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!
6 answers
I would also like to see this question answered, as I have one that is similar!
I really need to see an example of this one too!
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
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
Can you please differentiate the answer for a and b?
The answer for a works for b I think
the average value times the width is the integral
the average value times the width is the integral
(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)
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)
1 answer