What is the average value of 3-x^2 on [0,2]

is it:

a. 3
b. 4
c. 2
d. 5/3

I worked it out and got a. but I'm not sure if it's correct or not.

The average value of f(x) on the interval [a, b] is given by:
F(avg) = 1/(b - a) ∫ f(x) dx (from x=a to b).

So, the average value of f(x) = 3 - x^2 on [0, 2] is:
F(avg) = 1/(2 - 0) ∫ (3 - x^2) dx (from x=0 to 2)
= 1/2 ∫ (3 - x^2) dx (from x=0 to 2)
= (1/2)[3x - (1/3)x^3 (evaluated from x=0 to 2)], by the FTC
= (1/2)[(6 - 8/3) - (0 - 0)]
= 5/3.

Therefore, the answer is (D) 5/3.

I hope this helps!