Let f(x) be a function satisfying`f'(x)=f(x)` with`f(0) =1` and g(x) be a function that satisfies `f(x) + g(x) = x^2` . Then the value of the integral `int_0^1f(x) g(x) dx`, is

Options

A. `"lee" -(e^(2))/2 - 5/2`
B. `e+ (e^(2))/2-3/2`
C. `e- (e^(2))/2 - 3/2 `
D. ` e + (e^(2))/2 + 5/2`

Answer

C

Explanation

Let `f(x)=e^(x)`
Also , `f(x)+g(x)=x^(2)rArr g(x)=x^(2)-e^(x)`
Now `int_(0)^(1)f(x)g(x)dx = int_(0)^(1)e^(x)(x^(2)-e^(x))dx`
`=int_(0)^(1)x^(2)e^(x)dx-int_(0)^(1)e^(2x)dx`
`=[x^(2)e^(x)-int2xe^(x)dx]_(0)^(1)-(1)/(2)[e^(2x)]_(0)^(1)`
`=[x^(2)e^(x)-2xe^(x)+2e^(x)]_(0)^(1)-(1)/(2)(e^(2)-1)`
`=[(1-2+2)e^(1)-(0-0+2)e^(0)]-(1)/(2)e^(2)+(1)/(2)`
`=e-(e^(2))/(2)-(3)/(2)`