Calculus 2 : Indefinite Integrals

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #31 : Indefinite Integrals

\displaystyle \int 3x^2+2x+4 \;dx

Possible Answers:

\displaystyle x^3+x^2+C

\displaystyle 6x+2+C

\displaystyle x^3+x^2+4x+C

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

\displaystyle 6x+2

Correct answer:

\displaystyle x^3+x^2+4x+C

Explanation:

When integrating, we do the opposite of a derivative.  You increase the exponent by one and divide the function by that new power.  Since this is an indefinite integral, we have to add a \displaystyle +C at the end of the equation.

\displaystyle \int 3x^2+2x^2+4dx=x^3+x^2+4x+C.

Example Question #2402 : Calculus Ii

\displaystyle \int 5sin(x)-2cos(x)\;dx

Possible Answers:

\displaystyle -5cos(x)-2sin(x)+C

\displaystyle 5cos(x)-2sin(x)+C

\displaystyle 5cos(x)+2sin(x)+C

\displaystyle -5cos(x)+2sin(x)+C

Correct answer:

\displaystyle -5cos(x)-2sin(x)+C

Explanation:

The antiderivative of \displaystyle sin(x)=-cos(x).  The antiderivative of \displaystyle cos(x)=sin(x).  Remember, this is the opposite of a derivative.  Therefore, for our integral, we have:

\displaystyle \int 5sin(x)-2cos(x)\;dx=-5cos(x)-2sin(x)+C.

Example Question #2403 : Calculus Ii

2q

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

2a

Example Question #32 : Indefinite Integrals

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

Possible Answers:

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

\displaystyle 3e^{-x}(x-1)

\displaystyle 3e^{-x}(1+x)

\displaystyle 3e^{-x}(1-x)

Correct answer:

\displaystyle 3e^{-x}(1-x)

Explanation:

Using integration by parts:

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

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

Example Question #662 : Integrals

Euler's identity states that: 

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

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

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

Possible Answers:

\displaystyle \frac{1}{2}

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

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

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

Correct answer:

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

Explanation:

To determine the integral we just do \displaystyle u substitution: 

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

 

By the fundamental theorem of calculus: 

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

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

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

Example Question #33 : Indefinite Integrals

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

Doing integration by parts twice: 

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

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

 

 

Example Question #37 : Indefinite Integrals

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

Possible Answers:

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

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

\displaystyle -cos(x^2) +C

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

Correct answer:

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

Explanation:

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

Using \displaystyle u substitution,

\displaystyle u=x^2

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

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

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

Example Question #34 : Indefinite Integrals

Evaluate the following Definite Integral:

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

Possible Answers:

\displaystyle 2xCos(x)+C

\displaystyle 2xtan(x^2)+C

\displaystyle tan(x^2)+C

\displaystyle Sec(x^2)+C

\displaystyle tan(x)+C

Correct answer:

\displaystyle 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 \displaystyle 2x is the derviative of \displaystyle x^2, this urges us to use the u-substitution method.

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

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

Example Question #664 : Integrals

An identity of \displaystyle sin(x) is given by:

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

Determine :

 \displaystyle \int e^tsin(t) dt

Possible Answers:

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

\displaystyle 0

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

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

Correct answer:

\displaystyle \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: 

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

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

This reduces to:

\displaystyle \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

Example Question #36 : Indefinite Integrals

Calculate the following integral:

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

Possible Answers:

In progress

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

\displaystyle y=x^{2}lnx+C

\displaystyle y=xlnx+C

\displaystyle y=x^{2}+C

Correct answer:

\displaystyle 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: \displaystyle \int udv=uv-\int vdu

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

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

 

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

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