Results 1 to 6 of 6

Math Help - L'hopitals rule problem involving area and volume

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    24

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    Quote Originally Posted by helpplz View Post
    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:

    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?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2009
    Posts
    24
    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 :/
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2009
    Posts
    24
    Wouldn't it just be -e^-t?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    Quote Originally Posted by helpplz View Post
    Wouldn't it just be -e^-t?
    The Fundamental Theorem of Calculus states:

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

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

    In your case, -e^{-t} is F(b), but what is F(a)?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Oct 2009
    Posts
    24
    Oh geez. I forgot to plug in zero. F(a) would be -1. Thank you!!!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. L'Hopitals Rule
    Posted in the Calculus Forum
    Replies: 3
    Last Post: July 25th 2011, 01:21 PM
  2. l'hopitals rule problem
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 3rd 2010, 08:20 AM
  3. L'Hopitals rule problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 14th 2010, 07:22 PM
  4. Replies: 4
    Last Post: April 6th 2009, 01:27 AM
  5. Replies: 0
    Last Post: October 20th 2008, 07:12 PM

Search Tags


/mathhelpforum @mathhelpforum