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Calculus 2 : Integrals

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Example Questions

Example Question #671 : Finding Integrals

Evaluate the integral:

$\int \frac{dx}{\sqrt{x+2}}$

Possible Answers:

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

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

$\sqrt{x+2}+C$

$2\sqrt{x+2}$

Correct answer:

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

Explanation:

To evaluate the integral, we perform the following substitution:

$u=(x+2)^{\frac{1}{2}}$

$du=\frac{1}{2}(x+2)^{-\frac{1}{2}}dx$

The derivative was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Now, we rewrite the integrand and integrate:

$2\int du=2u+C$

The integral was performed using the following rule:

$\int dx=x+C$

Finally, replace u with our original x term:

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

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Example Question #1022 : Integrals

The Laplace Transform is an integral transform that converts functions from the time domain to the complex frequency domain . The transformation of a function f(t) into its complex frequency function F(s) is given by:

$F(s)= \int_{0}^{\infty}e^{-st}*f(t) dt$

Where s=a+bi , where and are constants and is the imaginary number.

Evaluate the Laplace Transform of the function f(t) at time t-a . Suppose that f(t)=0 when $t \leq a$ .

Possible Answers:

saF(s)

F(s-a)

$e^{-sa}F(s)$

$e^{sa}F(s)$

Correct answer:

$e^{-sa}F(s)$

Explanation:

The Laplace Transform will be given by:

$\int_{0}^{\infty}e^{-s(t-a)}*f(t-a) dt$

Since f(t-a)=0 when $t \leq a$ , we can change the integral to:

$\int_{a}^{\infty}e^{-st}*f(t-a) dt$

This is because when you change the lower bound of the integral, the exponent will only exist for values for which is defined.

Let U=t-a

t=U+a

$\frac{dU}{dt}=1$

This changes our integral to:

$\int_{a}^{\infty}e^{-s(U+a)}*f(U) dU$

We can now move the $e^{-sa}$ term out of the integral, which will give us:

$e^{-sa}\int_{a}^{\infty}e^{-sU}*f(U) dU=e^{-sa} F(s)$

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Example Question #673 : Finding Integrals

Evaluate the following integral using substitution:

$\int\frac{18x^2}{(3x^3+7)^2}dx$

Possible Answers:

$\frac{4}{9(3x^3+7)^3}+C$

$-\frac{4}{9(3x^3+7)^3}+C$

$-\frac{2}{3x^3+7}+C$

$-\frac{1}{9(3x^3+7)}+C$

$-\frac{4}{(3x^3+7)^3}+C$

Correct answer:

$-\frac{2}{3x^3+7}+C$

Explanation:

To evaluate this integral, we first make the following substitution:

u=3x^3+7

Differentiating this expression, we get:

$du=9x^2 dx\rightarrow x^2dx=\frac{1}{9}du$

Now, we can rewrite the original integral with our substitution and solve:

$\int\frac{18x^2}{(3x^3+7)^2}dx=\int\frac{18}{u^2}\frac{du}{9}=\int2u^{-2}du=-2u^{-1}+C$

Finally, we have to replace u with our earlier definition:

$-2u^{-1}+C=-\frac{2}{3x+7}+C$

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Example Question #1024 : Integrals

In exponentially decaying systems, often times the solutions to differential equations take on the form of an integral called Duhamel's Integral. This is given by:

$g(t)=\int_{0}^{t}e^{-s(t-b)}*f(b) db$

Where is a constant and f(b) is a function that represents an external force.

Suppose I introduce growth factors that effect my population at a rate of

$f(b)=e^{kb}$ . At what rate do I need in order for my population to grow? (Hint: Find and determine for what will increase in time.)

Possible Answers:

$k\geq -s$

k>0

k > s

s>k

Correct answer:

k>0

Explanation:

Start by substituting $f(b)=e^{kb}$ into the integral to get:

$g(t)=\int_{0}^{t}e^{-s(t-b)}*e^{kb} db$

We can combine this into one large term:

$g(t)=\int_{0}^{t}e^{-s(t-b)+kb} db$

u=-st+sb+kb

$\frac{du}{db}=s+k$

$db=\frac{u}{s+k}$

$g(t)=\frac{1}{s+k}\int_{0}^{t}e^{u} du$

Since $\int e^u=e^u$ .

$g(t)=\frac{1}{s+k}[e^u]_{0}^{t}=\frac{1}{s+k}[e^{-st+sb+kb}]_{0}^{t}= \frac{1}{s+k}[e^{kt}-e^{-st}]$

This can only grow when:

k > 0

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Example Question #2771 : Calculus Ii

Evaluate the integral with a substitution,

$\small \small \int\frac{ \sin(\ln x)}{x} dx$

Possible Answers:

$\small -\sin(\ln x)+\cos(\ln x)+C$

$\small \small - \cos(\ln x) + C$

$\small \small -\sin(\ln x) + C$

$\small \small \small \cos(\ln x)+\frac{1}{x}+C$

$\small \frac{\ln x}{x}+\sin(\ln x)$

Correct answer:

$\small \small - \cos(\ln x) + C$

Explanation:

$\small \small \int\frac{ \sin(\ln x)}{x} dx$

Let

$\small u = ln(x)$

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

$\small \small \small \small \int\frac{ \sin(\ln x)}{x} dx=\int\sin(u)du=-\cos(u)$

We can now convert this back to a function of $\small x$ by substituting $\small u = \ln x$ ,

$\small -\cos(u)=-\cos(\ln x)$

$\small \small \small \small \small \int\frac{ \sin(\ln x)}{x} dx = - \cos(\ln x)$

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Example Question #671 : Finding Integrals

Calculate the following integral: $\int \frac{x}{x+2}dx$

Possible Answers:

$x+2-\frac{1}{2}ln\left |x+2 \right |+C$

$2x-2ln\left |x+2 \right |+C$

$x+2-ln\left |x+2 \right |+C$

$x-2ln\left |x+2 \right |+C$

Correct answer:

$x-2ln\left |x+2 \right |+C$

Explanation:

$\int \frac{x}{x+2}dx$

Add 2 and subtract 2 from the numerator of the integrand: $\int \frac{x}{x+2}dx=\int \frac{x+2-2}{x+2}dx$ .

Simplify and apply the difference rule: $\int \frac{x+2-2}{x+2}dx=\int (\frac{x+2}{x+2}-\frac{x}{x+2})dx=\int dx-\int \frac{2}{x+2}dx$

Solve the first integral: $\int dx=x+C$ .

Make the following substitution to solve the second integral: u=x+2 du=dx

Apply the substitution to the integral: $\int \frac{2}{x+2}dx=2\int\frac{du}{u}$

Solve the integral: $2\int\frac{du}{u}=2ln\left | x+2\right |+C$

Combine the answers to the two integrals: $\int dx-\int \frac{2}{x+2}dx=x-2ln\left | x+2\right |+C$ .

Solution: $\int \frac{x}{x+2}dx=x-2ln\left | x+2\right |+C$

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Example Question #672 : Finding Integrals

Evaluate the Integral:

$\int \frac{sin(x)}{2+cos(x)}dx$

Possible Answers:

ln(cos(x)+2)+c

ln(cos(x)-2)+c

ln(cos(x))+2+c

-ln(cos(x)-2)+c

-ln(2+cos(x))+c

Correct answer:

-ln(2+cos(x))+c

Explanation:

We use substitution to solve the problem:

Let u=2+cos(x) and du=-sin(x)dx $\Rightarrow sinxdx=-du$

Therefore:

$\int \frac{sin(x)}{2+cos(x)}dx=-\int \frac{1}{u}du=-ln(u)+c=-ln(2+cos(x))+c$

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Example Question #1021 : Integrals

Evaluate

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

Possible Answers:

tan(x)+e^x +c

$e^{cosx}+c$

$e^{sinx}+c$

$e^{sinx} +cos(x) +c$

$e^{cosx}-sin(x)+c$

Correct answer:

$e^{sinx}+c$

Explanation:

Here we use substitution to solve for the integrand. Let u=sin(x) therefore du= cos(x)dx. Plug your values back in:

$\int cos(x)e^{sinx}dx = \int e^udu = e^u +c =e^{sinx}+c$

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Example Question #95 : Solving Integrals By Substitution

$\int_{1}^{2}\frac{x}{x^2+1}dx$

Possible Answers:

0.3496

0.2466

0.3266

0.4581

1.3466

Correct answer:

0.4581

Explanation:

To integrate this expression, you have to use u substitution. First, assign your u expression:

u=x^2+1

du=2xdx

$\frac{1}{2}du=xdx$

Now, plug everything back in so you can integrate:

$\frac{1}{2}\int_{1}^{2}\frac{1}{u}du$

Now integrate:

$\frac{1}{2}\ln\left | u \right |$

From here substitute the original variable back into the expression.

$\frac{1}{2}\ln|x^2+1|$

Evaluate at 2 and then 1.

Subtract the results:

$\frac{1}{2}\left(\ln|2^2+1|-\ln|1+1| \right )$

$\frac{1}{2}\left(\ln|5|-\ln|2|\right)$

$\frac{\ln5-\ln2}{2}=0.4581$

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Example Question #1022 : Integrals

Calculate the following integral:

$y=\int \sqrt{4-4x^{2}}dx$

Possible Answers:

$y=sin^{-1}x-sin(2sin^{-1}x)+C$

$y=sin^{-1}x+x\sqrt{1-x^2}+C$

$y=2sin^{-1}x-sin(2sin^{-1}x)+C$

$y=2sin^{-1}x-\frac{1}{2}sin(2sin^{-1}x)+C$

Correct answer:

$y=sin^{-1}x+x\sqrt{1-x^2}+C$

Explanation:

$y=\int \sqrt{4-4x^{2}}dx$

Factor out $\sqrt{4}$ from the integrand, and simplify:

$y=\int \sqrt{4}\sqrt{1-x^{2}}dx=\int 2\sqrt{1-x^{2}}dx=2\int \sqrt{1-x^{2}}dx$

Make the following substitution: $x=sin\theta$ $dx=cos\theta d\theta$

Plug the substitution into the integrand:

$y=2\int \sqrt{1-sin^{2}\theta }cos\theta d\theta$ .

Use the Pythagorean identity to make the following substitution, and simplify:

$\\y=2\int \sqrt{1-sin^{2}\theta }cos\theta d\theta\\=2\int \sqrt{cos^{2}\theta }cos\theta d\theta\\=2\int cos\theta cos\theta d\theta\\= 2\int cos^{2}\theta d\theta$

Apply the following identity to the integrand:

$cos^{2}\theta=\frac{1+cos(2 \theta )}{2}$ :

$y=2\int cos^{2}\theta=2\int \frac{1+cos(2) \theta }{2}d\theta$ .

Separate the integral into two separate integrals:

$y= 2\int \frac{1}{2}d\theta+2\int\frac{cos(2\theta )}{2}d\theta$ .

Solve the first integral:

$y=\theta+2\int\frac{cos(2\theta )}{2}d\theta$ .

Make the following substitution for the second integral:

$u=2\theta$ $du=2d\theta$ $\frac{1}{2}du=d\theta$ .

Apply the substitution, and solve the integral:

$2\int \frac{cos(2\theta )}{2}=2\int \frac{cosu}{2}\frac{1}{2}du=\frac{1}{2}sinu=\frac{1}{2}sin(2\theta )+C$ .

Combine answers for both integrals:

$y=2\int cos^{2}\theta=\theta+\frac{1}{2}sin(2\theta )+C$

Solve for $\theta$ :

$x=sin\theta$

$sin^{-1}x=\theta$

Plug values for $\theta$ back into solution to integral:

Recall that,

$sin(2\theta)=2sin(\theta)cos(\theta)$

and from above,

$\\x=sin(\theta) \\cos(\theta)=\sqrt{1-sin^2(\theta)}=\sqrt{1-x^2}$

Therefore,

$\\\theta+\frac{1}{2}sin(2\theta )+C \\=sin^{-1}+4sin(x)cos(x) \\=sin^{-1}(x)+x\sqrt{1-x^2}+C$ .

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Calculus 2 : Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #671 : Finding Integrals

Example Question #1022 : Integrals

Example Question #673 : Finding Integrals

Example Question #1024 : Integrals

Example Question #2771 : Calculus Ii

Example Question #671 : Finding Integrals

Example Question #672 : Finding Integrals

Example Question #1021 : Integrals

Example Question #95 : Solving Integrals By Substitution

Example Question #1022 : Integrals

All Calculus 2 Resources

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