+0  
 
0
45
2
avatar+703 

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 

 Jan 20, 2019
 #1
avatar+4015 
0

\(\bar{x} = \dfrac{1}{3-0} \displaystyle \int \limits_0^3~2x^2-3~dx = \\ \dfrac 1 3 \left(\left. \dfrac 2 3 x^3 - 3x \right |_0^3\right) = \\ \dfrac 1 3 \left(18-9\right) = \\ \dfrac 9 3 = 3\)

 

I suspect you omitted the factor of 1/3 used to normalize the integral

 

I assume you can do part (b)

 Jan 20, 2019
 #2
avatar+703 
0

yes i have, thank you for your help

YEEEEEET  Jan 20, 2019

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