Question
Let $$f\left( x \right)$$ be a function satisfying $$f'\left( x \right) = f\left( x \right)$$ with $$f\left( 0 \right) = 1$$ and $$g\left( x \right)$$ be a function that satisfies $$f\left( x \right) + g\left( x \right) = {x^2}.$$ Then the value of the integral $$\int\limits_0^1 {f\left( x \right)\,g\left( x \right)dx,} $$ is-
A.
$$e + \frac{{{e^2}}}{2} + \frac{5}{2}$$
B.
$$e - \frac{{{e^2}}}{2} - \frac{5}{2}$$
C.
$$e + \frac{{{e^2}}}{2} - \frac{3}{2}$$
D.
$$e - \frac{{{e^2}}}{2} - \frac{3}{2}$$
Answer :
$$e - \frac{{{e^2}}}{2} - \frac{3}{2}$$
Solution :
Given $$f'\left( x \right) = f\left( x \right) \Rightarrow \frac{{f'\left( x \right)}}{{f\left( x \right)}} = 1$$
Integrating $$\log f\left( x \right) = x + c \Rightarrow f\left( x \right) = {e^{x + c}}$$
$$\eqalign{
& f\left( 0 \right) = 1 \Rightarrow f\left( x \right) = {e^x} \cr
& \therefore \int\limits_0^1 {f\left( x \right)\,g\left( x \right)} dx = \int\limits_0^1 {{e^x}\left( {{x^2} - {e^x}} \right)} dx \cr
& = \int\limits_0^1 {{x^2}{e^x}dx} - \int\limits_0^1 {{e^{2x}}dx} \cr
& = \left[ {{x^2}{e^x}} \right]_0^1 - 2\left[ {x{e^x} - {e^x}} \right]_0^1 - \frac{1}{2}\left[ {{e^{2x}}} \right]_0^1 \cr
& = e - \left[ {\frac{{{e^2}}}{2} - \frac{1}{2}} \right] - 2\left[ {e - e + 1} \right] \cr
& = e - \frac{{{e^2}}}{2} - \frac{3}{2} \cr} $$