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: Evaluating definite integrals for the area of the regoin

  1. Sep 2, 2014 #1
    1. The problem statement, all variables and given/known data
    Evaluate the definite integral that gives the area of the region bounded by the graph of the function and the tangent line to the graph at the given point.

    [itex]f(x) = \frac{1}{x^2+1}[/itex] at the point (1,1/2)

    2. Relevant equations

    3. The attempt at a solution

    So far I have obtained that the tangent line is [itex]y = 1- \frac{1x}{2}[/itex]. I'm having difficultly integrating for the area.

    I've worked out that [itex]A = \int_1^0 [\frac{1}{x^2+1} - (\frac{-1x}{2} +1)]\,dx[/itex].
    I'm not really sure what to do from here. I know that i'm suppose to fit the first fraction into the integral of arctan, however I can not get the top to be equal to du. In addition, I thought I should use the power rule for the second part of the fraction to obtain x^2/2 but the answer in my book states that it's x^2/4 which is confusing me. Any help would be appreciated, but please let me know what rules you are using to obtain your solution so it's easier for me to understand. Thanks.
    Last edited: Sep 2, 2014
  2. jcsd
  3. Sep 2, 2014 #2


    User Avatar
    Science Advisor

    I'm not clear what region you mean. There are two regions between the tangent line and the function- but both are infinite. You have your integral from 0 to 1 (Actually, you have 1 to 0. What reason do you have for going from 1 down[/b to 0?) . Does the problem specify that the y-axis is also a boundary?

    You say "fit the first fraction into the integral of arctan". I guess that you mean that [tex]\int \frac{dx}{x^2+ 1}= arctan(x)+ C[/tex] which is true but I don't know what you mean by "get the top to be du". Why "du"? You already have dx. What's wrong with that?

    Finally, I presume you know that [itex]\int x dx= x^2/2+ C[/itex] but you already have a factor of 1/2. So you have [itex](1/2)\int x dx= (1/2)(x^2/2 + C)= x^2/4+ c[/itex].
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted