find the integral, from 0 to 1, of (2x^2) dx / (1 + x^2)

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wonky-faint

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the integral of (2x^2)dx/(1+x^2) from limits 0 to 1

i'm not sure where to go with this. I tried integration by parts, but i think this is supposed to use one of the easier methods, but i cant think of it right now. any help would be appreciated. thanks.
 
int 0 to 1 of 2x^2 dx /[x^2+1] divide
int 0 to 1 of 2 dx - 2dx/[x^2+1]

int 2x dx= x^2 evaluated at 1,0
int 2x dx = 1

int 2 dx/ [x^2+1] = 2 Tan^-1 x evaluated at 1,0
int 2 dx / [1+x^2] =2 Tan^-1 1
int 2 dx / [1+x^2] = 2 pi /4
int 2 dx / [1+x^2]= pi/2

int 0 to 1 2x^2 dx / [x^2+1] = 1-pi/4 answer

please check for errors

Arthur
 
arthur ohlsten said:
int 0 to 1 of 2x^2 dx /[x^2+1] divide
int 0 to 1 of 2 dx - 2dx/[x^2+1]

int 2x dx= x^2 evaluated at 1,0
int 2x dx = 1
That should be: \(\displaystyle \mbox{\int_0^1 2 dx = 2}\)

int 2 dx/ [x^2+1] = 2 Tan^-1 x evaluated at 1,0
int 2 dx / [1+x^2] =2 Tan^-1 1
int 2 dx / [1+x^2] = 2 pi /4
int 2 dx / [1+x^2]= pi/2

int 0 to 1 2x^2 dx / [x^2+1] = 1-pi/4 answer <- pi/2 was fine!

please check for errors

Arthur
Click to expand...
 
Re: find the integral

Hello, wonky-faint!

\(\displaystyle \L\int^{\;\;\;1}_0 \frac{2x^2}{1\,+\,x^2}\,dx\)
Click to expand...

Use long division . . .

\(\displaystyle \L\int^{\;\;\;1}_0\frac{2x^2}{1\,+\,x^2}\,dx\;=\;\int^{\;\;\;1}_0\left(2\,-\,\frac{2}{1\,+\,x^2}\right)\,dx \;= \;2x\,-\,2\arctan x\,\bigg]^1_0\)

Evaluate: \(\displaystyle \L\:\left[2(1)\,-\,2\arctan(1)\right] \,-\,\left[2(0)\,-\,2\arctan(0)\right]\)

. . \(\displaystyle \L=\;\left[2 - 2\left(\frac{\pi}{4}\right)\right]\,-\,[0 \,-\,0] \;= \;\fbox{2\,-\,\frac{\pi}{2}}\)

 
How do you make the integral sign?
I mean the German script S. I don't know how to do it
Arthur
 
arthur ohlsten said:
How do you make the integral sign?
I mean the German script S. I don't know how to do it
Arthur
Click to expand...
It is all about TeX!
If you type \(\displaystyle \int\limits_{ - 2}^1 {} f(x)dx \) then you will get \(\displaystyle \int\limits_{ - 2}^1 {} f(x)dx\)
 
Here's the trig sub way. Just for kicks.

Let \(\displaystyle \L\\x=tan{\theta} \;\ and \;\ dx=sec^{2}{\theta}d{\theta}\)

You have to change the limits of integration.

\(\displaystyle \L\\1=tan{\theta} \;\ and \;\ 0=tan{\theta}\)

That gives \(\displaystyle \L\\{\theta}=\frac{\pi}{4} \;\ and \;\ 0\)

You then have:

\(\displaystyle \L\\\int_{0}^{\frac{\pi}{4}}\frac{2(tan{\theta})^{2}}{1+(tan{\theta})^{2}}sec^{2}{\theta}d{\theta}\)

Simplifying gives:

\(\displaystyle \L\\2\int_{0}^{\frac{\pi}{4}}tan^{2}{\theta}d{\theta}\)

=\(\displaystyle \L\\2(tan{\theta}-{\theta})\Rightarrow{2(tan(\frac{\pi}{4})-\frac{\pi}{4})}=\frac{4-{\pi}}{2}\approx{0.429}\)
 
thank you. I will use it on next integral I do for a student.
Do you know how to make a sum sign,[the capital Greek letter sigma]?
Arthur
 
I believe we should tell the student that 2x^2/[1+x^2] is a improper fraction, and suggest it be reduced to a proper fraction. This will help him in the future with similiar problems
2x^2/[x^2+1] = 2 -2/[x^2+1] after division
Arthur
 
Hello Arthur:

Yes, it is worth learning a little LaTex if one is going to participate on the forums.

The sum sign is:

\(\displaystyle \L\\\sum_{n=1}^{\infty}\)

Click on 'quote' at the upper right corner of my post to see the LaTex code I used.

There are many tutorials on the web.
 
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