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Calculus 2 : Finding Integrals

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Example Questions

Example Question #311 : Finding Integrals

Possible Answers:

x^4+cos(x)+x^2+C

$\frac{x^4}{4}+cos(x)+\frac{x^2}{2}+C$

4x^4+sin(x)+2x^2+C

$\frac{x^4}{4}+cos(x)+\frac{x^2}{2}$

x^4+sin(x)+x^2+C

Correct answer:

$\frac{x^4}{4}+cos(x)+\frac{x^2}{2}+C$

Explanation:

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Example Question #34 : Indefinite Integrals

Find $\int( e^{-x}*3x) dx$

Possible Answers:

$3e^{-x}(1+x)$

$3e^{-x}(x-1)$

$-\frac{3}{2}x^2e^{-x}$

$3e^{-x}(1-x)$

Correct answer:

$3e^{-x}(1-x)$

Explanation:

Using integration by parts:

$\int( e^{-x}*3x) dx= 3x*-e^{-x}-\int 3e^{-x}dx= -3xe^{-x}+3e^{-x}$

$=-3xe^{-x}+3e^{-x}=3e^{-x}(1-x)$

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Example Question #2411 : Calculus Ii

Euler's identity states that:

$Re(e^{(a+bi)t})= e^{at}cos(bt)$

Also recall that: $Re(\frac{1}{1+i})=\frac{1}{2}$

Determine $re(\int_{0}^{2\pi} e^{(1+i)t} dt)$

Possible Answers:

$\frac{1}{2}(e^{2\pi}+1)$

$\frac{1}{2}$

$\frac{1}{2}(e^{2\pi}-1)$

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

Correct answer:

$\frac{1}{2}(e^{2\pi}-1)$

Explanation:

To determine the integral we just do substitution:

$\int e^{({1+i})t} dt= \frac{1}{1+i}e^{(1+i)t}=F(t)$

By the fundamental theorem of calculus:

$\int_{0}^{2\pi} e^{(1+i)t} dt = F(2\pi)-F(0)=\frac{1}{1+i}*(e^{(1+i)2\pi}-1)$

$re(\frac{1}{1+i}*(e^{(1+i)2\pi}-1))=\frac{1}{2}*(e^{2\pi}(cos(2\pi)-1))$

$=\frac{1}{2}[e^{2\pi}-1]$

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Example Question #35 : Indefinite Integrals

Determine: $\int x^2 e^x dx$

Possible Answers:

e^x(x^2-2x+2) +C

e^x(x^2+2x+2) +C

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

e^x(x^2-2x-2) +C

Correct answer:

e^x(x^2-2x+2) +C

Explanation:

Doing integration by parts twice:

$\int x^2e^x dx=x^2e^x-\int2xe^x dx=x^2e^x-(2xe^x-\int 2e^x dx)$

=x^2e^x-2xe^x+2x^x+C= e^x(x^2-2x+2) +C

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Example Question #37 : Indefinite Integrals

Determine $\int sin(x^2)*x dx$

Possible Answers:

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

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

-cos(x^2) +C

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

Correct answer:

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

Explanation:

$\int sin(x^2)*x dx$

Using substitution,

u=x^2

$\frac{\mathrm{d} }{\mathrm{d} x} u=2x$

$\int sin(x^2)*x dx =\frac{1}{2}\int sin(u) du=\frac{1}{2}(-cos(u)) +C$

$\int sin(x^2)*x dx= -\frac{1}{2}cos(x^2) +C$

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Example Question #34 : Indefinite Integrals

Evaluate the following Definite Integral:

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

Possible Answers:

2xCos(x)+C

2xtan(x^2)+C

tan(x^2)+C

Sec(x^2)+C

tan(x)+C

Correct answer:

tan(x^2)+C

Explanation:

Upon early inspection of this problem, two things may be seen immediately: a trigonometric function and a composite function. One may notice that is the derviative of x^2 , this urges us to use the u-substitution method.

Let u=x^2, du=(2x) dx , therefore the problem may be rewritten as:

$\int(sec^2(u)) du$ , this is a known trigonometric integral to be tan(u)+C , when plugging in for u, the final answer is: tan(x^2)+C .

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Example Question #35 : Indefinite Integrals

An identity of sin(x) is given by:

$sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$ , where is the imaginary number

Determine :

$\int e^tsin(t) dt$

Possible Answers:

$\frac{1}{2i}[\frac{1}{1+i}e^{t(1+i)}-\frac{1}{1-i}e^{t(i-1)}] +C$

$\frac{1}{2i}[\frac{1}{1+i}e^{t(1+i)}-\frac{1}{1-i}e^{t(1-i)}] +C$

$\frac{1}{2i}[\frac{1}{1+i}e^{t(1+i)}-\frac{1}{1-i}e^{t(i+1)}] +C$

Correct answer:

$\frac{1}{2i}[\frac{1}{1+i}e^{t(1+i)}-\frac{1}{1-i}e^{t(1-i)}] +C$

Explanation:

Using the definition above:

$\int e^tsin(t) dt=\frac{1}{2i}\int e^t(e^{it})-e^t(e^{-it}) dt$

$=\frac{1}{2i}[\int e^{t(1+i)})dx-\int e^{t(1-i)}dt]$

This reduces to:

$\frac{1}{2i}[\int e^{t(1+i)})dt-\int e^{t(1-i)}dt]= \frac{1}{2i}[\frac{1}{1+i}e^{t(1+i)}-\frac{1}{1-i}e^{t(1-i)}] +C$

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Example Question #36 : Indefinite Integrals

Calculate the following integral:

$y=\int x^{3}lnxdx$

Possible Answers:

In progress

$y=\frac{1}{4}x^{4}lnx-\frac{1}{16}x^{4}+C$

$y=x^{2}lnx+C$

y=xlnx+C

$y=x^{2}+C$

Correct answer:

$y=\frac{1}{4}x^{4}lnx-\frac{1}{16}x^{4}+C$

Explanation:

We can use integration by parts to solve this integral

Integration by parts states: $\int udv=uv-\int vdu$

Let u = lnx and dv= $x^{3}dx$

Thus, our integral becomes: $\int x^{3}lnxdx=\frac{1}{4}x^{4}(lnx)-\int \frac{1}{4}x^{4}(\frac{1}{x})dx$

Which simplifies to: $\frac{1}{4}x^{4}-\frac{1}{4}\int x^{3}dx$ , which equals : $\frac{1}{4}x^{4}lnx-\frac{1}{16}x^{4}+C$ , giving us our answer $y=\frac{1}{4}x^{4}lnx-\frac{1}{16}x^{4}+C$

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Example Question #41 : Indefinite Integrals

Calculate the following integral: $y=\int \frac{4}{16x^{2}+1}dx$

Possible Answers:

$y=\frac{4x}{16x^{2}+1}+C$

$y=ln\left |4x+1 \right |+C$

$y=tan^{-1}(4x)+C$

$y=4ln\left |16x^{2}+1 \right |+C$

Correct answer:

$y=tan^{-1}(4x)+C$

Explanation:

At first glance, it may look like we need to use partial fraction decomposition to solve this integral. However, this integral is much simpler than that. recall that

$\int \frac{1}{1+x^{2}}dx=tan^{-1}x+C$ . The integral we need to solve, is just the derivative of $tan^{-1}x$ , scaled by a factor of four. So, the solution to our integral is:

$y=tan^{-1}(4x)+C$

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Example Question #41 : Indefinite Integrals

$y=\int sinxcosx dx$

Possible Answers:

$y=sin^{2}xcos^{2}x+C$

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

$y=sin^{2}x+C$

$y=-sin^{2}x+C$

Correct answer:

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

Explanation:

We can solve this integral with u substitution. let u=sinx , so, du=cosxdx

Making this substitution, our integral looks like this:

$y=\intudu$ $\int udu$

So, $y=\frac{1}{2}u^{2}+C$

$y=\frac{1}{2}sin^{2}x+C$

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Calculus 2 : Finding Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #311 : Finding Integrals

Example Question #34 : Indefinite Integrals

Example Question #2411 : Calculus Ii

Example Question #35 : Indefinite Integrals

Example Question #37 : Indefinite Integrals

Example Question #34 : Indefinite Integrals

Example Question #35 : Indefinite Integrals

Example Question #36 : Indefinite Integrals

Example Question #41 : Indefinite Integrals

Example Question #41 : Indefinite Integrals

All Calculus 2 Resources

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