AP Calculus AB Exam Question:

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

Solution:

To find the average value of a function on a given interval, we use the formula:

Average value = 1/(b-a) * ∫(a to b) f(x) dx

In this case, the interval is [1, 4] and the function is f(x) = 2x^2 + 3x - 5. So, we have:

Average value = 1/(4-1) * ∫(1 to 4) (2x^2 + 3x - 5) dx

To find the integral of the function, we can use the power rule of integration. The power rule states that for any real number n (except -1), the integral of x^n with respect to x is (1/(n+1))x^(n+1).

∫(2x^2 + 3x - 5) dx = 2/3 * x^3 + 3/2 * x^2 - 5x

Now, we can evaluate the integral from 1 to 4:

∫(1 to 4) (2x^2 + 3x - 5) dx = [2/3 * x^3 + 3/2 * x^2 - 5x] from 1 to 4

= (2/3 * 4^3 + 3/2 * 4^2 - 5 * 4) - (2/3 * 1^3 + 3/2 * 1^2 - 5 * 1)

= (128/3 + 48 - 20) - (2/3 + 3/2 - 5)

= (128/3 + 48 - 20) - (2/3 + 9/2 - 15/1)

= (128/3 + 48 - 20) - (2/3 + 27/6 - 30/2)

= (128/3 + 48 - 20) - (2/3 + 27/6 - 60/6)

= (128/3 + 48 - 20) - (2/3 + 27/6 - 10)

= (128/3 + 48 - 20) - (2/3 + 27/6 - 20/2)

= (128/3 + 48 - 20) - (2/3 + 27/6 - 40/2)

= 128/3 + 48 - 20 - 2/3 - 27/6 + 80/6

= 128/3 + 48 - 20 - 2/3 - 9/2 + 40/2

= 128/3 + 48 - 20 - 2/3 - 27/6 + 20

= 128/3 + 48 - 20 - 2/3 - 9/2 + 20

= 128/3 - 2/3 + 48 - 9/2 + 20

= (128 - 2)/3 + 48 - (9/2) + 20

= 126/3 + 48 - (9/2) + 20

= 42 + 48 - (9/2) + 20

= 90 - (9/2) + 20

= 90 - (9/2) + 40/2

= 90 - (9/2) + 20

= 106 - (9/2)

= 106 - 4.5

= 101.5

Therefore, the average value of the function f(x) = 2x^2 + 3x - 5 on the interval [1, 4] is 101.5.