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!
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!
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
(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)
To find the average value of a function on an interval, you need to compute the definite integral of the function over that interval and divide it by the difference in the bounds of the interval.
(a) To find the average value of f(x) on the interval [0,3], we need to evaluate the definite integral of f(x) over the interval [0,3] and then divide the result by the length of the interval, which is 3 - 0 = 3.
First, let's find the definite integral of f(x) over the interval [0,3]:
∫[0,3] (x^2 - 2x - 3) dx.
To compute this integral, you need to find the antiderivative of the function, f(x), and evaluate it at the upper and lower limits of integration. The antiderivative of f(x) = x^2 - 2x - 3 can be found by using the power rule for integration: add 1 to the exponent and divide by the new exponent.
The antiderivative of x^2 is (1/3)x^3, the antiderivative of -2x is -x^2, and the antiderivative of -3 is -3x.
Now, evaluate the antiderivative at the upper and lower limits of integration:
(1/3)(3)^3 - (3)^2 - 3(3) - [(1/3)(0)^3 - (0)^2 - 3(0)].
Simplifying this gives:
(1/3)(27) - 9 - 9 - 0,
(1/3)(27) - 18.
Therefore, the definite integral of f(x) over [0,3] is (1/3)(9) = 3.
Now, to find the average value of f(x) on the interval [0,3], divide the result of the definite integral by the length of the interval:
Average value = (Definite integral of f(x) over [0,3]) / Length of interval
= 3 / 3
= 1.
So, the average value of f(x) on the interval [0,3] is 1.
(b) To find the value(s) of C guaranteed by the Mean Value Theorem for Integrals, we need to find the derivative of f(x), set it equal to the average value of f(x) over the interval [0,3], and solve for x.
The derivative of f(x) is given by:
f'(x) = d/dx (x^2 - 2x - 3)
= 2x - 2.
The Mean Value Theorem for Integrals states that there exists at least one value c in the interval [0,3] such that f'(c) = average value of f(x) over the interval [0,3].
Since we found earlier that the average value of f(x) over the interval [0,3] is 1, we can set f'(x) equal to 1:
2x - 2 = 1.
Solving this equation for x gives:
2x = 3,
x = 3/2.
Therefore, the value(s) of C guaranteed by the Mean Value Theorem for Integrals on the interval [0,3] is x = 3/2.
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