Calculus 2 : Indefinite Integrals

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #161 : Indefinite Integrals

Solve the indefinite integral.

\displaystyle \int sec^2x\,dx

Possible Answers:

\displaystyle cot\,x+c

\displaystyle cos\,x+c

\displaystyle tan\,x+c

\displaystyle sin\,x+c

Correct answer:

\displaystyle tan\,x+c

Explanation:

The antiderivative of \displaystyle sec^2x is \displaystyle tan\,x.

As such,
\displaystyle \int sec^2x\,dx=tan\,x+c

Example Question #2533 : Calculus Ii

\displaystyle \int x^3+2x

Possible Answers:

\displaystyle \frac {x^4}{4}+x^2+C

\displaystyle \frac {x^4}{4}-x^2

\displaystyle \frac {x^4}{4}+x^3+C

None of the Above

Correct answer:

\displaystyle \frac {x^4}{4}+x^2+C

Explanation:

Step 1: We will first take the antiderivative of \displaystyle x^3.. To do this, we add \displaystyle 1 to the exponent. We then divide the entire term by the new exponent..

So, \displaystyle x^3=\frac {x^{3+1}}{3+1}=\frac {x^4}{4}

Step 2: We will take the antiderivative of \displaystyle 2x. We will apply the same rules that we used in Step 1...

\displaystyle 2x=2 \cdot \frac {x^{1+1}}{1+1}=2 \cdot \frac {x^2}{2}=x^2

We will add the antiderivatives from both steps together..

(Recall that when taking the indefinite integral of a function, +C must be added to the integrated function as it represents any constants that may be present.)

The final answer is: 

\displaystyle \frac {x^4}{4}+x^2+C

Example Question #162 : Indefinite Integrals

\displaystyle \int \frac{3x+1}{x}dx

Possible Answers:

\displaystyle 3x+4ln\left | x \right |+C

\displaystyle 3x+ln\left | x \right |

\displaystyle 3x+ln\left | 3x \right |+C

\displaystyle ln\left | x \right |+C

\displaystyle 3x+ln\left | x \right |+C

Correct answer:

\displaystyle 3x+ln\left | x \right |+C

Explanation:

First, chop up the fraction into two separate terms:
\displaystyle \int 3+\frac{1}{x}dx

Now integrate:

\displaystyle 3x+ln\left | x \right |

Add a C because it is an indefinite integral:

\displaystyle 3x+ln\left | x \right |+C

Example Question #442 : Finding Integrals

\displaystyle \int \frac{5x^2+20x-10}{5}dx

Possible Answers:

\displaystyle \frac{x^3}{3}-x^2-2x+C

\displaystyle x^3+2x^2-2x+C

\displaystyle \frac{x^3}{3}+2x^2-2x+C

\displaystyle \frac{x^3}{3}+2x^2-x+C

\displaystyle \frac{x^3}{3}+2x^2-2x

Correct answer:

\displaystyle \frac{x^3}{3}+2x^2-2x+C

Explanation:

First, chop up the fraction into three simplified terms:
\displaystyle \int x^2+4x-2dx

Now integrate.

\displaystyle \frac{x^3}{3}+\frac{4x^2}{2}-2x=\frac{x^3}{3}+2x^2-2x

Now add a C because it is an indefinite integral:

\displaystyle \frac{x^3}{3}+2x^2-2x+C

Example Question #2542 : Calculus Ii

Calculate the following integral: \displaystyle \int cos^{2}x+x\sqrt{x^{2}+1}dx

Possible Answers:

\displaystyle \frac{1}{2}x+\frac{1}{4}sin(2x)+(\sqrt{x^{2}+1})^{3}+C

\displaystyle \frac{1}{4}sin(2x)+\frac{1}{3}(\sqrt{x^{2}+1})^{3}+C

\displaystyle \frac{1}{2}x+\frac{1}{4}sin(x)+\frac{1}{3}(\sqrt{x^{2}+1})^{3}+C

\displaystyle \frac{1}{2}x+\frac{1}{4}sin(2x)+\frac{1}{3}(\sqrt{x^{2}+1})^{3}+C

Correct answer:

\displaystyle \frac{1}{2}x+\frac{1}{4}sin(2x)+\frac{1}{3}(\sqrt{x^{2}+1})^{3}+C

Explanation:

Separate integral into two separate integrals:

\displaystyle \int cos^{2}x+x\sqrt{x^{2}+1}dx=\int cos^{2}xdx+\int x\sqrt{x^{2}+1}dx  

Solve \displaystyle \int cos^{2}xdx first. 

Use power-reducing identity to simplify integral: \displaystyle \int \frac{1+cos(2x)}{2}dx.

Factor  \displaystyle \frac{1}{2} out of the integral: \displaystyle \int \frac{1}{2}+\frac{cosx}{2}dx

Separate into two integrals: \displaystyle \frac{1}{2}\int 1dx+\frac{1}{2}\int cos (2x)dx=\frac{1}{2}x+\frac{1}{2}\int cos (2x)dx.

Use the following substitution for the second integral: \displaystyle u=2x \displaystyle du=2 \displaystyle \frac{1}{2}du=dx

Plug in substitution and solve:  \displaystyle \frac{1}{4}\int cosudu=\frac{1}{4}sinu+C=\frac{1}{4}sin(2x)+C  \displaystyle \frac{1}{4}sinu=\frac{1}{4}sin(2x)+C.

Therefore: \displaystyle \int cos^{2}x=\frac{1}{2}x+\frac{1}{4}sin(2x)+C

Solve \displaystyle \int x\sqrt{x^{2}+1}dx

Make the following substitution: \displaystyle u=x^{2}+1  \displaystyle du=2xdx   \displaystyle \frac{1}{2}du=dx. Plug in substitution and solve: \displaystyle \frac{1}{2}\int \sqrt{u}du=(\frac{2}{3})\frac{1}{2}(x^{2}+1)^{\frac{3}{2}}+C=\frac{1}{3}(x^{2}+1)^{\frac{3}{2}}+C

Combine answers to two original integrals: \displaystyle \int cos^{2}xdx+\int x\sqrt{x^{2}+1}dx=\frac{1}{2}x+\frac{1}{4}sin(2x)+\frac{1}{3}(x^{2}+1)^{\frac{3}{2}}+C

\displaystyle \int cos^{2}x+x\sqrt{x^{2}+1}dx=\frac{1}{2}x+\frac{1}{4}sin(2x)+\frac{1}{3}(x^{2}+1)^{\frac{3}{2}}+C

Example Question #791 : Integrals

Evaluate.

\displaystyle \int 44x^{43} \ dx

Possible Answers:

\displaystyle F(x) = 43x^{44 }+ C

\displaystyle F(x) = x^{43}+ C

\displaystyle F(x) = x^{44 }+ C

Answer not listed

\displaystyle F(x) = \frac{1}{44}x^{44 }+ C

Correct answer:

\displaystyle F(x) = x^{44 }+ C

Explanation:

\displaystyle \int f(x) dx

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

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

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

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

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

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

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

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

 

In this case, \displaystyle f(x) = 44x^{43}.

The antiderivative is  \displaystyle F(x) = x^{44 }+ C.

Example Question #792 : Integrals

Evaluate.

\displaystyle \int \frac{1}{4x-19} \ dx

Possible Answers:

\displaystyle F(x) = \frac{4x-19}{4} + C

\displaystyle F(x) = 4x-19 + C

\displaystyle F(x) = \frac{4}{4x-19} + C

\displaystyle F(x) = \frac{x-4}{4x-19} + C

Answer not listed.

Correct answer:

\displaystyle F(x) = \frac{4x-19}{4} + C

Explanation:

\displaystyle \int f(x) dx

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

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

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

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

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

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

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

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

 

In this case, \displaystyle f(x) = \frac{1}{4x-19}.

The antiderivative is  \displaystyle F(x) = \frac{4x-19}{4} + C.

Example Question #2541 : Calculus Ii

Integrate to lowest terms: \displaystyle \large \int \sqrt {x}-1

Possible Answers:

\displaystyle \large \large \frac {x^\frac {3}{2}-\frac {3}{2}x}{\frac {3}{2}}+C

\displaystyle \large \large \frac {2x^\frac {3}{2}+{3}x} {3}+C

\displaystyle \large \large \frac {2x^\frac {3}{2}-{3}x} {3}+C

None of the Above

Correct answer:

\displaystyle \large \large \frac {2x^\frac {3}{2}-{3}x} {3}+C

Explanation:

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Example Question #794 : Integrals

Evaluate.

\displaystyle \int e^{8x-43}\ dx

Possible Answers:

Answer not listed.

\displaystyle F(x) = \frac{8x}{e^{8x-43}} + C

\displaystyle F(x) = \frac{8}{e^{8x-43}} + C

\displaystyle F(x) = \frac{e^{8x-43}}{8} + C

Correct answer:

\displaystyle F(x) = \frac{e^{8x-43}}{8} + C

Explanation:

\displaystyle \int f(x) dx

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

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

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

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

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

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

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

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

 

In this case, \displaystyle f(x)= e^{8x-43}.

The antiderivative is  \displaystyle F(x) = \frac{e^{8x-43}}{8} + C.

Example Question #795 : Integrals

Evaluate.

\displaystyle \int \cos (12) \ dx

Possible Answers:

\displaystyle F(x) = \frac{\cos(12)}{x} + C

\displaystyle F(x) = \cos(12x) + C

\displaystyle F(x) = \cos(12)x + C

\displaystyle F(x) = \cos(12) + C

Answer not listed.

Correct answer:

\displaystyle F(x) = \cos(12)x + C

Explanation:

\displaystyle \int f(x) dx

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

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

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

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

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

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

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

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

 

In this case, \displaystyle f(x)= \cos(12).

The antiderivative is  \displaystyle F(x) = \cos(12)x + C.

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