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Calculus 3 : Calculus Review

Study concepts, example questions & explanations for Calculus 3

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

Example Questions

Example Question #312 : Calculus 3

Calculate $\int_1^{e^3} \frac{4}{x}dx$

Possible Answers:

4e^3

Correct answer:

Explanation:

The solution will be obtained by basic definitions of integrals, and properties of the natural logarithm, namely the fact that $\ln x^a= a \ln x$ and $\ln 1 = 0$ .

$\int_1^{e^3} \frac{4}{x}dx = 4 \int_1^{e^3} \frac{dx}{x}= 4 \left( \ln e^3 \right) = 4*3 \ln e= 12$ .

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Example Question #321 : Calculus 3

Find $\int 8x e^{x^2}dx$

Possible Answers:

$8x^2e^{x^3}+C$

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

Can not be determined

$8e^{x^2}+C$

Correct answer:

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

Explanation:

This integral can most easily be done by u-substitution. Begin by rewriting the integral as

$\int 8xe^{x^2}dx= 4\int 2xe^{x^2}dx$ .

Choose $u = x^2 \Rightarrow du = 2xdx$ , therefore

$\int 8x e^{x^2}dx= 4 \int 2x e^{x^2}dx = 4 \int e^udu= 4e^{x^2}+C$ .

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Example Question #322 : Calculus 3

Calculate $\int _0^{\pi/2} \sin \theta e^{2 \cos \theta} d \theta$

Possible Answers:

$\frac{e^2-1}{2}$

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

1-e^2

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

Correct answer:

$\frac{e^2-1}{2}$

Explanation:

This integral can be evaluated by using u-substitution. Choose

$u = 2 \cos \theta\Rightarrow du = -2 \sin \theta d \theta \Rightarrow -\frac{du}{2}= \sin \theta d \theta$ . This means

$\int _0^{\pi/2} \sin \theta e^{2 \cos \theta} d \theta= -\frac{1}{2}\int _{u_i}^{u_f}due^u= -\frac{1}{2}\left(e^0-e^2 \right )=\frac{e^2-1}{2}$ .

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Example Question #24 : Integration

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

Possible Answers:

$\frac{3x^6}{2}+ \frac{5x^5}{4}+\frac{x^3}{3}+C$

$\frac{9x^7}{7}+ \frac{6x^5}{5}+\frac{x^3}{3}+C$

$\frac{(3x^3+x)^3}{3}$

$\textup{Cannot be integrated.}$

$\frac{x^4}{12}+\frac{x^2}{6}+C$

Correct answer:

$\frac{9x^7}{7}+ \frac{6x^5}{5}+\frac{x^3}{3}+C$

Explanation:

In order to integrate, we will need to expand the binomial. Substitution will not work since there is no valid term to replace if we let u=3x^3+x .

Expand the binomial.

(3x^3+x)^2 = (3x^3+x)(3x^3+x)

(3x^3)(3x^3)+(3x^3)(x)+(x)(3x^3)+ x^2

= 9x^6+6x^4+x^2

The integral becomes:

$\int (3x^3+x)^2 \:dx = \int (9x^6+6x^4+x^2)\:dx$

Integrate each term.

The answer is: $\frac{9x^7}{7}+ \frac{6x^5}{5}+\frac{x^3}{3}+C$

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Example Question #324 : Calculus 3

Integrate: $\int \frac{x+3}{\sqrt{x-3}}\:dx$

Possible Answers:

$\frac{2}{3}\sqrt{x-3}\:[x+15]+C$

$\frac{3}{2}\sqrt{x+3}\:[x-15]+C$

$\frac{3}{2}\sqrt{x-3}\:[x+15]+C$

$\textup{The answer is not given.}$

$\frac{3}{2}ln{\sqrt{x-3}}+C$

Correct answer:

$\frac{2}{3}\sqrt{x-3}\:[x+15]+C$

Explanation:

Use substitution to solve this question.

Let u=x-3 , and by adding three on both sides, and we have x=u+3 .

Take the derivative of u with the respect to x.

du=dx

With the three equations, we can substitute the integral in terms of u.

$\int \frac{x+3}{\sqrt{x-3}}\:dx = \int \frac{(u+3)+3}{\sqrt{u}}\:du = \int \frac{u+6}{\sqrt{u}}\:du$

Rewrite the denominator as $u^{\frac{1}{2}}$ , and separate the integral into two integrals.

$\int \frac{u+6}{\sqrt{u}}\:du= \int \frac{u+6}{u^{\frac{1}{2}}}\:du=\int u^{\frac{1}{2}}\:du +6\int u^{-\frac{1}{2}}\:du$

Integrate the terms separately.

$\int u^{\frac{1}{2}}\:du = \frac{2}{3}u^{\frac{3}{2}}+C_1$

$6\int u^{-\frac{1}{2}}\:du = 6[2u^{\frac{1}{2}}] = 12u^{\frac{1}{2}}+C_2$

Add the two integrals. The constants can be combined as a single constant at the end.

$\frac{2}{3}u^{\frac{3}{2}}+C_1 +12u^{\frac{1}{2}}+C_2 = \frac{2}{3}u^{\frac{3}{2}}+12u^{\frac{1}{2}}+C_2$

We can also pull out a $\frac{2}{3}u^{\frac{1}{2}}$ as a common factor.

$\frac{2}{3}u^{\frac{1}{2}}\:[u+18]+C$

Substitute u=x-3 back into the answer.

$\frac{2}{3}(x-3)^{\frac{1}{2}}\:[(x-3)+18]+C$

Simplify the terms in the bracket.

$\frac{2}{3}(x-3)^{\frac{1}{2}}\:[x+15]+C$

The fractional exponent of one-half is a square root of the quantity.

The answer is: $\frac{2}{3}\sqrt{x-3}\:[x+15]+C$

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Example Question #135 : Calculus Review

Evaluate: $\int_{1}^{4} x^6+4x^3-x^2+7x+3$

Possible Answers:

$\frac {36903}{74}$

$-\frac {36903}{14}$

$\frac {36903}{14}$

2635

Correct answer:

$\frac {36903}{14}$

Explanation:

Step 1: Find the integration of each term:

$x^6\rightarrow \frac {x^7}{7}$
$4x^3\rightarrow x^4$
$-x^2\rightarrow\frac {-x^3}{3}$
$7x\rightarrow \frac {7x^2}{2}$
$3\rightarrow 3x$

Step 2: Re-write the function in terms of the integrated terms:

$\frac {x^7}{7}+x^4-\frac {x^3}{3}+\frac {7x^2}{2}+3x$

Step 3: Find the upper limit:

$\frac {4^7}{7}+4^4-\frac {4^3}{3}+\frac {7(4)^2}{2}+3(4)$

$=\frac {16384}{7}+256-\frac {64}{3}+56+12=2643.238$

Step 4: Find the Lower Limit:

$\frac {1^7}{7}+1^4-\frac {1^3}{3}+\frac {7(1)^2}{2}+3(1)$

$=\frac {1}{7}+1-\frac {1}{3}+\frac {7}{2}+3=7.309$
Step 5: Subtract the upper limit and the lower limit:

$2643.238-7.309=2635.9\approx \frac {36903}{14}$

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Example Question #323 : Calculus 3

Integrate

$\int{\ln x^3}dx$

Possible Answers:

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

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

$\ln \left(\frac{x^3}{4}\right)+C$

$3(x\ln x-x)+C$

Correct answer:

$3(x\ln x-x)+C$

Explanation:

We will be using integration by parts for this problem

Untitled

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Example Question #137 : Calculus Review

Evaluate the integral:

$\int \cos^2(2x)dx$

Possible Answers:

$\frac{1}{8}(u+\sin(u))+C$

$\frac{1}{8}(4x+\sin(4x))+C$

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

$\frac{1}{8}(4x-\sin(4x))+C$

Correct answer:

$\frac{1}{8}(4x+\sin(4x))+C$

Explanation:

To integrate, we can change the integrand using the double angle formula to something we can integrate:

$\int\frac{(1+\cos(4x))}{2}dx$

Next, we make a substitution:

u=4x, du=4dx

The derivative was found using the rule

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Now, we rewrite the integral in terms of u, and integrate:

$\frac{1}{8}\int(1+\cos(u))du=\frac{1}{8}(u+\sin(u))+C$

We used the following integration rules:

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

Finally, replace u with our original x term:

$\frac{1}{8}(4x+\sin(4x))+C$

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Example Question #138 : Calculus Review

Integrate:

$\int 5x\cos(4x)dx$

Possible Answers:

$\frac{5x\sin(4x)}{4}+\frac{5}{8}\cos(4x)+C$

$\frac{5x\sin(4x)}{4}+\frac{5}{8}\cos(4x)$

$\frac{5x\sin(4x)}{4}-\frac{5}{8}\cos(4x)+C$

$-\frac{5x\sin(4x)}{4}+\frac{5}{8}\cos(4x)+C$

Correct answer:

$\frac{5x\sin(4x)}{4}+\frac{5}{8}\cos(4x)+C$

Explanation:

To integrate, we must integrate by parts, using the formula

$\int udv=uv-\int vdu$

We now designate our u and dv, and differentiate and integrate, respectively:

u=5x, du=5dx

$dv=\cos(4x)dx, v=\frac{\sin(4x)}{4}$

The rules used are as follows:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\int \cos(x)dx=\sin(x)+C$

Next, we use the above formula:

$\frac{5x\sin(4x)}{4}-\frac{5}{4}\int \sin(4x)dx$

Finally, integrate:

$\frac{5x\sin(4x)}{4}+\frac{5}{8}\cos(4x)+C$

The following rule was used to integrate:

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

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Example Question #139 : Calculus Review

Evaluate the following integral:

$\int\sin^3 x\cot^2 x dx$

Possible Answers:

$-\frac{1}{3}\cos^3 x+C$

$\frac{1}{3}\cos^3 x+C$

$-\cos x+C$

$-\frac{1}{3}\cos^3 x$

Correct answer:

$-\frac{1}{3}\cos^3 x+C$

Explanation:

The integrand can be simplified to make the problem much easier to solve. You must use the fact that $\cot^2 x=\frac{\cos^2 x}{\sin^2 x}$ in order to rewrite the integrand. After rewriting, the problem is now:

$\int\sin x\cos^2 x dx$

Here, a simple u-substitution can be used:

$u=\cos x$ , $du=-\sin x dx$ , $dx=-\frac{du}{\sin x}$

Once again, the integrand can be rewritten with values of u:

$\int\sin x\cdot u^2\cdot -\frac{du}{\sin x}=-\int u^2du=-\frac{1}{3}u^3 +C$

The integral was taken by using the following rule:

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

Once u has been replaced with x, the final answer is:

$-\frac{1}{3}\cos^3 x+C$

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Calculus 3 : Calculus Review

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #312 : Calculus 3

Example Question #321 : Calculus 3

Example Question #322 : Calculus 3

Example Question #24 : Integration

Example Question #324 : Calculus 3

Example Question #135 : Calculus Review

Example Question #323 : Calculus 3

Example Question #137 : Calculus Review

Example Question #138 : Calculus Review

Example Question #139 : Calculus Review

All Calculus 3 Resources

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