# Thread: L'hopitals rule problem involving area and volume

1. ## L'hopitals rule problem involving area and volume

The question says: "let A(t) be the area of the region in the first quadrant enclosed by the coordinate axes, the curve y=e^-x and the line x=t>0. Let V(t) be the vlume of the solid generated by revolving the region about the x-axis. Find the following limits: lim t-> infinity A(t), lim t-> infinity V(t)/A(t)."

Okay, so the first thing I did was integrate e^-x from 0 to t. The antiderivative of e^-x is -e^-x. Would that be A(t)? Because, when I took the limit as t->infinity, the value of -e^-x would get closer to 0. But the answer in the back of my book says 1.

Similarly, I used the disc method and then integrated the function pi(e^-x)^2 from 0 to t. I got: -2pi(e^-2x). However, I didn't try to take the limit of this because I had gotten the previous one wrong. Where is my mistake in all of this?

2. Originally Posted by helpplz
The question says: "let A(t) be the area of the region in the first quadrant enclosed by the coordinate axes, the curve y=e^-x and the line x=t>0.
Ok, so then you have the following:

$\displaystyle A(t) = \int _0 ^t e^{-x} dx$.

When you take the antiderivative, you will get that A(t) is a difference of two numbers (due to the fundamental theorem of calculus), one which is a function of t and another which is not. As a function of t, what is A(t) after you take the antiderivative?

3. Right, that's exactly what I thought too, and that's what I integrated. I'm thinking the problem is that I don't know where the t comes into play :/

4. Wouldn't it just be -e^-t?

5. Originally Posted by helpplz
Wouldn't it just be -e^-t?
The Fundamental Theorem of Calculus states:

If $\displaystyle F'(x)$ is the derivative of $\displaystyle F(x)$, then

$\displaystyle \int _a ^b F'(x) dx = F(b) - F(a)$.

In your case, $\displaystyle -e^{-t}$ is F(b), but what is F(a)?

6. Oh geez. I forgot to plug in zero. F(a) would be -1. Thank you!!!