=1/(2-(-1) ∫((1/2)x^2 - x + 3) dx from x = -1 to 2
=1/3 [ (1x-x^3/3 ] from x = -1 to 2
=1/3[2-8/3)-(1-1/3)]
1/3[-2/3-2/3] =0
Find the average value of the function f over the interval [-1, 2].
f(x)=1-x^2
3 answers
It still says it is wrong. =/
I get zero.
∫[-1,2] 1-x^2 dx
= x - x^3/3 [-1,2]
= (2 - 8/3) - (-1 + 1/3)
= 0
yeah, I know, we have to divide by 3 ...
∫[-1,2] 1-x^2 dx
= x - x^3/3 [-1,2]
= (2 - 8/3) - (-1 + 1/3)
= 0
yeah, I know, we have to divide by 3 ...
1 answer