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
For Further Reading
* calc check: average value - bobpursley, Saturday, June 30, 2007 at 6:06am
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.
-------------------
How did you get u=8? wouldn't it be u=12 when x=2?
To find this don't I just plug in 2 for f(x), f(2)=(2)^2*Sqrt(1+(2)^2)) = 12
and then I would evaluate:
(3/2)[(2/3)(x+1)^(3/2)] at 0 and 12?
Please let me know if this is correct. Thanks.
NEVERMIND...dumb mistake on my part. I don't know what I was think. I know where the 0 and 8 come from. Thanks.
To find the average value of a function on an interval, you can use the formula:
f_ave = (1/(b-a)) * integral from a to b of f(x) dx
Let's apply this formula to find the average value of the function f(x) = x^2 * sqrt(1+x^3) on the interval [0,2].
First, we substitute the limits of integration into the formula:
f_ave = (1/(2-0)) * integral from 0 to 2 of x^2 * sqrt(1+x^3) dx
Next, we can simplify the integral by substituting u = x^3. Therefore, du = 3x^2 dx.
Substituting u and du into the integral:
f_ave = (1/2) * integral from 0 to 2 of sqrt(1+u) * 3 du
Now, let's evaluate the integral.
Using the power rule for integration, we have:
integral of sqrt(1+u) du = (2/3) * (1+u)^(3/2) + C, where C is the constant of integration.
Substituting the limits of integration:
f_ave = (1/2) * [(2/3) * (1+u)^(3/2)] evaluated from 0 to 2
When u = 0, we have:
f_ave = (1/2) * [(2/3) * (1+0)^(3/2)]
When u = 2, we have:
f_ave = (1/2) * [(2/3) * (1+2)^(3/2)]
Calculating the values:
f_ave = (1/2) * [(2/3) * (1)^(3/2)] - (1/2) * [(2/3) * (3)^(3/2)]
Simplifying:
f_ave = (1/2) * (2/3) - (1/2) * (2/3) * (3)^(3/2)
f_ave = 1/3 - (1/2) * (2/3) * (3)^(3/2)
f_ave ≈ 4.1962
Therefore, the average value of the function f(x) = x^2 * sqrt(1+x^3) on the interval [0,2] is approximately 4.1962.
Yes, you made a mistake in your calculations. The correct limits of integration when changing variables from x to u are u=0 when x=0, and u=8 when x=2. So the corrected calculation would be:
f ave = (3/2)*integral of 0 to 8: sqrt(1+u) du
To evaluate this integral, you can use the power rule of integration:
= (3/2) * [(2/3)*(u+1)^(3/2)] evaluated from 0 to 8
= (3/2) * [(2/3)*(8+1)^(3/2) - (2/3)*(0+1)^(3/2)]
= (3/2) * [(2/3)*(9)^(3/2) - (2/3)*(1)^(3/2)]
= (3/2) * [(2/3)*27 - (2/3)*1]
= (3/2) * [18 - 2]
= (3/2) * 16
= 24
So the average value of the function f(x) = x^2 sqrt(1+x^3) on the interval [0,2] is 24.