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!

Find the area of the region given the boundaries

  1. Apr 3, 2017 #1
    1. The problem statement, all variables and given/known data
    The curve of y= 4/(x^2+4), the x-axis, and the vertical lines x = -2 and x = 2

    2. Relevant equations


    3. The attempt at a solution
    slow.jpg
     
  2. jcsd
  3. Apr 4, 2017 #2

    Math_QED

    User Avatar
    Homework Helper

    Hello beefiestcrib55!

    :welcome:

    To begin, every member here will appreciate if you type out your work. Most members won't even bother if you post images.

    There are a couple of things wrong about your approach. You set up the integral correctly, but then a couple of things went wrong:

    1) Reflect about your answer! Can an area be negative?
    2) You need to perform a substitution, you cannot (without experience) evaluate the integral into a primitive function in one step!
    3)##\int \frac{1}{x^2 + 1} dx = \arctan(x) + c ##, not ##\arctan(x^2)+c##
    4) Although your bounds should be changed by the substitution, why would you fill in ##1## as upperbound if you have ##2## as upperbound?
     
  4. Apr 4, 2017 #3
    Specifically, ##\int \frac{dx}{x^2+a^2}=\frac{1}{a}tan^{-1}(\frac{x}{a})## or in his case, ##\int \frac{dx}{\frac{x^2}{a^2}+1}=\frac{1}{a}tan^{-1}(\frac{x}{a})##. Also, ##tan^{-1}(0)≠\frac{\pi}{2}##. For future reference, ##\lim_{x \rightarrow \infty} {tan^{-1}(x)}=\frac{\pi}{2}##, and ##\lim_{x \rightarrow -\infty} {tan^{-1}(x)}=-\frac{\pi}{2}##.
     
    Last edited: Apr 4, 2017
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Find the area of the region given the boundaries
Loading...