help help help

[chica2006]

Hi,
I need help with the following question. Not sure where I went wrong.

Evaluate:
[integral] 4 0 (x-6)^2dx

here is what i have:

[integral] 1(x-6)^2dx
=1/2 [integral] (x-6)^2(1xdx)
=1/2 [integral] u^2du
=1/2*u^3/3+C
=1/6 (x-6)^3+C

-->
[integral] 4 0 (x-6)^2 = [1/2(x-6)^3]4 0
=1/6[(4-6)^3 - (0-6)^3]
=1/2[(-2)^3-(-6)^3]
=1/2(-8-(-216))
=1/2(-224)
= -112

*what did I do wrong???? aghhh helllpppppp

 

[stapel]

Evaluate:
[integral] 4 0 (x-6)^2dx

Does the above mean the following?

. . . . .\(\displaystyle \large{\int_4^0{\,(x\,-\,6)^2\,}dx}\)

Have you considered multiplying out the square, and then integrating term-by-term?

[integral] 1(x-6)^2dx

What is the purpose of the "1" in front of the parentheses?

Thank you.

Eliz.

 

[tkhunny]

From where does the ½ keep coming?

\(\displaystyle \L\,\int{x^{2}\,dx\,=\,\frac{1}{3}x^{3}\,+\,C}\)

There's no ½ in there.

 

[chica2006]

Wow ... I think maybe i just got it. the antiderivitive of x^2 is basically multiplying 1/3 by x^3 which would be x^3/3. but say in this problem it is (x-6)^2 so I would multiply (x-6) by 1/3 and change the power to 3. So it would be 1/3(x-6)^3 ... is this right?????

 

[pka]

[the] problem it is (x-6)^2 so I would multiply (x-6) by 1/3 and change the power to 3. So it would be 1/3(x-6)^3 ... is this right?????

Yes that is correct.