Let f(x) be a function defined on the interval [0, 4] by the equation:
f(x) = x^2 + 2x + 1
(a) Find the average value of f(x) over the interval [0, 4].
(b) Use the Mean Value Theorem to show that there exists at least one value c in the interval (0, 4) such that f'(c) is equal to the average rate of change of f(x) over [0, 4].
(a) To find the average value of f(x) over the interval [0, 4], we need to calculate the definite integral of f(x) over that interval, and then divide it by the length of the interval.
The definite integral of f(x) from 0 to 4 can be calculated as follows:
∫[0, 4] (f(x) dx) = ∫[0, 4] (x^2 + 2x + 1) dx
Expanding and integrating each term separately:
= ∫[0, 4] (x^2) dx + ∫[0, 4] (2x) dx + ∫[0, 4] (1) dx = (1/3)x^3 + x^2 + x |[0, 4] = [(1/3)(4)^3 + (4)^2 + (4)] - [(1/3)(0)^3 + (0)^2 + (0)] = [(1/3)(64) + 16 + 4] - 0 = [64/3 + 48/3 + 12/3] - 0 = (124/3) - 0 = 124/3
The length of the interval [0, 4] is 4 - 0 = 4.
Therefore, the average value of f(x) over the interval [0, 4] is:
( Average value ) = ( Definite Integral ) / ( Length of Interval ) = (124/3) / 4 = 31/3
Hence, the average value of f(x) over the interval [0, 4] is 31/3.
(b) According to the Mean Value Theorem, if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one value c in the open interval (a, b) such that:
f'(c) = ( f(b) - f(a) ) / ( b - a )
In this problem, we have already found the average rate of change of f(x) over the interval [0, 4], which is 31/3. We can now use the Mean Value Theorem to show the existence of a value c in the interval (0, 4), for which f'(c) is equal to the average rate of change.
The derivative of f(x) = x^2 + 2x + 1 is:
f'(x) = 2x + 2
Now, we can write the Mean Value Theorem equation as:
f'(c) = ( f(4) - f(0) ) / (4 - 0)
Substituting the values, we have:
2c + 2 = ( 4^2 + 2(4) + 1 - 0^2 - 2(0) - 1 ) / 4 2c + 2 = ( 16 + 8 + 1 - 0 - 1 ) / 4 2c + 2 = ( 24 ) / 4 2c + 2 = 6 2c = 6 - 2 2c = 4 c = 4/2 c = 2
Hence, using the Mean Value Theorem, we have found that there exists at least one value c in the interval (0, 4) such that f'(c) is equal to the average rate of change of f(x) over [0, 4]. In this case, c = 2.