Evaluate \int_{0}^{z} f(x)f(z-x)dx
where f(x)=1 for 0

Question

Evaluate \int_{0}^{z} f(x)f(z-x)dx
where f(x)=1 for 0<x<1 and f(x)=0 elsewhere.
Please show how to handle the inequalities.

anonymous · Answer

$$\int\limits_{0}^{z}f(x)f(z-x)dx$$ is the integral.

anonymous · Answer

What's z?

anonymous · Answer

z is an indeterminate. The integral is a function of z.

anonymous · Answer

I'm... as your profile picture says :(

anonymous · Answer

Integration by Parts, maybe?

anonymous · Answer

The integral is z for 0<=z<=1 and (2-z) for 1<z<=2. This is the answer. I dont know how to arrive here!

anonymous · Answer

Me neither... sounds horrible :D
@AravindG 
Ideas? :)

anonymous · Answer

I know I have to do some juggling with the inequalities. But  :(
Main idea is to concentrate on the intervals where the product f(x)f(z-x) is 1, only in these intervals, the integral is a nonzero value.

aravindg · Answer

hmm....give me a minute

anonymous · Answer

Is there a specific function f?

aravindg · Answer

well for interval  0<x<1 f(x) is a constant funtion . i bet that can help us here

anonymous · Answer

Its specific. f(x) is 1 only when $$x \in (0,1)$$, 0 elsewhere

aravindg · Answer

*function

anonymous · Answer

Oh, so there IS a value for f(x)
I thought f was an arbitrary function >.>

anonymous · Answer

Please read the question! I mentioned earlier!

aravindg · Answer

yep think of it graphically

anonymous · Answer

Right.  I'm so sorry :)

aravindg · Answer

for case  0<x<1

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