Let `f(x)` be a differentiable function such that `f(x)=x^2 +int_0^x e^-t f(x-t) dt` then `int_0^1 f(x) dx=`

Options

A. `1/4`
B. `-1/12`
C. `5/12`
D. `12/7`

Answer

C

Explanation

`f(x)=x^(2)+int_(0)^(x)e^(-t)(x-t)dt`………….1
`x^(2)+int_(0)^(x)e^(-x-t)f(x-(x-t))dt`
[Using `int_(a)^(b)f(x)dx=int_(a)^(b)f(a+b-x)dx`]
`=x^(2)+e^(-x)int_(0)^(x)e^(t)f(t)dt`……………..2
Differentiating w.r.t.`x` we get
`f'(x)=2x-e^(-x)int_(0)^(x)e^(t)f(t)dt+e^(-x)e^(x)f(x)`
`=2x-e^(-x)int_(0)^(x)e^(t)f(t)dt+f(x)`
`=2x+x^(2)` [using equation 2]
`:. f(x)=(x^(3))/3+x^(2)+c`
Also `f(0)=0` [from equation 1]
or `f(x)=(x^(3))/3+x^(2)`
or `f'(x)=x^(2)+2x`
Thus `f'(x)=` has real roots. Hence `f(x)` is non monotonic.
Hence `f(x)` is many one but range is `R` and hence, is surjective
`int_(0)^(1)f(x)dx=int_(0)^(1)((x^(3))/3+x^(2))dx`
`=[(x^(4))/12+(x^(3))/3]_(0)^(1)`
`=1/12+1/3`
`=5/12`