Let f(x) be a function satisfying f'(x)=f(x) with f(0)=1 and g(x) be a function that satis…

Question

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 $\displaystyle\int_0^1 f(x)g(x)\,dx$, is:

Accepted Answer

Given $$f'\left( x \right) = f\left( x \right) \Rightarrow {{f'\left( x \right)} \over {f\left( x \right)}} = 1$$

Integrating log

$$f\left( x \right) = x + c \Rightarrow f\left( x \right) = {e^{x + c}}$$

$$f\left( 0 \right) = 1 \Rightarrow f\left( x \right) = {e^x}$$

$$\therefore$$ $$\int\limits_0^1 {f\left( x \right)g\left( x \right)} dx$$

$$ = \int\limits_0^1 {{e^x}} \left( {{x^2} - {e^x}} \right)dx$$

$$ = \int\limits_0^1 {{x^2}} {e^x}dx - \int\limits_0^1 {{e^{2x}}} dx$$

$$ = \left[ {{x^2}{e^x}} \right]_0^1 - 2\left[ {x{e^x} - {e^x}} \right]_0^1 - {1 \over {20}}\left[ {{e^{2x}}} \right]_0^1$$

$$ = e - \left[ {{{{e^2}} \over 2} - {1 \over 2}} \right] - 2\left[ {e - e + 1} \right]$$

$$ = e - {{{e^2}} \over 2} - {3 \over 2}$$

Solutions · 1