Asked by Liam

Consider f(x)=x^2-2x-3

(a). Find the average value of f(x) on the interval [0,3]

(b). Find the value(s) of C in [0,3] guaranteed by the Mean Value Theorem for Integrals.

Any help is appreciated!
Answered by Mitchell
I would also like to see this question answered, as I have one that is similar!
Answered by Drin
I really need to see an example of this one too!
Answered by Anonymous
0 at x = -1 and x = 3 so NEGATIVE the whole way
integral = x^3/3 - x^2 - 3 x
at 3 minus at 0
at 3 = 9 - 9 - 9 = -9
at 0 = -3
-9 - -3 = -6
from - 1 tto +3 is 4
so
-10.6666/4
about - 2 2/3
Answered by Drin
Can you please differentiate the answer for a and b?
Answered by Anonymous
The answer for a works for b I think
the average value times the width is the integral
Answered by oobleck
(a) ∫[0,3] x^2-2x-3 dx = -9
so the average value is -9/3 = -3


(b) you want y'(c) = -1
y' = 2x-2, so
2c-2 = -1
c = 1/2 which is in the interval (0,3)
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