Mean = E (x)

$$=\int\limits_{-\infty}^{\infty}x.f(x).dx$$

$=\int\limits_0^1x.6(x-x^2)dx\\ =6\int\limits_0^1(x^2-x^3)dx\\ =6 \Bigg[\dfrac {x^3}3-\dfrac {x^4}4\Bigg]_0^1 =6\Bigg[\Bigg(\dfrac {1^3}3-\dfrac {1^4}4\Bigg)-(0-0)\Bigg] \\ =0.5\\ Consider \\ E(x^2)=\int\limits_{-\infty}^{\infty}x^2.f(x)dx \\ =\int\limits_0^1 x^2.6(x-x^2) dx\\ =6\int\limits_0^1[x^3-x^4]dx\\ =6\Bigg[\dfrac {x^4}4-\dfrac {x^5}5\Bigg]_0^1\\ =6\Bigg[\Bigg(\dfrac {1^4}4-\dfrac {1^5}5\Bigg)-(0-0)\Bigg] \\ =0.3$

Variance $= E (x^2) – [E(x)]^2\\ = 0.3 – 0.5^2 \\ = 0.05$

Hence, Mean $= 0.5$ and variance $= 0.05$