Thread: 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

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:



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

CB

Ok thanks

Originally Posted by auonline

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

Another way:

Originally Posted by VonNemo19

Another way:

Ah, the easy way!

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, becomes . That is NOT some "random hard to understand process". That is a very logical, standard method known as "integration by substitution".

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.