Could someone answer this question so I understand it. Thanks
For the function f(x)=-6x^-2-2
Find the average value of f(x) on the interval [0,4].
The average value of f(x) on [0,4] would be
[∫f(x)dx from 1 to 4]/(4-1)
To find the average value of a function f(x) on a given interval [a, b], you need to calculate the definite integral of f(x) over that interval and then divide it by the length of the interval (b - a).
In this case, you are given the function f(x) = -6x^-2 - 2 and the interval [0, 4].
To find the definite integral of f(x) over the interval [0, 4], we need to find the antiderivative of f(x) first. Let's start by finding the antiderivative of -6x^-2.
To find the antiderivative of -6x^-2, we add 1 to the power and divide by the new power. The power of -6x^-2 is -2, so by adding 1 and dividing by the new power, we get:
∫(-6x^-2) dx = -6 * (x^-2 + 1) / -1 = 6/x + C
Now, we can calculate the definite integral of f(x) = -6x^-2 - 2 over the interval [0, 4]:
∫[-6x^-2 - 2] dx = [6/x + 2x] from 0 to 4
To evaluate the integral at the upper and lower limits, we substitute the values into the expression:
[6/4 + 2(4)] - [6/0 + 2(0)]
However, there is an issue with the lower limit of integration. The function f(x) = -6x^-2 - 2 is not defined at x = 0 due to the division by zero. This means that we need to compute the integral from a small positive value ε (a close approximation of zero) instead.
Let's consider ε = 0.0001:
[6/4 + 2(4)] - [6/0.0001 + 2(0.0001)] = 1.5 - 60002
So, the definite integral over the interval [0.0001, 4] is 1.5 - 60002.
To find the average value of f(x) on the interval [0, 4], we divide this definite integral by the length of the interval: 4 - 0.0001 = 3.9999.
Therefore, the average value of f(x) on the interval [0, 4] is approximately (1.5 - 60002) / 3.9999.