1u edited ur ans...........
ya thats rit
but thats not a good way.....u shud hav mentioned taht u edited it...
Find f(x) which is a continuos fuction such that f(x)=\frac{e^{2x}}{2(e-1)}\int_{0}^{1}e^{-y}f(y)dy+\int_{0}^{\frac{1}{2}}f(y)dy+\int_{0}^{\frac{1}{2}}\sin^{2}(\pi y)dy
for practice.
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62Somehow no one seems to have got hold of this one.....
Hint: there are three constants here. [1]
Reconginze them and replace them by a, b, and c
29i am getting the answer f(x) = \frac{1}{2}\left(e^{2x}+e \right)
actually the value of definite Integrals is independent of x....\int_{0}^{1}{e^{-y}}f(y)dy = a = (e-1)....... \int_{0}^{1/2}{f(y)dy} = b = \frac{2e-1}{4}........ \int_{0}^{1/2}{sin^{2}(\Pi y)dy} = c = 1/4
ans edited..
che now it's ok .... :)
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