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Calculus 1 : Integral Expressions

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Example Questions

Example Question #71 : Integral Expressions

Find the indefinite integral.

$\int4x^2-e^{3x}+\sec (x)^2 dx$

Possible Answers:

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

$x^3-\frac{e^{3x}}{3}+\tan (x)+C$

None of these

$\frac{2}{3}x^3-\frac{e^{3x}}{3}+\tan (x)+C$

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

Correct answer:

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

Explanation:

For this integral you need to know an integration rule for each part of the problem.

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

Using this rule we can get the integral of the first part

$\int4x^2dx=\frac{4}{3}x^3$ .

For the second part we must know that

$\int e^xdx=e^x$

$\int-e^{3x}dx=-\frac{e^{3x}}{3}$ .

and finally the last part follows the rule exactly

$\int \sec (x)^2dx=\tan (x)$ .

We must also include a C for the constant of integration as the integral is indefinite.

Thus the final answer is

$\frac{4}{3}x^3-\frac{e^{3x}}{3}+\tan (x)+C$ .

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Example Question #2151 : Calculus

Give the integral expression for the velocity of an object falling from initial velocity V(0)=G towards the earth with acceleration $-32\frac{ft}{s^2}$ , between and seconds after it was released.

Possible Answers:

$V(t)=\int_{0}^{4}(-32+G)dt$

$V(t)=[\int_{0}^{4}(G)dt] -32$

$V(t)=[\int_{4}^{0}(-32)dt] +G$

$V(t)=[\int_{0}^{4}(-32)dt] +G$

Correct answer:

$V(t)=[\int_{0}^{4}(-32)dt] +G$

Explanation:

We know that the definite integral of acceleration between 0 and 4 seconds is the velocity. We know that the acceleration due to gravity is $-32\frac{ft}{s^2}$ . We also know that the inititial velocity is . This means that when t=0, V(t)=G . Therefore, the velocity function can be written as

$V(t)=[\int_{0}^{4}(-32)dt] +G$

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Example Question #71 : Writing Equations

Evaluate the following integral:

$\int \sin(x)e^{2cos(x)}dx$

Possible Answers:

$e^{2\cos(x)}+C$

$e^{\2sin(x)}+C$

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

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

Correct answer:

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

Explanation:

To integrate, we must make the following substitution:

$u=2\cos(x), du=-2\sin(x)dx$

We used the following rule to derivate:

$\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

Now, rewrite the integral and integrate:

$-\frac{1}{2}\int e^u du=-\frac{1}{2}e^u+C$

We used the following rule for the integration:

$\int e^xdx=e^xdx+C$

Now, replace u with the original x-containing term to finish:

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

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Example Question #71 : Integral Expressions

Evaluate the following integral:

$\int x^3\cos(x^4)dx$

Possible Answers:

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

$\frac{1}{4}\sin(x^4)$

$\sin(x^4)+C$

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

Correct answer:

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

Explanation:

To integrate, we must make the following subsitution:

u=x^4, du=4x^3 dx

We used the following rule to derivate:

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

Now, after rearranging, we get the following integral:

$\frac{1}{4}\int \cos(u) du$

and after integrating we get

$\frac{1}{4}\sin(u)+C$

We used the following rule to integrate:

$\int \cos(x)dx=\sin(x)+C$

To finish, we plug our x term back in place of u:

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

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Example Question #72 : Writing Equations

Evaluate the following integral:

$\int (x^4+2x^2+\sec^2(x))dx$

Possible Answers:

$x^5+\frac{2x^3}{3}+\tan(x)+C$

$\frac{x^5}{5}+\frac{2x^3}{3}-\tan(x)+C$

$\frac{x^5}{5}+\frac{x^3}{3}+\tan(x)+C$

$\frac{x^5}{5}+\frac{2x^3}{3}+\tan(x)+C$

Correct answer:

$\frac{x^5}{5}+\frac{2x^3}{3}+\tan(x)+C$

Explanation:

After integrating, we get

$\frac{x^5}{5}+\frac{2x^3}{3}+\tan(x)+C$

The integration was performed using the following rules:

$\int \sec^2(x)=\tan(x)+C$ , $\int x^ndx=\frac{1}{n+1}x^{n+1}+C$

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Example Question #2152 : Calculus

Evaluate the following integral:

$\int (4x^2\sin(x^3)+e^x)dx$

Possible Answers:

$\frac{4}{3}\cos(x^3)+e^x+C$

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

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

$\frac{4}{3}\sin(x^3)+e^x+C$

Correct answer:

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

Explanation:

In order to evaluate this integral, we must split the integrand into two seperate integrals:

$\int4x^2\sin(x^3)dx+\int e^xdx$

For the first integral, we need to make the following substitution:

u=x^3, du=3x^2dx

We found the derivative using the following rule:

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

Now, rewrite the integral in terms of u and solve:

$\frac{4}{3}\int \sin(u)du=-\frac{4}{3}\cos(u)+C$

We integrated using the following rule:

$\int \sin(x)=-\cos(x)+C$

For the second integral, we just integrate:

$\int e^xdx=e^x+C$

and we use the rule

$\int e^xdx=e^x+C$

Now, just add together the results of the two integrals (the integral of the sum is equal to the sum of the integrals):

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

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Example Question #73 : Writing Equations

Evaluate the following integral:

$\int \frac{1}{\sqrt{x+4}}e^{\sqrt{x+4}}dx$

Possible Answers:

$\ln(\sqrt{x+4})+C$

$2e^{\sqrt{x+4}}+C$

$-2e^{\sqrt{x+4}}+C$

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

Correct answer:

$2e^{\sqrt{x+4}}+C$

Explanation:

To integrate, we must make the following substitution:

$u=\sqrt{x+4}, du=\frac{1}{2}(x+4)^{-\frac{1}{2}}dx$

We found the derivative using the following rule:

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

Now, rewrite the integral in terms of u, and solve:

$2\int e^udu=2e^{u}+C$

We used the following integration rule:

$\int e^xdx=e^xdx+C$

Now, replace u with the original term containing x:

$2e^{\sqrt{x+4}}+C$

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Example Question #71 : Writing Equations

Evaluate the following integral:

$\int (x^2+x^3+2x)dx$

Possible Answers:

$\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^2}{2}+C$

$x^3+\frac{x^4}{4}+x^2+C$

$\frac{x^3}{3}+\frac{x^{4}}{4}+x^2+C$

$\frac{3x^3}{3}+\frac{x^4}{4}+x^2+C$

Correct answer:

$\frac{x^3}{3}+\frac{x^{4}}{4}+x^2+C$

Explanation:

When we integrate, we get

$\frac{x^3}{3}+\frac{x^{4}}{4}+x^2+C$

which was found using the rule

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

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Example Question #2155 : Calculus

Evaluate the following indefinite integral:

$\int(8x^{3}+3x^{2}+2x)dx$

Possible Answers:

$2x^{4}+3x^{3}+2x+C$

$2x^{5}+x^{4}+3x^{2}+C$

$2x^{4}+x^{3}+x^{2}+C$

$x^{4}+x^{3}+x^{2}+C$

Correct answer:

$2x^{4}+x^{3}+x^{2}+C$

Explanation:

To evaluate the integral the integral, use the inverse power rule:

$\int x^{n}dx=\frac{x^{n+1}}{n+1} n\neq1$

Applying that rule to this problem gives us the following for the first term:

$\int 8x^{3}=\frac{8x^{3+1}}{3+1}=\frac{8x^{4}}{4}=2x^{4}$

The following for the second term:

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

And the following for the third term:

$\int 2x=\frac{2x^{1+1}}{1+1}=\frac{2x^{2}}{2}=x^{2}$

We can combine these terms and add our "C" to get the final answer:

$2x^{4}+x^{3}+x^{2}+C$

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Example Question #72 : Writing Equations

Evaluate the indefinite integral:

$\int 2cos(x)dx$

Possible Answers:

cos(2x)+C

2sin(x)+C

$\frac{1}{2}sin(x)+C$

sin(2x)+C

Correct answer:

2sin(x)+C

Explanation:

To evaluate this integral, we have to remember the trig laws. When taking the integral of cosine, our answer is always sine. In this case we do the following:

Move the 2 outside of the integral:

$2\int cos(x)dx$

Evaluate the integral of cosine and add our "C"

2sin(x)+C

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Calculus 1 : Integral Expressions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #71 : Integral Expressions

Example Question #2151 : Calculus

Example Question #71 : Writing Equations

Example Question #71 : Integral Expressions

Example Question #72 : Writing Equations

Example Question #2152 : Calculus

Example Question #73 : Writing Equations

Example Question #71 : Writing Equations

Example Question #2155 : Calculus

Example Question #72 : Writing Equations

All Calculus 1 Resources

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