1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Area Between Curves

  1. May 5, 2005 #1
    Hey guys, I am curious if I am setting this up right. Could you take a look and make sure I am on the right path?

    I have three questions.

    1) I am trying to find the area bound by [tex]y=x^2[/tex] and [tex]y=4x+5[/tex]
    Upper function [tex]y=4x+5[/tex]
    Lower function [tex]y=x^2[/tex]
    For my integral I have [tex]\int^{5}_{-1}4x-5-x^2[/tex]

    I then get [tex]\left[\frac{-x^{3}}{3}+2x^{2}+5x\right]^{5}_{-1}[/tex]

    and my answer is 110/3 units squared

    Is my answer correct? Am I doing it right? I just figured out you can do it this way. My teacher has been teaching us to use squares, triangles, and other geometric methods (not talking about riemann sums) rather than just pure integration...

    Ok, second question..
    Find the area bounded by y=x-2, y=4, the x-axis, and the y-axis.
    Would I set this problem up like [tex]\int^{6}_{2}x-2-\int^{2}_{0}x-2[/tex] this?

    Last question..
    I need to find the derivative of [tex]ln(x-1)^{2}[/tex]
    I am confused as to whether they are saying to square x-1 or the natural log.

    Thanks guys
     
  2. jcsd
  3. May 5, 2005 #2
    On your first question you set up the problem correctly, but the integral of -5 is -5x, you put +5x, maybe that's a typo. I didn't check the exact answer, but you definitely have the right idea.

    For the second question, remember that the area is the integral of the top minus the integral of the bottom. So set it up like this:

    [tex]\int_{0}^{6} 4 - (x-2)dx[/tex]

    They only want the first quadrant which is why the lower bound for the integral is zero.

    As to the third question, that's rather ambiguous, I could interpret that either way.
     
  4. May 5, 2005 #3
    On the second question are they wanting me to find the area under x-2 or the area above it? It seems it could be looked at either way.
     
  5. May 5, 2005 #4
    From the graph and your bounds (x and y axes) it looks like you are trying to find the area in the first quadrant, above y=x-2 and below y=4. The area below x-2 is unbounded on the x axis, so your best bet is to go for the one that is contained the most.
     
  6. May 5, 2005 #5
    Alright, I will go with that. Thanks!
     
  7. May 5, 2005 #6
    ln(x-1)^2 take derivative, you must use chain rule
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Area Between Curves
  1. Area between curves (Replies: 3)

  2. Area between curves (Replies: 1)

  3. Find Area Under A Curve. (Replies: 10)

Loading...