# Double check

• Oct 24th 2009, 10:01 PM
xwrathbringerx
Double check
Umm for this question:

If f(x) = x^2(x-3) calculate the area between the X-axis, the lines x=4 and x=6 and the curve y=sqrt(f(x))

I keep on getting [12(sqrt(3)-1)]/5

BUT the txtbk answer is [12(4sqrt(3)-1)]/5

• Oct 24th 2009, 10:08 PM
Jameson
Quote:

Originally Posted by xwrathbringerx
Umm for this question:

If f(x) = x^2(x-3) calculate the area between the X-axis, the lines x=4 and x=6 and the curve y=sqrt(f(x))

I keep on getting [12(sqrt(3)-1)]/5

BUT the txtbk answer is [12(4sqrt(3)-1)]/5

I can confirm your setup if you post what you did.
• Oct 24th 2009, 10:29 PM
xwrathbringerx
Sori it's a little messy
• Oct 24th 2009, 10:53 PM
Jameson
I'm trying to read through your work, but in the mean time can you tell me how you got the integral $A=\int_{4}^{6}x\sqrt{x-3}dx$?
• Oct 24th 2009, 11:18 PM
Jameson
Ok, so let's just do this the easy way. We need to figure out if you made a wrong calculation or a conceptual error.

I don't have a calculator on me, so you should evaluate the definite integral you start with from your work and get the decimal answer if you have to. Then convert the two possible answers you have to decimals and see which one is right. If you get the book answer, your work has an error. If you get your answer, then either the book is wrong or you set up the problem wrong.

This is why I want you to explain your setup.