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Calculus 2 : Finding Integrals

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Example Questions

Example Question #581 : Finding Integrals

Evaluate:

$\int_{-2}^{2} 6^{x}\; \mathrm{d}x$

Possible Answers:

$72 \ln 6$

$\frac{1,295 \ln 6 }{36}$

$\frac{1,295}{36 \ln 6}$

$\frac{72}{\ln 6}$

Correct answer:

$\frac{1,295}{36 \ln 6}$

Explanation:

$6^{x} = (e^{\ln 6} )^{x} = e^{x \ln 6}$ , so the integral can be rewritten as $\int_{-2}^{2} e^{x \ln 6} \; \mathrm{d}x$ .

Substitute $u = x \ln 6$ .

Therefore, $\frac{\mathrm{d} u}{\mathrm{d} x} = \ln 6$ , $\mathrm{d} u = \ln 6 \cdot \mathrm{d} x$ , and $\frac{\mathrm{d} u }{\ln 6}= {d} x$ .

The bounds of integration become $u = 2 \ln 6 = \ln 36$ and $u = - 2 \ln 6 = \ln \frac{1}{36}$ .

$\int_{-2}^{2} e^{x \ln 6} \; \mathrm{d}x$

$= \int_{\ln \frac{1}{36}}^{\ln 36} e^{u} \;\cdot \frac{1}{\ln 6} \mathrm{d}u$

$= \frac{1}{\ln 6} \int_{\ln \frac{1}{36}}^{\ln 36} e^{u} \; \mathrm{d}u$

$= \left.\begin{matrix} \frac{1}{\ln 6} e^{u} \\ \textrm{ } \end{matrix}\right|$ $\begin{matrix} \ln 36 \\ \\\ln \frac{1}{36} \end{matrix}$

$= \frac{1}{\ln 6} e ^{\ln 36}- \frac{1}{\ln 6} e ^{\ln \frac{1}{36}}$

$= \frac{1}{\ln 6} \cdot 36 - \frac{1}{\ln 6} \cdot \frac{1}{36}$

$= \frac{1}{\ln 6} \cdot \frac{1, 296}{36} - \frac{1}{\ln 6} \cdot \frac{1}{36} = \frac{1}{\ln 6} \cdot \frac{1, 295}{36} = \frac{1,295}{36 \ln 6}$

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Example Question #581 : Finding Integrals

Evaluate:

$\int_{\frac{\pi}{6} }^{\frac{2\pi}{3}} \sin x \cos x \; \mathrm{d} x$

Possible Answers:

$\frac{1}{4}$

$\frac{\sqrt {3}-1}{8}$

$\frac{\sqrt {2}}{4}$

$\frac{\sqrt {3}}{4}$

Correct answer:

$\frac{1}{4}$

Explanation:

Substitute $u = \sin x$ . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = \cos x$ and $\mathrm{d} u = \cos x \; \mathrm{d} x$ .

The bounds of integration become

$u = \sin \frac{\pi }{6} = \frac{1}{2}$

and

$u = \sin \frac{2\pi }{3} = \frac{ \sqrt{3}}{2}$ .

The integral can be rewritten as

$\int_{\frac{\pi}{6} }^{\frac{2\pi}{3}} \sin x \cos x \; \mathrm{d} x$

$= \int_{\frac{1}{2} }^{\frac{ \sqrt{3}}{2}} u \; \mathrm{d}u$

$= \frac{u ^{2}}{2} \left|\begin{matrix}\frac{ \sqrt{3}}{2}\\ \\\frac{1}{2} \end{matrix}\right.$

$= \frac{\left $ \frac{ \sqrt{3} } {2} \right $ ^{2}}{2} - \frac{\left $ \frac{ 1 } {2} \right $ ^{2}}{2}$

$= \frac{3}{8} - \frac{1}{8} = \frac{1}{4}$

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Example Question #583 : Finding Integrals

Evaluate:

$\int_{-\ln 5}^{\ln 5} \sqrt{e ^{x} }\; \mathrm{d} x$

Possible Answers:

$\frac{8 \sqrt{5}}{5}$

$8 \sqrt{5}$

$4 \sqrt{5}$

$\frac{4 \sqrt{5}}{5}$

Correct answer:

$\frac{8 \sqrt{5}}{5}$

Explanation:

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

$\int_{-\ln 5}^{\ln 5} e^{\frac{1}{2}x}\; \mathrm{d} x$ .

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

Also, the bounds of integration become

$\frac{1}{2} \ln 5 = \ln \sqrt{5}$

and

$- \frac{1}{2} \ln 5 =- \ln \sqrt{5} = \ln \frac{1}{\sqrt{5}} = \ln \frac{\sqrt{5}}{5}$ .

The integral can be rewritten as

$\int_{ \ln \frac{\sqrt{5}}{5}}^{ \ln \sqrt{5}} e^{u}\; \cdot 2 \; \mathrm{d} u$

$= 2 \int_{ \ln \frac{\sqrt{5}}{5}}^{ \ln \sqrt{5}} e^{u} \mathrm{d} u$

$= \left.\begin{matrix} 2e^{u} \\ \textrm{ } \end{matrix}\right|\begin{matrix} \ln \sqrt{5}\\ \ln \frac{\sqrt{5}}{5} \end{matrix}$

$= 2e^{\ln \sqrt{5}} - 2e^{ \ln \frac{\sqrt{5}}{5}}$

$= 2 \sqrt{5} - \frac{2 \sqrt{5}}{5} = \frac{8 \sqrt{5}}{5}$

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Example Question #584 : Finding Integrals

Give the indefinite integral:

$\int \frac{\mathrm{d}x}{4x^{2}+9}$

Possible Answers:

$\frac{ 1}{6} \tan^{-1} \frac{2} { 3}x + C$

$\frac{ 1}{6} \sec^{-1} \frac{2} { 3}x + C$

$\frac{ 1}{6} \sec^{-1} \frac{4} {9}x + C$

$\frac{ 1}{6} \tan^{-1} \frac{4} {9}x + C$

$\frac{ 1}{6} \cot^{-1} \frac{2} { 3}x + C$

Correct answer:

$\frac{ 1}{6} \tan^{-1} \frac{2} { 3}x + C$

Explanation:

Substitute $x =\frac{ 3}{2} \tan \theta$ .

Then $2x = 3 \tan \theta$ ; also, $\frac{\mathrm{d} x}{\mathrm{d} \theta} =\frac{ 3}{2} \sec^{2} \theta$ and $\mathrm{d} x =\frac{ 3}{2} \sec^{2} \theta\; \mathrm{d} \theta$ .

We can rewrite this:

$\int \frac{\mathrm{d}x}{4x^{2}+9}$

$=\int \frac{\mathrm{d}x}{\left (2x \right )^{2}+9}$

$=\int \frac{\frac{ 3}{2} \sec^{2} \theta\; \mathrm{d} \theta}{\left (3 \tan \theta \right )^{2}+9}$

$=\frac{ 3}{2} \int \frac{ \sec^{2} \theta\; \mathrm{d} \theta}{ 9 \tan ^{2} \theta +9}$

$=\frac{ 3}{2} \cdot \frac{1}{9}\int \frac{ \sec^{2} \theta\; \mathrm{d} \theta}{ \tan ^{2} \theta +1}$

$=\frac{ 1}{6} \int \frac{ \sec^{2} \theta\; \mathrm{d} \theta}{ \sec^{2} \theta}$

$=\frac{ 1}{6} \int \mathrm{d} \theta$

$=\frac{ 1}{6} \left ( \theta +C \right ) =\frac{ 1}{6} \theta +\frac{ 1}{6} C =\frac{ 1}{6} \theta + C$

Note the absorption of $\frac{1}{6}$ into the constant.

Now, since $x =\frac{ 3}{2} \tan \theta$ ,

$\tan \theta =\frac{2} { 3}x$

and

$\theta =\tan^{-1} \frac{2} { 3}x$ .

Therefore the answer becomes

$\frac{ 1}{6} \theta + C = \frac{ 1}{6} \tan^{-1} \frac{2} { 3}x + C$ .

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Example Question #2682 : Calculus Ii

Determine the indefinite integral:

$\int \frac{4x+ 10}{x^2+4x +3} \; \mathrm{d}x$

Possible Answers:

$\ln\left ( x^4+10x^3+36x^2+56x+32 \right )+ C$

$\ln\left ( x^4+10x^3+36x^2+54x+27 \right )+ C$

$\ln\left ( x^4+14x^3+72x^2+160x+128 \right )+ C$

$\ln\left ( x^4+6x^3+12x^2+10x+3 \right )+ C$

$\ln\left ( 4x^3+26x^2+52x+30 \right )+ C$

Correct answer:

$\ln\left ( x^4+6x^3+12x^2+10x+3 \right )+ C$

Explanation:

The integrand can be rewritten using the method of partial fractions.

For some A,B ,

$\frac{4x+ 10}{x^2+4x +3} = \frac{4x+ 10}{(x+1)(x+3)} = \frac{A}{x+ 1}+ \frac{B}{x+3}$

$= \frac{A(x+3)}{(x+1)(x+3)}+ \frac{B(x+ 1)}{(x+1)(x+3)}$

4x+ 10 =A(x+3)+B(x+ 1) = Ax+ 3A + Bx +B = (A +B)x + (3A + B)

Therefore,

A + B = 4

3A + B = 10

It can be quickly seen that A = 3,B = 1 , and

$\frac{4x+ 10}{x^2+4x +3} = \frac{3}{x+ 1}+ \frac{1}{x+3}$ .

The integral can be rewritten as

$\int \frac{4x+ 10}{x^2+4x +3} \; \mathrm{d}x$

$=\int \left ( \frac{3}{x+ 1}+ \frac{1}{x+3} \right )\; \mathrm{d}x$

$=3 \int \frac{1}{x+ 1} \; \mathrm{d}x+\int \frac{1}{x+3} \; \mathrm{d}x$

Substitute $u = x+1, \widetilde{u} = x+ 3$ . Then $\mathrm{d}u = \mathrm{d}\widetilde{u} = \mathrm{d}x$ , and the expression becomes

$=3 \int \frac{\mathrm{d}u}{u} +\int \frac{\mathrm{d}\widetilde{u}}{\widetilde{u}}$

$=3 \ln u + C_{1} +\ln \widetilde{u} + C_{2}$

$=3 \ln (x+1) +\ln (x+3)+ C$

$= \ln\left [ (x+1)^3 (x+3) \right ]+ C$

$= \ln\left ( x^4+6x^3+12x^2+10x+3 \right )+ C$

Note the absorption of the two constants in the third-to-last step.

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Example Question #1 : Solving Integrals By Substitution

Solve the following integral.

$\int xsin(x^2)dx$

Possible Answers:

-cos(x^2)+c

$-\frac{1}{2}cos(x^2)+c$

$-\frac{1}{2}cos(u)+c$

$\frac{1}{2}cos(x^2)+c$

$-\frac{1}{2}sin(x^2)+c$

Correct answer:

$-\frac{1}{2}cos(x^2)+c$

Explanation:

To solve this problem we need to use u-substitution. The key to knowing that is by noticing that we have both an and an x^2 term, and that hypothetically if we could take the derivate of the x^2 term it could cancel out the term. Let's take a closer look.

Let's choose our in this problem to be x^2 . We would need to calculate the term since we're switching from the x variable to the u variable.

u=x^2

$\frac{du}{dx}=2x,dx=\frac{du}{2x}$

Let's take a look back at the orginal problem and begin to make substitutions.

$\int xsin(x^2)dx=\int xsin(u)\frac{du}{2x}$

Notice here that the x's will cancel out, leaving us with an integral with entirely the u variable.

$\frac12 \int sin(u)du$

This is a nice simple integral that results in

$-\frac12 cos(u)+c$ . Now all we need to do is replace that u with the original variable. u=x^2

$-\frac12 cos(x^2)+c$

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Example Question #1 : Solving Integrals By Substitution

Evaluate:

$\int \frac{2x-3}{2x^2-6x}dx$

Possible Answers:

$\frac{1}{2}ln\left | 2x^2-6x\right |+C$

$ln\left | x^2-3x\right |+C$

$\frac{1}{4}ln\left | 2x^2-6x\right |+C$

$\frac{1}{2}ln\left | x^2-3x\right |+C$

$\frac{1}{4}ln\left | x^2-3x\right |+C$

Correct answer:

$\frac{1}{2}ln\left | x^2-3x\right |+C$

Explanation:

The problem $\int \frac{2x-3}{2x^2-6x}dx$ is a U-substitution question. The term might not be easily seen, but the term must equal to $2x-3\: dx$ .

Factor the denominator by taking as the common factor.

2x^2-6x= 2(x^2-3x)

Rewrite the integral.

$\int \frac{2x-3}{2x^2-6x}dx =\int \frac{2x-3}{2(x^2-3x)}dx=\frac{1}{2}\int \frac{2x-3}{(x^2-3x)}dx$

u=x^2-3x

$du=2x-3\: dx$

$\frac{1}{2}\int\frac{1}{u}du=\frac{1}{2}ln\left | u\right |+C$

Resubstitute .

$\frac{1}{2}ln\left | x^2-3x\right |+C$

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Example Question #1 : Solving Integrals By Substitution

$z(x)=\int (8x+2)e^{2x^2+x+17}dx$

What is z(x)?

Possible Answers:

$(4x+1)e^{4x+1}+c$

$e^{2x^2+x+17}+c$

$2e^{2x^2+x+17}+c$

$e^{4x+1}+c$

$(8x+2)e^{2x^2+x+17}+c$

Correct answer:

$2e^{2x^2+x+17}+c$

Explanation:

Although the integral may look tough, we see a possible relationship between the polynomial of the exponential and the polynomial in front of the exponential.

Let's make our

u=2x^2 + x+ 17

$\frac{du}{dx}=4x+1$

$dx=\frac{du}{4x+1}$

Now let's see the original integral to make the substitutions.

$\int (8x+2)e^{2x^2+x+17}dx=\int (8x+2)e^{u}\frac{du}{4x+1}$ .

At first sight it may seem that nothing cancels out, but at a closer look we can see that the (8x+2) term can actually be simplified to 2(4x+1) , which can now cancel with the denominator.

$\int2e^udu$ is our new integral, which simply leads to

2e^u+c . Now substitute u for what we had earlier.

$2e^{2x^2+x+17}+c$

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Example Question #1 : Solving Integrals By Substitution

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

Possible Answers:

$\sqrt{x+2}\left[\frac{2}{3}x-\frac{8}{3}\right]+C$

xln|x+2|+C

$\sqrt{x+2}[ln|x+2|]+C$

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

$\sqrt{x+2}\left[\frac{1}{2}x-\frac{4}{3}\right]+C$

Correct answer:

$\sqrt{x+2}\left[\frac{2}{3}x-\frac{8}{3}\right]+C$

Explanation:

To evaluate $\int \frac{x}{\sqrt{x+2}}\: dx$ , use U-substitution.

Let u=x+2 , which also means x=u-2 . Take the derivative and find .

du=dx

Rewrite the integral in terms of and , and separate into two integrals.

$\int \frac{x}{\sqrt{x+2}}\: dx= \int \frac{u-2}{\sqrt u}\:du=\int u^\frac{1}{2} \:du-2\int u^{-\frac{1}{2}}\:du$

Evaluate the two integrals.

$\frac{2}{3}u^{\frac{3}{2}}-4u^{\frac{1}{2}}+C$

Re-substitute .

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

Pull out the common factor $(x+2)^{\frac{1}{2}}$ .

$=(x+2)^{\frac{1}{2}}\left[\frac{2}{3}(x+2)-4\right]+C$

$=(x+2)^{\frac{1}{2}}\left[\frac{2}{3}x+\frac{4}{3}-\frac{12}{3}\right]+C$

$=\sqrt{x+2}\left[\frac{2}{3}x-\frac{8}{3}\right]+C$

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Example Question #2 : Solving Integrals By Substitution

Solve the following integral:

$\small \int_{0}^{\pi/2 } \frac{2sin\theta}{\sqrt{cos\theta }}d\theta$ .

Possible Answers:

$\small 4$

$\small 1$

$\small 3$

$\small 2$

$\small 0$

Correct answer:

$\small 4$

Explanation:

In order to solve this, we must use $\small u$ -substitution.

Because $\small \frac{\mathrm{d} }{\mathrm{d} \theta }(cos\theta )=-sin\theta$ , we should let $\small u=cos\theta$ so the $\small sin\theta$ can cancel out.

We can now change our integral to $\small \small \small \int_{0}^{\pi /2} \frac{2sin\theta }{\sqrt{u}}d\theta$ .

We know that $\small u=cos\theta$ , so $\small \small \frac{\mathrm{d} u}{\mathrm{d} \theta }=-sin\theta$ , which means $\small d\theta=\frac{du}{-sin\theta }$ .

We can substitue that in for $\small d\theta$ in the integral to get $\small \small \small \small \int \frac{2sin\theta }{\sqrt{u}}\frac{du}{-sin\theta }$ .

The $\small sin\theta$ can cancel to get $\small \small \small \small \small \int \frac{-2 }{\sqrt{u}}du$ .

The limits of the integral have been left off because the integral is now with respect to $\small u$ , so the limits have changed. This integral can now be solved using the power rule $\small \small \small \int au^{n}du=\frac{a}{n+1}u^{n+1}+c$

which will give you $\small \small \small -4\sqrt{u}+c$ .

Now we can substitute $\small cos\theta$ back in for $\small u$ to get $\small -4\sqrt{cos\theta }+c$ . We can bring back our limits now and evaluate because we're in terms of $\small \theta$ again, which is what our original limits were with respect to.

$\small \small \small -4\sqrt{cos(\pi/2) }+c\left-\small \small (-4\sqrt{cos(0) }+c\left)=c-(-4+c)=c+4-c=4$

TIP: whenever doing $\small u$ -substitution, don't use the limits until you're in terms of the original variable.

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Calculus 2 : Finding Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #581 : Finding Integrals

Example Question #581 : Finding Integrals

Example Question #583 : Finding Integrals

Example Question #584 : Finding Integrals

Example Question #2682 : Calculus Ii

Example Question #1 : Solving Integrals By Substitution

Example Question #1 : Solving Integrals By Substitution

Example Question #1 : Solving Integrals By Substitution

Example Question #1 : Solving Integrals By Substitution

Example Question #2 : Solving Integrals By Substitution

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