Find k and then E(X) if X has the p.d.f

Since the total probability is 1

$∫_{-∞}^∞f(x)$ = 1

$∫_0^2kx(2-x) \ dx = k∫_0^2(2x-x^2 )\ dx = k[x^2-\frac{x^3}{3}]_0^2 = k . \frac{4}{3} = 1$

∴ k = $\frac{3}{4}$

∴ f(x) = $ \begin{cases} \frac{3}{4}x(2-x) \,\,\, if \ 0\leq x \leq 2, \ k \gt 2 \\ 0 \,\,\,\, elsewhere \end{cases} $

By definition, $E(X) =∫_0^2xf(x) \ dx = ∫_0^2x.\frac{3}{4} x(2-x) \ dx = ∫_0^2 \frac{3}{4}(2x^2-x^3) \ dx$

= $ \frac{3}{4} [\frac{2x^3}{3}-\frac{x^4}{4}]_0^2$ = 1