# Another Integral Question

• Jan 17th 2010, 02:42 PM
Lolcats
Another Integral Question
$\displaystyle \int \frac{{{2x}^2+2}}{x^2}dx$

There's an upper limit of 4 and lower limit of 1.
• Jan 17th 2010, 02:44 PM
VonNemo19
Quote:

Originally Posted by Lolcats
$\displaystyle \int \frac{{{2x}^2+2}}{x^2}dx$

There's an upper limit of 4 and lower limit of 1.

Note that $\displaystyle \frac{2x^2+2}{x^2}=2+2x^{-2}$

So, can you find $\displaystyle \int_1^4(2+2x^{-2})dx$ ?
• Jan 17th 2010, 02:44 PM
General
Quote:

Originally Posted by Lolcats
$\displaystyle \int \frac{{{2x}^2+2}}{x^2}$

There's an upper limit of 4 and lower limit of 1.

Hello :) .
$\displaystyle \frac{{{2x}^2+2}}{x^2}=\frac{2x^2}{x^2}+\frac{2}{x ^2}$
$\displaystyle =2+2x^{-2}$
Now,
$\displaystyle \int_{1}^{4} \frac{{{2x}^2+2}}{x^2} dx$
$\displaystyle =2\int_{1}^{4}dx+2\int_{1}^{4}x^{-2}dx$.

Can you complete it ?
• Jan 17th 2010, 02:48 PM
Lolcats
Jeez these are easy problems I'm screwing up. Thanks !
• Jan 17th 2010, 02:53 PM
Lolcats
Sorry to be a bother, the answer should be 4 shouldn't it?
• Jan 17th 2010, 02:55 PM
General
Quote:

Originally Posted by Lolcats
Sorry to be a bother, the answer should be 4 shouldn't it?

No.
See this:
integrate [ 2 + 2 x^(-2) ] from x=1 to x=4 - Wolfram|Alpha

• Jan 17th 2010, 02:58 PM
VonNemo19
Quote:

Originally Posted by Lolcats
Sorry to be a bother, the answer should be 4 shouldn't it?

no

$\displaystyle 15/2$
• Jan 17th 2010, 03:02 PM
Lolcats
K well i'd integrate and get 2x-(2/x), then insert the upper limit so:
2(4) + 2/4 = 4 then minus that from 2(1) + 2/1 = 0

Sorry i'm replying slow I have a seperated shoulder atm and can only type with 1 hand.
• Jan 17th 2010, 03:03 PM
General
Quote:

Originally Posted by Lolcats
K well i'd integrate and get 2x-(2/x), then insert the upper limit so:
2(4) + 2/4 = 4 then minus that from 2(1) + 2/1 = 0

Sorry i'm replying slow I have a seperated shoulder atm and can only type with 1 hand.

See the red.
• Jan 17th 2010, 03:06 PM
Lolcats
Wops.. thats embarrasing. Thanks again.