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Calculus 3 : Calculus Review

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #121 : Calculus Review

The physical interpretation of the integral of a function f (x) , denoted by F (x) , is what?

Possible Answers:

No physical significance.

The average value of the tangent lines on f (x) .

Area under the function f (x) .

The points of inflection of the function f (x) .

Correct answer:

Area under the function f (x) .

Explanation:

By definition, the integral of a fucntion f (x) is the summation of an infinite number of small areas, thus giving the total area of the function with respect to the axis of integration.

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Example Question #122 : Calculus Review

Calculate the integral of the function f (x) given below.

$f (x) = x^4 - \cos (2x)$

Possible Answers:

$\int f (x)dx= \frac{x^5}{5}-\frac{1}{2} \sin (2x)$

$\int f (x)dx= \frac{x^5}{5}-\frac{1}{2} \sin (2x)+C$

$\int f (x)dx= \frac{x^5}{5}-\frac{1}{2} \cos^2 (2x)+C$

$\int f (x)dx= x^5-\sin (2x)+C$

Correct answer:

$\int f (x)dx= \frac{x^5}{5}-\frac{1}{2} \sin (2x)+C$

Explanation:

We have a seperable integral, meaning each term can be integrated independently, written as

$\int f (x)dx= \int \left(x^4 - \cos (2x)\right )dx = \int x^4 dx - \int \cos(2x)dx$ .

The first term is a power-rule integral, while the second is a trigonometric intergral. One should recall

$\int x^n dx = \frac{1}{n+1}x^{n+1}+C$

and $\int \cos (ax)dx = \frac{1}{a} \sin(ax)+C$

Putting these facts together leads us to the final answer of

$\int f (x)dx= \frac{x^5}{5}-\frac{1}{2} \sin (2x)+C$ .

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Example Question #161 : How To Find Position

If the acceleration function of an object is a(t)= 2t+11 , what is the position of the object at t=1 ? Assume the initial velocity and position is zero.

Possible Answers:

$\frac{35}{6}$

$\frac{13}{2}$

$\frac{11}{5}$

Correct answer:

$\frac{35}{6}$

Explanation:

To find the position function from the acceleration function, integrate a(t)= 2t+11 twice.

$v(t)=\int a(t)dt=t^2+11t$

When integrating, remember to increase the exponent of the variable by one and then divide the term by the new exponent. Do this for each term.

$s(t)=\int v(t)dt=\int\int a(t)dt = \frac{1}{3}t^3+\frac{11}{2}t^2$

Solve for s(t=1) .

$\frac{1}{3}(1)^3+\frac{11}{2}(1)^2=\frac{1}{3}+\frac{11}{2} = \frac{2}{6}+\frac{33}{6}= \frac{35}{6}$

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Example Question #123 : Calculus Review

Evaluate $\int_0^1 2t\sqrt{1+t^2}dt$ .

Possible Answers:

$\frac{3}{4}$

None of the other answers

$\frac{1}{2}$

$\frac{1}{2}(2^{3/2}-1)$

Correct answer:

None of the other answers

Explanation:

The correct answer is $\frac{2}{3}(2^{3/2}-1)$ .

We can evaluate this integral using -substitution.

Let u =1+t^2 , then du = 2tdt , hence we have

$\int_0^1 2t\sqrt{1+t^2}dt$ (Don't forget to change the bounds of integration)

$=\int_1^2 \sqrt{u} du$

$= \int_1^2 u^{1/2}du$

$=[\frac{2u^{3/2}}{3}]^2_1$

$=\frac{2\times{2}^{3/2}}{3} -\frac{2}{3}$

$=\frac{2}{3}(2^{3/2}-1)$ .

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Example Question #14 : Integration

Evaluate $\int_0^t e^ax dx$ , where is any constant.

Possible Answers:

None of the other answers

$\frac{t^2e^a}{2}$

$\frac{e^t}{2}$

t^2e^a

Correct answer:

$\frac{t^2e^a}{2}$

Explanation:

Since is a constant, so is e^a , therefore we can factor it out of the integral.

$\int_0^te^ax dx$

$=e^a\int_0^txdx$

$=e^a[\frac{x^2}{2}]^t_0$

$=e^a(\frac{t^2}{2}-0)$

$=\frac{t^2e^a}{2}$

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Example Question #122 : Calculus Review

Calculate $\int \cos (x) e^{sin(x)}dx$ .

Possible Answers:

$e^{-\sin(x)}+C$

$e^{\sin(x)}+C$

$e^{\sin (x)}$

$e^{\cos(x)}+C$

Correct answer:

$e^{\sin(x)}+C$

Explanation:

This integral can be found using u-substitution. Consider

$u = \sin(x), du = \cos(x)$ . This means we can rewrite our integral as

$\int e^{u}du=e^u+C$ , by definition of the integral of an exponential.

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Example Question #125 : Calculus Review

Calculate $\int \ln x dx$

Possible Answers:

Can not be determined.

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

$x \ln x-x$

e^x+C

Correct answer:

$x \ln x-x$

Explanation:

This integral can be done using integration by parts. Consider

$\int 1 \cdot \ln x dx$ .

Choose $u = \ln x, du = \frac{dx}{x}$ and dv = dx, v = x .

Using the definition of integration by parts,

$\int u dv = uv - \int v du$ ,

$uv - \int v du= x \ln x - \int dx= x \ln x - x$ .

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Example Question #126 : Calculus Review

Calculate $\int_0^{\infty} 2xe^{-x} dx$

Possible Answers:

$\frac{1}{2}$

$+ \infty$

Correct answer:

Explanation:

This can be a challenging integral using standard methods. However, it is easy if we use integration by-parts, given as

$\int udv = uv - \int v du$

Choose

$u = x, dv = e^{-x} \Rightarrow du = dx, v = -e^{-x}$ .

From the definition,

$2\int _0^{\infty}xe^{-x}dx= 2\left(-x*e^{-x} \left | _0 ^{\infty}+\int _0^{\infty}e^{-x}dx\right)=2\left(-e^{-x} \left| _0^{\infty}\right) = 2$

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Example Question #127 : Calculus Review

Calculate $\int \cos \theta \left( 2 \sin \theta+5\right )^3 d \theta$

Possible Answers:

$\frac{1}{8}\left(2\sin \theta +5 \right )^4+C$

$\frac{\cos \theta}{8}\left(2\sin \theta +5 \right )^4+C$

$\frac{1}{4}\left(2\sin \theta +5 \right )^4+C$

$\frac{1}{8}\left(2\sin \theta +5 \right )^4- \cos \theta + C$

Correct answer:

$\frac{1}{8}\left(2\sin \theta +5 \right )^4+C$

Explanation:

This integral is most easily found by implementing u-substitution. Choose

$u = 2 \sin \theta + 5 \Rightarrow \frac{du}{2} = \cos \theta d \theta$ , which means we can rewrite the integral in a more familiar form

$\int \cos \theta \left( 2 \sin \theta + 5\right )^3 d \theta = \frac{1}{2}\int u^3 du= \frac{1}{2}*\frac{1}{4}u^4+C= \frac{1}{8}\left(2 \sin \theta + 5 \right )^4 + C$

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Example Question #128 : Calculus Review

Calculate $\int \frac{6x+15}{\left(x^2+5x+10 \right )^3} dx$

Possible Answers:

$\frac{3}{2\left(x^2+5x+10 \right )^2}+C$

$-\frac{3x}{2\left(x^2+5x+10 \right )^2}+C$

$-\frac{3}{2\left(x^2+5x+10 \right )^2}+C$

$-\frac{x}{2\left(x^2+5x+10 \right )^3}+C$

Correct answer:

$-\frac{3}{2\left(x^2+5x+10 \right )^2}+C$

Explanation:

This integral is most easily done by using u-substitution. Initially rewrite our integral as

$\int \frac{6x+15}{\left(x^2+5x+10 \right )^3} dx=\int \frac{3(2x+5)}{\left(x^2+5x+10 \right )^3} dx$ , then choose

u = x^2 + 5x + 10, du = (2x+5)dx . Therefore

$\int \frac{6x+15}{\left(x^2+5x+10 \right )^3} dx= 3\int u^{-3}du= -\frac{3}{2}u^{-2}+C=-\frac{3}{2\left(x^2+5x+10 \right )^2}+C$

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Calculus 3 : Calculus Review

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #121 : Calculus Review

Example Question #122 : Calculus Review

Example Question #161 : How To Find Position

Example Question #123 : Calculus Review

Example Question #14 : Integration

Example Question #122 : Calculus Review

Example Question #125 : Calculus Review

Example Question #126 : Calculus Review

Example Question #127 : Calculus Review

Example Question #128 : Calculus Review

All Calculus 3 Resources

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