Improper integrals with infinite discontinuities

[nweis84]

Homework Statement

$$\int$$$$\frac{13}{(x-8)^2}$$dx

Homework Equations

it is integrated from 7 to 9 and i am aware that there is an infinite discontinuity at x=8 so we have to take the limit from both sides individually.

The Attempt at a Solution

The only thing I can think that I might be doing wrong is just integrating it wrong but it seems like such an easy integration. The integration that I come up with:

-$$\frac{13}{x-8}$$


as a goes to 8 [$$\frac{-13}{a-8}$$+$$\frac{13}{7-8}$$]+[$$\frac{-13}{9-8}$$+$$\frac{13}{a-8}$$]

and this comes out to be -26 which has to be wrong because the graph is above the positive x-axis and also it is not one of my answer choices.

 

[Hurkyl]

as a goes to 8 [$$\frac{-13}{a-8}$$+$$\frac{13}{7-8}$$]+[$$\frac{-13}{9-8}$$+$$\frac{13}{a-8}$$]

I thought you said you were going to take the limit of both sides individually? This looks more like you threw both sides together first, and then took a limit.

 

[nweis84]

I'm not really sure how to write it in there with this program but I did take the limits individually when I did the work on my paper. The first bracket is the limit from 7 to a as a approaches 8 from the left and the second bracket is the limit from a to 9 as a approaches 8 from the right. Did i do that right?

 

[nweis84]

then don't you just add them in the end?

 

[Hurkyl]

What were the limits on the two sides then? What two numbers did you add together to get -26?

 

[nweis84]

both of them came out to -13

 

[Hurkyl]

Interesting. Could you show how you calculated that?

 

[nweis84]

lim as a --> 8- [ $$\frac{-13}{x-8}$$ ]

from 7 to a =

[$$\frac{-13}{(a-8)}$$+$$\frac{13}{(7-8)}$$]

[ 0 + $$\frac{13}{-1}$$]

= -13

 

[nweis84]

its the same basically for the other side

 

[Hurkyl]

[ 0 + $$\frac{13}{-1}$$]

The 13/-1 makes sense. How did you get that 0?

 

[nweis84]

well doesn't the 8- with the negative sign to the right of the number mean +8 but that it is approaching from the left side?
or are you saying that it means I should have plugged in a negative 8 for a?

or does that mean that that factor goes to infinity?

 

[Hurkyl]

or does that mean that that factor goes to infinity?

Bingo. ($+\infty$, I believe? You should check)

 

[nweis84]

thanks for your help It's just been a while since calc I and i kinda forgot how to do limits

and yes one of the answer choices is that the integral diverges