1. 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

2. 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:

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

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

CB

3. Ok thanks

4. Originally Posted by auonline

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

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

5. Originally Posted by VonNemo19
Another way:

$\displaystyle \lim_{n\to\infty}\sum_{i=1}^n\left[\frac{1}{[(-1+\frac{2i}{n})-2]^2}+1\right]\left(\frac{2}{n}\right)$
Ah, the easy way!

6. 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, $\displaystyle \int_{-1}^1 ((x-2)^2 + 1)dx$ becomes $\displaystyle \int_{-3}^{-1} (u^{-2}+ 1) du$. That is NOT some "random hard to understand process". That is a very logical, standard method known as "integration by substitution".

7. Originally Posted by HallsofIvy
Ah, the easy way!
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.