Posted by COFFEE on Monday, July 2, 2007 at 2:58pm.
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)
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.
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