# Integration- Area under curves

• Jul 12th 2009, 11:47 PM
auonline
Integration- Area under curves
I think this question is pretty basic, but I would be very happy if you could help :)

Find the area under the curve y = (x-2)^-2 + 1, from x = -1 to x = 1 :)

Do I integrate y = (x-2)^-2 + 1?? How teacher uses some random hard to understand process where she finds the derivate and uses that to adjust the integral o.O

Thanks!!
Lachlan
• Jul 13th 2009, 02:18 AM
CaptainBlack
Quote:

Originally Posted by auonline
I think this question is pretty basic, but I would be very happy if you could help :)

Find the area under the curve y = (x-2)^-2 + 1, from x = -1 to x = 1 :)

Do I integrate y = (x-2)^-2 + 1?? How teacher uses some random hard to understand process where she finds the derivate and uses that to adjust the integral o.O

Thanks!!
Lachlan

Yes the required area is:

$A=\int_{x=-1}^1 \frac{1}{(x-2)^2}+1 \; dx$

what method you use to evaluate it is up to you and/or your teacher.

CB
• Jul 13th 2009, 02:25 AM
auonline
Ok thanks :D
• Jul 13th 2009, 04:58 AM
VonNemo19
Quote:

Originally Posted by auonline

Find the area under the curve y = (x-2)^-2 + 1, from x = -1 to x = 1 :)

Another way:

$\lim_{n\to\infty}\sum_{i=1}^n\left[\frac{1}{[(-1+\frac{2i}{n})-2]^2}+1\right]\left(\frac{2}{n}\right)$
• Jul 13th 2009, 05:35 AM
HallsofIvy
Quote:

Originally Posted by VonNemo19
Another way:

$\lim_{n\to\infty}\sum_{i=1}^n\left[\frac{1}{[(-1+\frac{2i}{n})-2]^2}+1\right]\left(\frac{2}{n}\right)$

Ah, the easy way! (Rofl)
• Jul 13th 2009, 05:50 AM
HallsofIvy
Quote:

Originally Posted by auonline
I think this question is pretty basic, but I would be very happy if you could help :)

Find the area under the curve y = (x-2)^-2 + 1, from x = -1 to x = 1 :)

Do I integrate y = (x-2)^-2 + 1?? How teacher uses some random hard to understand process where she finds the derivate and uses that to adjust the integral o.O

Thanks!!
Lachlan

If you let u= x-2, differentiating both sides give du= dx (was that what you meant by "finds the derivate"); when x= -1 , u= -1-2= -3, and when x= 1, u= 1-2= -1. Now replace each of those things in your integral by its value in terms of u, $\int_{-1}^1 ((x-2)^2 + 1)dx$ becomes $\int_{-3}^{-1} (u^{-2}+ 1) du$. That is NOT some "random hard to understand process". That is a very logical, standard method known as "integration by substitution".
• Jul 13th 2009, 01:16 PM
VonNemo19
Quote:

Originally Posted by HallsofIvy
Ah, the easy way! (Rofl)

I think so. I realize that this is not as quick as u-substitution, but it is good to see things from all sides when trying to tackle a problem.