The average value of a function f(x) on the interval [a, b] is given by the formula:

fave = 1/(b - a) ∫ from a to b f(x) dx

Here, f(x) = 3x^2 + 8x and the interval is [-1, 5]. So, a = -1 and b = 5.

First, we need to compute the integral of f(x) from -1 to 5.

∫ from -1 to 5 f(x) dx = ∫ from -1 to 5 (3x^2 + 8x) dx

To compute the integral, we use the power rule for integration, which states that the integral of x^n dx is (1/(n+1))x^(n+1).

So, the integral of 3x^2 dx is (3/3)x^3 = x^3 and the integral of 8x dx is (8/2)x^2 = 4x^2.

Therefore, ∫ from -1 to 5 (3x^2 + 8x) dx = [x^3 + 4x^2] evaluated from -1 to 5.

= [(5)^3 + 4*(5)^2] - [(-1)^3 + 4*(-1)^2]
= [125 + 100] - [-1 + 4]
= 225 - 3
= 222

Then, we substitute a = -1 and b = 5 into the formula for fave.

fave = 1/(5 - (-1)) * 222
= 1/6 * 222
= 37

So, the average value of the function f(x) = 3x^2 + 8x on the interval [-1, 5] is 37.

Question

The average value of a function f(x) on the interval [a, b] is given by the formula:

fave = 1/(b - a) ∫ from a to b f(x) dx

Here, f(x) = 3x^2 + 8x and the interval is [-1, 5]. So, a = -1 and b = 5.

First, we need to compute the integral of f(x) from -1 to 5.

∫ from -1 to 5 f(x) dx = ∫ from -1 to 5 (3x^2 + 8x) dx

To compute the integral, we use the power rule for integration, which states that the integral of x^n dx is (1/(n+1))x^(n+1).

So, the integral of 3x^2 dx is (3/3)x^3 = x^3 and the integral of 8x dx is (8/2)x^2 = 4x^2.

Therefore, ∫ from -1 to 5 (3x^2 + 8x) dx = [x^3 + 4x^2] evaluated from -1 to 5.

= [(5)^3 + 4*(5)^2] - [(-1)^3 + 4*(-1)^2]
= [125 + 100] - [-1 + 4]
= 225 - 3
= 222

Then, we substitute a = -1 and b = 5 into the formula for fave.

fave = 1/(5 - (-1)) * 222
= 1/6 * 222
= 37

So, the average value of the function f(x) = 3x^2 + 8x on the interval [-1, 5] is 37.

Knowee AI · Accepted Answer