If f(x) satisfies the relation f(x) = ex + 01 (y + xex) f(y) dy, then e + f(0) is equal to .
2
Let A = 01 y f(y) dy and B = 01 f(y) dy, which are constants. Then f(x) = ex + A + Bxex, so f(0) = 1 + A.
Integrating f from 0 to 1,
B = 01 ex dx + A + B 01 xex dx
and 01 xex dx = [xex - ex]01 = 1,
B = (e - 1) + A + B
A = 1 - e
so f(0) = 1 + A = 2 - e.
e + f(0) = 2
Answer: 2
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