Find the average value of the function f over the indicated interval [0, 8].
f(x) = 7/(x + 1)
I keep getting the wrong answer.
As I showed you earlier, it is
∫[0,8] 7/(x+1) dx
----------------------
(8-0)
= 1/8 (7log(x+1) [0,8])
= 7/8 (log9-log1)
= 7/8 log9
what were your solution steps?
The thing is that I did get 7/8(log9-log1) I got 7/8(log8-log1).
but thanks for your help.
To find the average value of a function over an interval, you need to evaluate the definite integral of the function over that interval and then divide it by the length of the interval. Let's go through the steps to find the average value of the function f(x) = 7/(x + 1) over the interval [0, 8]:
Step 1: Evaluate the definite integral of f(x) over the interval [0, 8]
The definite integral of f(x) over the interval [0, 8] can be calculated as follows:
∫[0, 8] 7/(x + 1) dx
Step 2: Simplify the integral
To simplify the integral, we can use the natural logarithm function:
= 7ln|x + 1| evaluated from 0 to 8
Step 3: Evaluate the integral at the upper and lower limits
At x = 8:
= 7ln|8 + 1|
= 7ln9
At x = 0:
= 7ln|0 + 1|
= 7ln1
= 0
Step 4: Calculate the average value
The average value of f(x) over the interval [0, 8] can be found by dividing the definite integral by the length of the interval. In this case, the length of the interval is 8 - 0 = 8.
Average value = (1/8) * ∫[0, 8] 7/(x + 1) dx
Substituting the evaluated integral values:
Average value = (1/8) * (7ln9 - 0) = 7/8 * ln9
Thus, the average value of the function f(x) = 7/(x + 1) over the interval [0, 8] is (7/8) ln(9).