Find the average value of the function "f(x) = x^2 sqrt(1+x^3)" on the interval [0,2].
and this is what i did.. please check for mistakes. thanks :D
f(x) = x^2 sqrt(1+x^3), [0,2]
f ave = (1/(b-a))*inegral of a to b for: f(x) dx
f ave = (1/(2-0))*integral of 0 to 2 for: x^2 sqrt(1+x^3) dx
..let u = x^3 & du = 3x^2 dx
f ave = (1/2)*integral of 0 to 2 for: sqrt(1+u)*3 du
f ave = (3/2)*integral of 0 to 2 for: sqrt(1+u) du
= (3/2)[(2/3)(x+1)^(3/2)] from 0 to 2
= (3/2)[(2/3)((2)+1)^(3/2)] - (3/2)[(2/3)((0)+1)^(3/2)]
= (5.1962 - 1)
= 4.1962
When you change variables, you have to change limits of integration. When
x=0, u=0; when x=2, u=8
That will change the answer.
The approach you used to find the average value of the function is correct, but there is a mistake in the calculation of the integral.
Let's go through the steps to find the average value of the function f(x) = x^2 sqrt(1+x^3) on the interval [0,2] again:
1. Start with the formula for the average value of a function:
f_ave = (1/(b-a)) * integral from a to b of f(x) dx
2. Substituting the values into the formula:
f_ave = (1/(2-0)) * integral from 0 to 2 of x^2 sqrt(1+x^3) dx
3. To simplify the integral, let's make the substitution u = 1+x^3, then du = 3x^2 dx:
f_ave = (1/2) * integral from 0 to 2 of sqrt(u) (1/3) du
4. Now we can integrate:
f_ave = (1/2) * (1/3) * integral from 0 to 2 of sqrt(u) du
f_ave = (1/6) * integral from 0 to 2 of u^(1/2) du
f_ave = (1/6) * (2/3) * u^(3/2) evaluated from 0 to 2
f_ave = (1/9) * (2^(3/2) - 0^(3/2))
f_ave = (1/9) * (2^(3/2))
f_ave = 2^(1/2)/9
The correct average value of the function f(x) = x^2 sqrt(1+x^3) on the interval [0,2] is 2^(1/2)/9, which is approximately 0.1046.