Value of integration

1. Nov 18, 2011

songoku

1. The problem statement, all variables and given/known data
If $$f(x) = x^2 - x - \int_0^1 f(x) dx$$, find $$\int_0^2f(x) dx$$

2. Relevant equations

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

Is it possible the answer in numerical value? If yes, please guide me. Thanks

2. Nov 18, 2011

canis89

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. Nov 18, 2011

ehild

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

ehild

4. Nov 18, 2011

CompuChip

I managed to get something slightly prettier by writing
$$\int_0^2 f(x) dx = I + \int_1^2 (x^2 - x - I) dx$$
where
$$I = \int_0^1 f(y) dy$$
is a constant.

5. Nov 18, 2011

songoku

Ah I get it. Integrating f(x) from 0 to 1 never crosses my mind. Thanks a lot