Value of integration

  • Thread starter songoku
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  • #1
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Homework Statement


If [tex] f(x) = x^2 - x - \int_0^1 f(x) dx [/tex], find [tex]\int_0^2f(x) dx[/tex]


Homework Equations





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
 

Answers and Replies

  • #2
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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.
 
  • #3
ehild
Homework Helper
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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
 
  • #4
CompuChip
Science Advisor
Homework Helper
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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.
 
  • #5
1,547
101
Ah I get it. Integrating f(x) from 0 to 1 never crosses my mind. Thanks a lot
 

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