Let f be a derivable function satisfying the equation ∫x0f(t)dt+∫x0t.f(x−t)dt=e−x−1
f’(0) has the value equal to
2
0
1−1e
1+1e
Put x = 0 in (2) f′(0)+f(0)=1⇒f′(0)=2
∫10f(x)dx is equal to
Let f(x) be a function satisfying f′(x)=f(x) with f(0) = 1 and g(x) be the function satisfying f(x)+g(x)=x2 The value of integral ∫10f(x)g(x) dx is equal to [AIEEE 2003; DCE 2005]