Explanation

(c)
Let $F(x) = e^{x}$
Also, $f(x) + g(x) = x^{2}\ \Rightarrow \ g(x) = x^{2} - e^{x}$
Now, $\int_{0}^{1}{f(x)g(x)dx = \int_{0}^{1}{e^{x}\left( x^{2} - e^{x} \right)dx}}$
$= \int_{0}^{1}{x^{2}e^{2}\ dx - \int_{0}^{1}{e^{2x}dx}}$
$= \lbrack x^{2}e^{x} - \int_{}^{}{2xe^{x}dx\left. \ \ \right\rbrack_{0}^{1}} - \frac{1}{2}\left\lbrack e^{2x} \right\rbrack_{0}^{1}$
$= \left\lbrack x^{2}e^{x} - 2xe^{x} + 2e^{x} \right\rbrack_{0}^{1} - \frac{1}{2}(e^{2} - 1)$
$= \left\lbrack (1 - 2 + 2)e^{1} - (0 - 0 + 2)e^{0} \right\rbrack - \frac{1}{2}e^{2} + \frac{1}{2}$
$= e - \frac{e^{2}}{2} - \frac{3}{2}$