AP Calculus AB Exam Question:

Let f(x) be the function defined by f(x) = 2x^2 + 5x - 3 on the interval [1, 4]. Find the average value of f(x) over this interval.

Answer:

To find the average value of a function f(x) over a closed interval [a, b], we use the formula:

Average value = (1 / (b - a)) * ∫[a, b] f(x) dx

Given that f(x) = 2x^2 + 5x - 3 and the interval is [1, 4], we can calculate the average value as follows:

First, let's find the definite integral of f(x) over the given interval:

∫[1, 4] f(x) dx = ∫[1, 4] (2x^2 + 5x - 3) dx = [ (2/3)x^3 + (5/2)x^2 - 3x ] evaluated from 1 to 4 = [(2/3)(4^3) + (5/2)(4^2) - 3(4)] - [(2/3)(1^3) + (5/2)(1^2) - 3(1)] = [(2/3)(64) + (5/2)(16) - 12] - [(2/3)(1) + (5/2)(1) - 3] = [(128/3) + (40/2) - 12] - [(2/3) + (5/2) - 3] = (128/3 + 80/3 - 36/3) - (2/3 + 15/6 - 18/6) = (192/3 - 18/3) - (4/3 - 3/6) = (174/3) - (5/6) = 174/3 - 10/6 = (348 - 10) / 6 = 338 / 6 = 169/3

Now, let's calculate the average value by multiplying the integral by (1 / (b - a)):

Average value = (1 / (4 - 1)) * ∫[1, 4] f(x) dx = (1 / 3) * (169/3) = 169/9

Hence, the average value of f(x) over the interval [1, 4] is 169/9.