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September 2, 2014

September 2, 2014

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

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.

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