin(2x)+C$ $\frac{1}{4}sinu=\frac{1}{4}sin(2x)+C$ .

Therefore: $\int cos^{2}x=\frac{1}{2}x+\frac{1}{4}sin(2x)+C$

Solve $\int x\sqrt{x^{2}+1}dx$

Make the following substitution: $u=x^{2}+1$ du=2xdx $\frac{1}{2}du=dx$ . Plug in substitution and solve: $\frac{1}{2}\int \sqrt{u}du=(\frac{2}{3})\frac{1}{2}(x^{2}+1)^{\frac{3}{2}}+C=\frac{1}{3}(x^{2}+1)^{\frac{3}{2}}+C$

Combine answers to two original integrals: $\int cos^{2}xdx+\int x\sqrt{x^{2}+1}dx=\frac{1}{2}x+\frac{1}{4}sin(2x)+\frac{1}{3}(x^{2}+1)^{\frac{3}{2}}+C$

$\int cos^{2}x+x\sqrt{x^{2}+1}dx=\frac{1}{2}x+\frac{1}{4}sin(2x)+\frac{1}{3}(x^{2}+1)^{\frac{3}{2}}+C$

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Example Question #2543 : Calculus Ii

Evaluate.

$\int 44x^{43} \ dx$

Possible Answers:

$F(x) = 43x^{44 }+ C$

$F(x) = x^{44 }+ C$

$F(x) = \frac{1}{44}x^{44 }+ C$

Answer not listed

$F(x) = x^{43}+ C$

Correct answer:

$F(x) = x^{44 }+ C$

Explanation:

$\int f(x) dx$

In order to evaluate this integral, first find the antiderivative of f(x).

If f(x) = a then F(x) = ax + C

If f(x) = x a then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then F(x) = ln(x) + C

If f(x) = e ^x then F(x) = e^x + C

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, $f(x) = 44x^{43}$ .

The antiderivative is $F(x) = x^{44 }+ C$ .

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Example Question #2544 : Calculus Ii

Evaluate.

$\int \frac{1}{4x-19} \ dx$

Possible Answers:

Answer not listed.

$F(x) = \frac{x-4}{4x-19} + C$

$F(x) = \frac{4}{4x-19} + C$

F(x) = 4x-19 + C

$F(x) = \frac{4x-19}{4} + C$

Correct answer:

$F(x) = \frac{4x-19}{4} + C$

Explanation:

$\int f(x) dx$

In order to evaluate this integral, first find the antiderivative of f(x).

If f(x) = a then F(x) = ax + C

If f(x) = x a then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then F(x) = ln(x) + C

If f(x) = e ^x then F(x) = e^x + C

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, $f(x) = \frac{1}{4x-19}$ .

The antiderivative is $F(x) = \frac{4x-19}{4} + C$ .

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Example Question #162 : Indefinite Integrals

Integrate to lowest terms: $\large \int \sqrt {x}-1$

Possible Answers:

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

$\large \large \frac {2x^\frac {3}{2}+{3}x} {3}+C$

None of the Above

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

Correct answer:

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

Explanation:

Screen shot 2016 01 02 at 10.25.11 am

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Example Question #2546 : Calculus Ii

Evaluate.

$\int e^{8x-43}\ dx$

Possible Answers:

$F(x) = \frac{e^{8x-43}}{8} + C$

$F(x) = 8e^{8x-43}} + C$

Answer not listed.

$F(x) = \frac{8x}{e^{8x-43}} + C$

$F(x) = \frac{8}{e^{8x-43}} + C$

Correct answer:

$F(x) = \frac{e^{8x-43}}{8} + C$

Explanation:

$\int f(x) dx$

In order to evaluate this integral, first find the antiderivative of f(x).

If f(x) = a then F(x) = ax + C

If f(x) = x a then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then F(x) = ln(x) + C

If f(x) = e ^x then F(x) = e^x + C

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, $f(x)= e^{8x-43}$ .

The antiderivative is $F(x) = \frac{e^{8x-43}}{8} + C$ .

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Example Question #2547 : Calculus Ii

Evaluate.

$\int \cos (12) \ dx$

Possible Answers:

Answer not listed.

$F(x) = \cos(12) + C$

$F(x) = \cos(12)x + C$

$F(x) = \cos(12x) + C$

$F(x) = \frac{\cos(12)}{x} + C$

Correct answer:

$F(x) = \cos(12)x + C$

Explanation:

$\int f(x) dx$

In order to evaluate this integral, first find the antiderivative of f(x).

If f(x) = a then F(x) = ax + C

If f(x) = x a then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then F(x) = ln(x) + C

If f(x) = e ^x then F(x) = e^x + C

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, $f(x)= \cos(12)$ .

The antiderivative is $F(x) = \cos(12)x + C$ .

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Example Question #171 : Indefinite Integrals

Evaluate.

$\int \sin(10x-3) \ dx$

Possible Answers:

$F(x) = -\frac{\cos(10x-3)}{10} + C$

$F(x) = -10\cos(10x-3) + C$

$F(x) = -\frac{10}{\sin(10x-3)} + C$

Answer not listed.

$F(x) = -\frac{\cos(10x)}{\sin(10x-3)} + C$

Correct answer:

$F(x) = -\frac{\cos(10x-3)}{10} + C$

Explanation:

$\int f(x) dx$

In order to evaluate this integral, first find the antiderivative of f(x).

If f(x) = a then F(x) = ax + C

If f(x) = x a then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then F(x) = ln(x) + C

If f(x) = e ^x then F(x) = e^x + C

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, $f(x)= \sin(10x-3)$ .

The antiderivative is $F(x) = -\frac{\cos(10x-3)}{10} + C$ .

Report an Error

Example Question #172 : Indefinite Integrals

Evaluate.

$\int_{1}^{2}\cos (12) \ dx$

Possible Answers:

2.34

0.98

5.32

Answer not listed.

-0.53

Correct answer:

0.98

Explanation:

$\int_{a}^{b}f(x) \ dx$

In order to evaluate this integral, first find the antiderivative of f(x).

In this case, $f(x)= \cos(12)$ .

The antiderivative is $F(x) = \cos(12)x$ .

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative:

$\int_{1}^{2}\cos (12) \ dx = \left ( \cos(12)x \right )_{1}^{2}$

$= \cos(12)(2) - \left ( \cos(12)(1) \right )$

= 0.98

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Example Question #173 : Indefinite Integrals

Evaluate.

$\int_{1}^{2} \sin(10x-3) \ dx$

Possible Answers:

5.8

2.9

1.3

0.1

Answer not listed

Correct answer:

Answer not listed

Explanation:

$\int_{a}^{b}f(x) \ dx$

In order to evaluate this integral, first find the antiderivative of f(x).

In this case, $f(x)= \sin(10x-3)$ .

The antiderivative is $F(x) = -\frac{\cos(10x-3)}{10}$ .

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative:

$\int_{1}^{2} \sin(10x-3) \ dx = \left ( -\frac{\cos(10x-3)}{10} \right )_{1}^{2}$

$= -\frac{\cos(10(2)-3)}{10} - \left ( -\frac{\cos(10-3)}{10} \right )$

= 0.004

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #442 : Finding Integrals

Example Question #2542 : Calculus Ii

Example Question #2543 : Calculus Ii

Example Question #2544 : Calculus Ii

Example Question #162 : Indefinite Integrals

Example Question #2546 : Calculus Ii

Example Question #2547 : Calculus Ii

Example Question #171 : Indefinite Integrals

Example Question #172 : Indefinite Integrals

Example Question #173 : Indefinite Integrals

All Calculus 2 Resources

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