Answer:60
Explanation:
To find the average value of a function on a given interval, we need to find the definite integral of the function over that interval and divide it by the width of the interval.
In this case, we are given the function f(x) = 6x³ and the interval 1 ≤ x ≤ 3. To find the definite integral of f(x) over the interval [1, 3],
we can use the power rule of integration.
First, we integrate each term of the function. The integral of x^n is (1/(n+1)) * x^(n+1).
Integrating 6x³ term by term, we get: ∫(6x³) dx = 6 * ∫(x³) dx = 6 * (1/4) * x^4 + C
Now, we evaluate the definite integral by substituting the upper and lower limits of the interval: ∫[1, 3] (6x³) dx = [6 * (1/4) * x^4] evaluated from 1 to 3 = 6 * (1/4) * 3^4 - 6 * (1/4) * 1^4 = 6 * (1/4) * 81 - 6 * (1/4) * 1 = 6 * 81/4 - 6/4 = 486/4 - 6/4 = 480/4 = 120
The definite integral of f(x) = 6x³ over the interval [1, 3] is equal to 120.
To find the average value of the function, we divide the definite integral by the width of the interval: Average value = (1/(b - a)) * ∫[a, b] f(x) dx
In this case, a = 1 and b = 3, so the width of the interval is 3 - 1 = 2.
Average value = (1/2) * 120 = 60
Therefore, the average value of the function f(x) = 6x³ on the interval 1 ≤ x ≤ 3 is 60.