(1 point) This is similar to a problem from the Hughes-Hallett textbook. Use the Fundamental Theorem of Calculus to evaluate the average value of f(x)=9x^(2)+2 on the interval x=0 to x=8
To evaluate the average value of a function using the Fundamental Theorem of Calculus, follow these steps:
Step 1: Find the definite integral of the function over the given interval.
Given function: f(x) = 9x^2 + 2
The definite integral of the function is given by:
∫(0 to 8) [9x^2 + 2] dx
Step 2: Evaluate the definite integral.
To evaluate the integral, we can use the power rule of integration:
∫x^n dx = (1/(n+1)) * x^(n+1) + C, where C is the constant of integration.
Applying the power rule to the integral, we have:
∫[9x^2 + 2] dx = (1/3) * x^3 + 2x + C
Now, we evaluate the definite integral from x = 0 to x = 8:
(1/3) * (8^3) + 2(8) - [(1/3) * (0^3) + 2(0)]
= (1/3) * 512 + 16 - 0
= 170.67 + 16
= 186.67
Step 3: Calculate the average value.
To find the average value of the function, divide the definite integral by the length of the interval.
Given interval: x = 0 to x = 8
Length of the interval = 8 - 0 = 8
Average value = (1/length) * definite integral
= (1/8) * 186.67
= 23.3325
Therefore, the average value of the function f(x) = 9x^2 + 2 on the interval x = 0 to x = 8 is approximately 23.3325.
To evaluate the average value of a function using the Fundamental Theorem of Calculus, follow these steps:
Step 1: Find the definite integral of the function over the given interval.
Step 2: Divide the result from Step 1 by the length of the interval.
Let's apply these steps to the given function f(x) = 9x^2 + 2 on the interval [0, 8].
Step 1: Find the definite integral of f(x) from 0 to 8.
∫[0,8] (9x^2 + 2) dx
To integrate the function, we use the power rule for integration:
∫ x^n dx = (1/(n+1)) * x^(n+1) + C
Applying this to the terms of f(x):
∫ 9x^2 dx = (9/(2+1)) * x^(2+1) + C = 3x^3 + C1
∫ 2 dx = 2x + C2
Combining both integrals:
∫[0,8] (9x^2 + 2) dx = [3x^3 + 2x] evaluated from 0 to 8
= [3(8^3) + 2(8)] - [3(0^3) + 2(0)]
= 3(512) + 16 - 0
= 1536 + 16
= 1552
Step 2: Divide the result from Step 1 by the length of the interval.
The length of the interval [0, 8] is 8 - 0 = 8.
Average Value = (1/Length of Interval) * ∫[0,8] (9x^2 + 2) dx
= (1/8) * 1552
= 194
Therefore, the average value of the function f(x) = 9x^2 + 2 on the interval [0, 8] is 194.
The average value of f(x) on the interval [a,b] is just
∫[a,b] f(x) dx / (b-a)
so plug in your numbers.
Come back with your work if you get stuck.
If this problem is only worth 1 point, it shouldn't be too hard ...