+845 
for this question i integrated the equation and subbed in 3 which gave me 9 as an answer
but my answer should be 3 please tell me what i did wrong, many thanks
+6251 \(\bar{x} = \dfrac{1}{3-0} \displaystyle \int \limits_0^3~2x^2-3~dx = \\ \dfrac 1 3 \left(\left. \dfrac 2 3 x^3 - 3x \right |_0^3\right) = \\ \dfrac 1 3 \left(18-9\right) = \\ \dfrac 9 3 = 3\)
I suspect you omitted the factor of 1/3 used to normalize the integral
I assume you can do part (b)
