Solve Value of Integration: Find Integral from 0 to 2

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Homework Statement


If f(x) = x^2 - x - \int_0^1 f(x) dx, find \int_0^2f(x) dx


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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
 
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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.
 
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
 
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.
 
Ah I get it. Integrating f(x) from 0 to 1 never crosses my mind. Thanks a lot
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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