Find the average value of f(x) on the interval [-1, 3] when f'(x)=3x^2-6x and f(2)=0.
the average value of f(x) on [a,b] is
∫[a,b] f(x) dx
-----------------------
b-a
f' = 3x^2-6x
f = x^3-3x^2+4
so, you want
∫[-1,3] x^3-3x^2+4 dx
--------------------------
3 - (-1)
which I'm sure you can do.
To find the average value of f(x) on the interval [-1, 3], we can use the Mean Value Theorem for Definite Integrals.
The Mean Value Theorem states that if f(x) is continuous on the interval [a, b] and differentiable on the interval (a, b), then there exists at least one value c in (a, b) such that the average value of f(x) on [a, b] is equal to f'(c).
In this case, f'(x) = 3x^2 - 6x.
To find the average value of f(x), we need to find the value of c in the interval (-1, 3) such that f'(c) is equal to the average rate of change of f(x) on the interval [-1, 3].
The average rate of change of f(x) on the interval [-1, 3] can be found by evaluating f(3) - f(-1) divided by 3 - (-1), which simplifies to f(3) - f(-1) divided by 4.
Given that f(2) = 0, we can substitute this value into the formula:
f(3) - f(-1) = f(3) - f(2)
Now, to find f(3), we need to integrate f'(x) from a point to 3. Since we know f'(x), we can integrate it:
∫(3x^2 - 6x) dx
= x^3 - 3x^2 + C
To find the constant C, we can use the given condition f(2) = 0:
f(2) = (2)^3 - 3(2)^2 + C
0 = 8 - 12 + C
C = 12 - 8
C = 4
So, f(x) = x^3 - 3x^2 + 4.
Now, to find f(3), we substitute x = 3 into the equation:
f(3) = (3)^3 - 3(3)^2 + 4
f(3) = 27 - 27 + 4
f(3) = 4
Now, we can calculate the average value of f(x) on the interval [-1, 3] by finding the value of c such that f'(c) is equal to the average rate of change:
Average value of f(x) = f'(c) = 3c^2 - 6c
Setting this equal to 4, we have:
3c^2 - 6c = 4
3c^2 - 6c - 4 = 0
Using quadratic formula, we can find the roots of this equation:
c = [6 ± sqrt(6^2 - 4(3)(-4))] / (2 * 3)
c = [6 ± sqrt(36 + 48)] / 6
c = [6 ± sqrt(84)] / 6
c = (6 ± 2sqrt(21)) / 6
Simplifying further, we get:
c = 1 ± (1/3)sqrt(21)
Therefore, the average value of f(x) on the interval [-1, 3] is given by f'(c), which is 3c^2 - 6c:
Average value of f(x) = f'(c) = 3(1 ± (1/3)sqrt(21))^2 - 6(1 ± (1/3)sqrt(21))
To find the average value of a function on a closed interval, you need to calculate the definite integral of the function over the interval and then divide the result by the length of the interval.
The given function is f'(x) = 3x^2 - 6x. To find the function f(x), we integrate f'(x) with respect to x.
∫(3x^2 - 6x) dx = x^3 - 3x^2 + C
Since f(2) = 0, we can find the value of C using this information.
f(2) = 2^3 - 3(2)^2 + C = 0
8 - 12 + C = 0
-4 + C = 0
C = 4
Therefore, the function f(x) is given by f(x) = x^3 - 3x^2 + 4.
To find the average value of f(x) on the interval [-1, 3], we need to calculate the definite integral of f(x) over this interval and then divide the result by the length of the interval.
∫[-1, 3] (x^3 - 3x^2 + 4) dx
To evaluate this definite integral, we can find the antiderivative of f(x) and then substitute the limits of integration.
F(x) = ∫(x^3 - 3x^2 + 4) dx = (1/4)x^4 - x^3 + 4x + K
Using the Fundamental Theorem of Calculus, we evaluate F(3) - F(-1).
F(3) - F(-1) = [(1/4)(3)^4 - (3)^3 + 4(3)] - [(1/4)(-1)^4 - (-1)^3 + 4(-1)]
= [81/4 - 27 + 12] - [1/4 + (-1) + (-4)]
= 81/4 - 15 - 9/4
= 81/4 - 60/4 - 9/4
= 12/4
= 3
Since the length of the interval [-1, 3] is 4, we divide 3 by 4 to find the average value.
Average value of f(x) on the interval [-1, 3] = 3/4