Find the average value of the function f(x)= 8/ (3+x)^2 on the interval [0,2]
To find the average value of a function on an interval, you need to evaluate the definite integral of the function over that interval, and then divide the result by the length of the interval.
In this case, the function is f(x) = 8/(3+x)^2 and the interval is [0, 2].
Step 1: Evaluate the definite integral
To find the definite integral, you need to first find the antiderivative of the function and then evaluate it at the interval's bounds.
The antiderivative of f(x) = 8/(3+x)^2 can be obtained by using the u-substitution method. Let u = 3+x, then du = dx.
So, the antiderivative becomes ∫8/u^2 du, which simplifies to -8/u + C, where C is the constant of integration.
Now, evaluate the antiderivative at the upper and lower bounds of the interval:
F(2) - F(0) = [-8/(3+2) + C] - [-8/(3+0) + C]
= [-8/(5) + C] - [-8/(3) + C]
= -8/5 + 8/3 - (-8/3 + C)
= -8/5 + 8/3 + 8/3 - C
= -8/5 + 16/3 - C
Step 2: Calculate the interval's length
The length of the interval [0, 2] is obtained by subtracting the lower bound from the upper bound: 2 - 0 = 2.
Step 3: Divide the integral result by the interval's length
To find the average value, divide the result from step 1 by the length of the interval from step 2:
(-8/5 + 16/3 - C) / 2
Simplifying further, we get:
(-16/10 + 32/6 - C) / 2
(-8/5 + 16/3 - C) / 2
So, the average value of the function f(x) = 8/(3+x)^2 on the interval [0,2] is (-8/5 + 16/3 - C) / 2.