Integration Parts

$\int f^2(x) \ dx = xf^2(x) - 2\int x f(x) f'(x) \ dx \ - \ \mathrm{Integration \ by \ parts}$ $\implies \int_{a}^{b} f^2(x) \ dx = bf^2(b) - af^2(a) - 2\int_{a}^{b} x f(x) f'(x) \ dx$ $\because f(a)=f(b) = 0$ $\implies \int_{a}^{b} f^2(x) \ dx = - 2\int_{a}^{b} x f(x) f'(x) \ dx$ $\implies 1 = - 2\int_{a}^{b} x f(x) f'(x) \ dx$ $\implies \boxed{\int_{a}^{b} x f(x) f'(x) \ dx = -\frac{1}{2}}$

By integration by parts :

$\begin{aligned} \blue{\int_a^b x f(x) f'(x) dx} & = xf^2(x) \ \big|_a^b - \int_a^b f^2 (x) dx - \blue{\int_a^b xf(x)f'(x) dx} \\ \blue{2 \int_a^b x f(x) f'(x) dx} & = bf(b) - af(a) - 1 = 0 - 0 - 1 \\ \implies \int_a^b x f(x) f'(x) dx & = - \frac 12 = \boxed{-0.5} \end{aligned}$