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Calculus 2 : Calculus II

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #174 : Indefinite Integrals

Evaluate.

$\int 3 \ dx$

Possible Answers:

F(x) = C

$F(x) = \frac{1}{3}x+ C$

F(x) = 3x^ 2+ C

F(x) = 3x+ C

Answer not listed

Correct answer:

F(x) = 3x+ C

Explanation:

$\int f(x) dx$

In order to evaluate this integral, first find the antiderivative of f(x).

If f(x) = a then F(x) = ax + C

If f(x) = x a then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then F(x) = ln(x) + C

If f(x) = e ^x then F(x) = e^x + C

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, f(x) = 3 .

The antiderivative is F(x) = 3x+ C .

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Example Question #801 : Integrals

Evaluate $\int \frac{3x}{1+x^4}dx$

Possible Answers:

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

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

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

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

None of the other answers

Correct answer:

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

Explanation:

This integral can be evaluated using -substitution.

u = x^2

du=2x dx

Then we can proceed as follows

$\int \frac{3x}{1+x^4}dx$ (Start)

$= \frac{3}{2}\int \frac{2x}{1+x^4}dx$ (Factor out a 3/2 , leaving a in the numerator)

$=\frac{3}{2}\int \frac{1}{1+u^2}du$ (Substitute the equations for u, du )

$=\frac{3}{2}\tan^{-1}(u)+C$ (Integrate, recall that $\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}$ )

$=\frac{3}{2}\tan^{-1}(x^2)+C$ (Substitute x^2 back in)

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Example Question #176 : Indefinite Integrals

$\int (e^x+4)dx$

Possible Answers:

xe^x+4+C

e^x

e^x+4x+C

e^x+C

e^x+4+C

Correct answer:

e^x+4x+C

Explanation:

The anti-derivative of e^x is simply e^x . This is the perfect function that no matter if you take the derivative or the anti-derivative, it always returns back the original function. The integral of the constant returns the constant multiplied by the integration variable. Then, since this is an indefinite integral, we must include the integration constant,

$\int(e^x+4)dx= e^x+4x+C.$

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Example Question #177 : Indefinite Integrals

$\int cos(3x)dx$

Possible Answers:

sin(3x) +C

3sin(3x)+C

$\frac{-sin(3x)}{3} +C$

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

-sin(3x) +C

Correct answer:

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

Explanation:

The indefinite integral is a reverse chain rule. Remember, anti-derivatives are the exact opposite of the derivative of this function. So, let's start there:

cos(3x)'=-3sin(3x).

Therefore, to undo this answer, we would have to get rid of the negative sign and divide by the chain rule part. We also add our integration constant. You would get the same result (but longer time) if you used u-substitution.

$\int cos(3x)dx= \frac{sin(3x)}{3}+C$ .

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Example Question #178 : Indefinite Integrals

Evaluate the following indefinite integral:

$\int (x^4-2x^2+6)dx$

Possible Answers:

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

$\frac{x^3}{4}-\frac{x}{4}+\frac{1}{6}+C$

4x^3+4x+C

$\frac{x^6}{30}-\frac{x^4}{6}+3x^2+C$

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

Correct answer:

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

Explanation:

This integral requires use of the power rule for antiderivatives, which simplifies as follows:

$\int (x^4-2x^2+6)dx=\int x^4dx-2\int x^2 dx+6\int dx$

$=(\frac{1}{5}\cdot x^5)-2 (\frac{1}{3}\cdot x^3)+6(x)+C=\frac{x^5}{5}-\frac{2x^3}{3}+6x+C$

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Example Question #179 : Indefinite Integrals

Evaluate the following indefinite integral:

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

Possible Answers:

$3\csc(3x^3)\cot({3x^3})+C$

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

$9\csc( 3x^2)+C$

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

$\frac{\tan(3x^2)}{9}+C$

Correct answer:

$\frac{\tan(3x^2)}{9}+C$

Explanation:

Make the substitution:

u=3x^3

where

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

Substituting this into the original expression:

$\int x^2\sec^2(3x^3)dx=\int\frac{\sec^2(u)}{9}du=\frac{\tan(u)}{9}+C=\frac{\tan(3x^3)}{9}+C$

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Example Question #180 : Indefinite Integrals

Evaluate the following indefinite integral:

$\int x\ln(x)dx$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

To solve this with integration by parts, we rewrite the expression in the form

$\int u \: dv$

where

$u=\ln(x), \: \: \:dv=x\, dx$

and

$du=\frac{1}{x}dx,\: \: \:v=\frac{x^2}{2}$

To integrate, apply the formula for integration by parts:

$\int u \: dv=uv-\int vdu=\frac{1}{2}x^2\ln(x)-\int (\frac{x^2}{2}\cdot \frac{1}{x})dx$

$=\frac{x^2}{2}\ln(x)-\frac{1}{2}\int xdx=\frac{x^2\ln(x)}{2}-\frac{x^2}{4}+C=\frac{x^2}{4}(2\ln(x)-1)+C$

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Example Question #802 : Integrals

Evaluate the following indefinite integral:

$\int\sqrt{9-x^2}\, dx$

Possible Answers:

$\frac{\sqrt{9-x^2}}{2}+\frac{9}{2}\sin^{-1}(x)+C$

$\frac{x\sqrt{9-x^2}}{2}+\frac{9}{2}\cos^{-1}(\frac{x}{3})+C$

$\frac{x\sqrt{9-x^2}}{2}+\frac{9}{2}\sin^{-1}(\frac{x}{3})+C$

$\frac{\sqrt{9-x^2}}{2}+\frac{9}{2}\cos^{-1}(x)+C$

$\frac{x\sqrt{9-x^2}}{2}+\frac{9}{2}\cos^{-1}(x)+C$

Correct answer:

$\frac{x\sqrt{9-x^2}}{2}+\frac{9}{2}\sin^{-1}(\frac{x}{3})+C$

Explanation:

Make the trigonometric substitution:

$x=3\sin(a),\: \: \:dx=3\cos(a)da$

Applying this to the expression given:

$\int\sqrt{9-x^2}\, dx=\int3\cos(a)\sqrt{9(1-\sin^2a)}\, da=\int9\cos^2(a)da$

$=9[\frac{1}{2}\cos(a)\sin(a)+\frac{a}{2}=\frac{x\sqrt{9-x^2}}{2}+\frac{9}{2}\sin^{-1}(\frac{x}{3})+C$

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Example Question #181 : Indefinite Integrals

Evaluate the following indefinite integral:

$\int\frac{1}{x^2\sqrt{x^2-25}}\,dx$

Possible Answers:

$\frac{\sqrt{x^2-25}}{5x}+C$

$\frac{x^2-25}{5x}+C$

$\frac{\sqrt{x^2-25}}{25x}+C$

$\frac{\sqrt{x^2-25}}{25}+C$

$\frac{\sqrt{x^2-25}}{5}+C$

Correct answer:

$\frac{\sqrt{x^2-25}}{25x}+C$

Explanation:

Make the trigonometric substitution:

$x=5\sec(a),\,\,\,dx=5\tan(a)\sec(a)da$

Applying this substitution to the expression given, we can integrate:

$\int\frac{1}{x^2\sqrt{x^2-25}}\,dx=\int\frac{5\tan(a)\sec(a)}{25\sec^2(a) \sqrt{25\sec^2(a)-25}} da$

$=\int\frac{5\tan(a)}{125\sec(a)\tan(a)}da=\int\frac{da}{25\sec(a)}=\int\frac{\cos(a)}{25}da$

$=\frac{sin(a)}{25}+C=\frac{\sqrt{x^2-25}}{25x}+C$

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Example Question #461 : Finding Integrals

Evaluate the following indefinite integral:

$\int \frac{2x-14}{x^2-2x-8}\,dx$

Possible Answers:

$3\ln(x+2)-\ln(x-4)+C$

$3\ln(x+2)+3\ln(x-4)+C$

$\frac{x^2-14x}{\frac{1}{3}x^3-x^2-8x}+C$

$-\ln(x+2)-3\ln(x-4)+C$

$5\ln(x+2)+3\ln(x-4)+C$

Correct answer:

$3\ln(x+2)-\ln(x-4)+C$

Explanation:

This integral can be solved by using partial fractions. First, we have to factor the denominator write the fraction as a sum of two fractions:

$\int \frac{2x-14}{x^2-2x-8}\,dx=\int \frac{2x-14}{(x+2)(x-4)}\,dx=\int(\frac{A}{x+2}+\frac{B}{x-4})dx$

Next, we can solve for A and B:

2x-14=A(x-4)+B(x+2)

When we let x=4:

$8-14=0+6B\rightarrow B=-1$

When x=-2:

$-4-14=-6A\rightarrow A=-1$

Replacing A and B in the integral, we can now solve it:

$\int(\frac{3}{x+2}-\frac{1}{x-4})dx=3\ln(x+2)-\ln(x-4)+C$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #174 : Indefinite Integrals

Example Question #801 : Integrals

Example Question #176 : Indefinite Integrals

Example Question #177 : Indefinite Integrals

Example Question #178 : Indefinite Integrals

Example Question #179 : Indefinite Integrals

Example Question #180 : Indefinite Integrals

Example Question #802 : Integrals

Example Question #181 : Indefinite Integrals

Example Question #461 : Finding Integrals

All Calculus 2 Resources

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