Find the average value of f(x) = x^3 + 3 for 0 ≤ x ≤ 2 and find all values of x* described in the Mean Value
Theorem for Integrals.
To find the average value of a function f(x) over the interval [a, b], you need to calculate the definite integral of f(x) from a to b and then divide it by the length of the interval (b - a). The average value of f(x) is denoted as f_avg and can be obtained using the following formula:
f_avg = (1 / (b - a)) * ∫[a, b] f(x) dx
In this case, we want to find the average value of f(x) = x^3 + 3 for the interval 0 ≤ x ≤ 2.
Step 1: Calculate the definite integral of f(x) from 0 to 2
∫[0, 2] (x^3 + 3) dx = [x^4/4 + 3x] evaluated from 0 to 2
= (2^4/4 + 3*2) - (0^4/4 + 3*0)
= (16/4 + 6) - (0 + 0)
= (4 + 6) - 0
= 10
Step 2: Calculate the length of the interval
The length of the interval [0, 2] is given by (2 - 0) = 2.
Step 3: Divide the integral result by the length of the interval
f_avg = (1 / (2 - 0)) * 10
= 1/2 * 10
= 5
Therefore, the average value of f(x) = x^3 + 3 over the interval 0 ≤ x ≤ 2 is 5.
Now, let's find all the values of x* that satisfy the Mean Value Theorem for Integrals. According to the theorem, there exists at least one value x* in the interval [0, 2] such that the instantaneous rate of change of f(x) at x* is equal to the average rate of change of f(x) over the same interval.
The instantaneous rate of change of f(x) can be found by taking the derivative of f(x). In this case, f(x) = x^3 + 3, so f'(x) = 3x^2.
Since the average rate of change is equal to the slope of the secant line connecting the endpoints, we have:
Average rate of change = (f(2) - f(0)) / (2 - 0)
Substituting the values:
Average rate of change = (2^3 + 3 - 0^3 - 3) / 2
= (8 + 3 - 0 - 3) / 2
= 8/2
= 4
Now, we need to find the value of x* such that f'(x*) = 4. Equating f'(x) = 4, we get:
3x^2 = 4
Solving this equation gives us:
x^2 = 4/3
Taking the square root of both sides:
x = ±√(4/3)
So, the values of x* that satisfy the Mean Value Theorem for Integrals are x = √(4/3) and x = -√(4/3).