Calculus examination

[Guest]

Hi! I was wondering if anyone could help me with some study questions i am having trouble with - my exam is in a few days, so needless to say i'm a little worried!

I have to integrate the following:

a. (x^2+2x+5)^(1/2)dx, limit from 0 to 1.

b. xarctan(x^2)dx

c. 1/(1-e^x)dx

d. x/{1+x^(1/3)}dx

I really hope you guys can help me. I'm pretty lost, and there are no solutions available anywhere! Thanks!

b

 

[jsbeckton]

If you have no clue how to do any of these and your exam is in a few days then you certainally should be worried. Is this a calc 1 or calc 2 class, if you attend class and these questions are going to be on your exam, one can only assume that you have been working on these problems for some time now and should have several examples in your notes.

People here cannot teach you calculus, they are here to help you out when you get stuck, they like to see some effort by you instead of just a list of problems.

That being said I'll show you an example of one similar to the first one.

\(\displaystyle \begin{array}{l}
\int {(3 - 2x - x^2 )^{\frac{1}{2}} dx} \\
\\
3 - 2x - x^2 = - (x^2 + 2x + 1) + 4 = 4 - (x + 1)^2 {\rm }let \\
\\
x + 1 = 2\sin \theta ,where - \frac{\Pi }{2} \le \theta \le \frac{\Pi }{2}.{\rm }Then{\rm dx = 2cos}\theta {\rm d}\theta {\rm and} \\
\\
\int {\sqrt {3 - 2x - x^{\rm 2} } } dx = \int {\sqrt {4 - (x + 1)^2 } } dx = \int {\sqrt {4 - 4\sin ^2 \theta } } {\rm 2cos}\theta {\rm d}\theta \\
\\
= 4\int {\cos ^2 } \theta {\rm d}\theta = 2\int {(1 + \cos 2\theta )d} \theta \\
\\
= 2\theta + \sin 2\theta + C = 2\theta + 2\sin \theta \cos \theta + C \\
\\
= 2\sin ^{ - 1} \left( {\frac{{x + 1}}{2}} \right) + 2\frac{{x + 1}}{2}\sqrt {\frac{{3 - 2x - x^2 }}{2}} + C \\
\\
= 2\sin ^{ - 1} \left( {\frac{{x + 1}}{2}} \right) + \frac{{x + 1}}{2}\sqrt {3 - 2x - x^2 } + C \\
\end{array}\)

I know it seems long and complicated, but you need to see it worked out in person with someone explaining it along the way, its one of the toughest concepts in calc 2 IMO. It involves a right triangle and trig substitution, check you book for pictures and examples, its not that bad once you understand the triangle concept.

By the way can anyone tell me ho to space out the lines when transfering from teXaide to this forum, I did it spaced out but when I transfered it, it is obviously not spaced, and how do I make it bigger? I tried changing the font to no avial.

 

[soroban]

Hello, bubblicious!

These are quite challenging . . . don't feel bad if you didn't get them.

\(\displaystyle b)\;\int x\cdot\arctan(x^2)dx\)

Integrate by parts:

Let \(\displaystyle \;u\,=\,\arctan(x^2)\;\;\;\;dv\,=\,x\,dx\)

Then \(\displaystyle \;du\,=\,\frac{2x}{1\,+\,x^4}\,dx\;\;\;\;v\,=\,\frac{1}{2}x^2\)

And we have: \(\displaystyle \L\;\frac{1}{2}x^2\cdot\arctan(x^2)\,-\,\int\frac{x^3}{1\,+\,x^4}\,dx\)

. . Then let \(\displaystyle \;u\,=\,1\,+\,x^4\)

\(\displaystyle \L c)\;\int\frac{dx}{1\,-\,e^x}\)

This is a classic puzzler . . . There are several ways to integrate it
. . all of them super-sneaky! .Here's one of them:

Multiply top and bottom by \(\displaystyle e^{-x}:\;\;\L\int \frac{e^{-x}}{e^{-x}\,+\,1}\,dx\)

. . Then let \(\displaystyle \;u = e^{-x}\,+\,1\)

\(\displaystyle \L d)\;\int \frac{x}{1\,+\,x^{\frac{1}{3}}}\,dx\)

The obvious substitution is: \(\displaystyle u = x^{1/3}\) . . . Let's see how it works out.

We have: \(\displaystyle \;u\,=\,x^{\frac{1}{3}}\;\;\Rightarrow\;\;x\,=\,u^3\;\;\Rightarrow\;\;dx\,=\,3u^2\,du\)

Substitute: \(\displaystyle \L\;\int\frac{u^3}{1\,+\,u}(3u^2\,du)\;=\;3\int\frac{u^5}{u\,+\,1}\,du\) . . . . good!

Long division: \(\displaystyle \L\;3\int\left(u^4\,-\,u^3\,+\,u^2\,-\,u\,+\,1\,-\,\frac{1}{u+1}\right)\,du\)

 

[Daniel_Feldman]

Hello, bubblicious!

These are quite challenging . . . don't feel bad if you didn't get them.

\(\displaystyle b)\;\int x\cdot\arctan(x^2)dx\)

Integrate by parts:

Let \(\displaystyle \;u\,=\,\arctan(x^2)\;\;\;\;dv\,=\,x\,dx\)

Then \(\displaystyle \;du\,=\,\frac{2x}{1\,+\,x^4}\,dx\;\;\;\;v\,=\,\frac{1}{2}x^2\)

And we have: \(\displaystyle \L\;\frac{1}{2}x^2\cdot\arctan(x^2)\,-\,\int\frac{x^3}{1\,+\,x^4}\,dx\)

. . Then let \(\displaystyle \;u\,=\,1\,+\,x^4\)

Actually, this could be done even if you are unfamiliar with integration by parts.

u=x^2

du=2xdx

xdx=(1/2)du

So the integral simplifies to (1/2)\(\displaystyle \;\int\arctan(u)du\)


Now just use integral of arctan.