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Calculus 3 : Integration

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Example Questions

Example Question #301 : Calculus 3

A particle's velocity in two dimensions is described by the functions:

v_x(t)=t^2+sin(t)

v_y(t)=tcos(t^2)+3

If the particle has an initial position of (0,0) , what will its position be at time $t=2\pi$ ?

Possible Answers:

$(\frac{8\pi^3}{3}+1,6\pi-1)$

$(\frac{8\pi^3}{3},6\pi)$

$(\frac{8\pi^3-1}{3},6\pi-1)$

$(\frac{8\pi^3}{3}-1,6\pi)$

$(\frac{4\pi^3}{3},2\pi)$

Correct answer:

$(\frac{8\pi^3}{3},6\pi)$

Explanation:

Position can be found by integrating velocity with respect to time:

$p(t)=\int v(t)dt$

For velocities:

v_x(t)=t^2+sin(t)

v_y(t)=tcos(t^2)+3

The position functions are:

$x(t)=\frac{t^3}{3}-cos(t)+C_x$

$y(t)=\frac{sin(t^2)}{2}+3t+C_y$

These constants of integration can be found by using the given initial conditions:

$x(0)=\frac{0^3}{3}-cos(0)+C_x=0$

C_x=1

$y(0)=\frac{sin(0^2)}{2}+3(0)+C_y=0$

C_y=0

Which gives the definite integrals:

$x(t)=\frac{t^3}{3}-cos(t)+1$

$y(t)=\frac{sin(t^2)}{2}+3t$

$x(2\pi)=\frac{(2\pi)^3}{3}-cos(2\pi)+1$

$x(2\pi)=\frac{8\pi^3}{3}$

$y(2\pi)=\frac{sin((2\pi)^2)}{2}+3(2\pi)$

$y(2\pi)=6\pi$

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Example Question #305 : Calculus 3

Calculate antiderivative of $f(x)=\frac{3x^2+2x+1}{x^3+x^2+x+1}$

Possible Answers:

$\ln(x^3+x^2+x+20)+C$

$\ln(x^3+x^2+1)+C$

$\ln(x^3+x^2+x+1)+C$

$\ln(x^3+x^2+x)+C$

Correct answer:

$\ln(x^3+x^2+x+1)+C$

Explanation:

We can calculate the antiderivative $\small f(x)=\frac{3x^2+2x+1}{x^3+x^2+x+1}$

$\small \small \int \frac{3x^2+2x+1}{x^3+x^2+x+1}dx$

by using $\small u$ -substitution. We set $\small u=x^3+x^2+x+1$ , so we get $\small du=3x^2+2x+1$ , so the integral becomes

$\small \small \small \int \frac{3x^2+2x+1}{x^3+x^2+x+1}dx=\int \frac{du}{u}=\ln u$

So we just plug $\small u=x^3+x^2+x+1$ to get

$\small \small \small \small \small \int \frac{3x^2+2x+1}{x^3+x^2+x+1}dx=\int \frac{du}{u}=\ln u=\ln (x^3+x^2+x+1)$

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Example Question #306 : Calculus 3

Calculate antiderivative of $\small f(x)=ex$ .

Possible Answers:

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

$\small e^x+C$

$\small \small e^2x+C$

$\small 2e^x+C$

Correct answer:

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

Explanation:

We can calculate the antiderivative $\small \small f(x)=ex$

$\small \small \small \int ex dx$

by using the power rule for antiderivatives:

$\small \int x^a=\frac{x^{a+1}}{a+1}+C$

In this case $\small a=1$ , so we have

$\small \small \small \small \small \int ex dx=e\frac{x^2}{2}+C=\frac{ex^2}{2}+C$ .

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Example Question #121 : Calculus Review

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

Possible Answers:

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

Area under the function f (x) .

No physical significance.

The average value of the tangent lines on 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:

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

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

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

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

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 #121 : Calculus Review

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:

$\frac{35}{6}$

$\frac{11}{5}$

$\frac{13}{2}$

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 #312 : Calculus 3

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

Possible Answers:

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

None of the other answers

$\frac{1}{2}$

$\frac{3}{4}$

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 #122 : Calculus Review

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

Possible Answers:

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

t^2e^a

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

None of the other answers

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 #123 : Calculus Review

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

Possible Answers:

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

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

$e^{\sin (x)}$

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

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 #14 : Integration

Calculate $\int \ln x dx$

Possible Answers:

$x \ln x-x$

Can not be determined.

e^x+C

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

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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Calculus 3 : Integration

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #301 : Calculus 3

Example Question #305 : Calculus 3

Example Question #306 : Calculus 3

Example Question #121 : Calculus Review

Example Question #122 : Calculus Review

Example Question #121 : Calculus Review

Example Question #312 : Calculus 3

Example Question #122 : Calculus Review

Example Question #123 : Calculus Review

Example Question #14 : Integration

All Calculus 3 Resources

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