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!

Value of integration

  1. Nov 18, 2011 #1
    1. The problem statement, all variables and given/known data
    If [tex] f(x) = x^2 - x - \int_0^1 f(x) dx [/tex], find [tex]\int_0^2f(x) dx[/tex]


    2. Relevant equations



    3. The attempt at a solution
    I found [tex]\int_0^2f(x) dx = \frac{2}{3} - 2 \int_0^1 f(x) dx[/tex]

    Is it possible the answer in numerical value? If yes, please guide me. Thanks
     
  2. jcsd
  3. Nov 18, 2011 #2
    Yes, it is. In fact, you can find the integral of f from 0 to 1 just by integrating both sides of the equation defining f.
     
  4. Nov 18, 2011 #3

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The definite integral is a number: [tex]\int_0^1{f(x)dx}=A[/tex]
    The first equation can be written as [tex]f(x)=x^2-x-A[/tex].
    Integrate it from x=0 to x=1: you get an equation for A.

    ehild
     
  5. Nov 18, 2011 #4

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    I managed to get something slightly prettier by writing
    [tex]\int_0^2 f(x) dx = I + \int_1^2 (x^2 - x - I) dx[/tex]
    where
    [tex]I = \int_0^1 f(y) dy[/tex]
    is a constant.
     
  6. Nov 18, 2011 #5
    Ah I get it. Integrating f(x) from 0 to 1 never crosses my mind. Thanks a lot
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Value of integration
  1. Value of integral (Replies: 1)

Loading...