1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Area Under 3 Curves PtII

  1. Feb 24, 2006 #1
    hi! Thanks for the advice, I managed to find the intersection points, but I seem to have run into a problem.
    I set up the integral as follows:

    (4-x^2) (for 1 to 2) + (4 - 1/x) (for 1/4 to 1)
    this worked out to
    ((8-8/3) - (4 - 1/3)) + (4 - (1 - ln1/4))
    (16/3 - 11/3) + (3 + ln1/4)
    5/3 + 3 + ln 1/4
    = 14/3 + ln1/4

    but the answer at the back says that the correct response is
    14/3 - ln4... What did I do wrong?
    Thanks so much,
  2. jcsd
  3. Feb 24, 2006 #2
    The (4-x^2) integral gives you the total area in that interval. However, the 1/x curve "intersects" the area. Therefore, you must subtract that part from the first integral. If you plot the curves, you will see what I am talking about.
  4. Feb 24, 2006 #3
    By the way, there is no need to start a second thread about the same problem. You could have just as well continued in your first thread.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook