Change of Variables for easier integration/ improper integral

[delgeezee]

\(\displaystyle 2\pi \int_0^4 \! x(4-x)^{-1/3} dx\)


Sh*t don't make sense!

We will start out with U-sub


sub U= 4-x
dU = -1
The book says it turns it this
\(\displaystyle 2\pi \int_0^4 \! (4-U)(U)^{-1/3} du\)

But I think it should be

\(\displaystyle 2\pi - \int_0^4 \! (4-U)(U)^{-1/3} du\)

 

[daon2]

\(\displaystyle 2\pi \int_0^4 \! x(4-x)^{-1/3} dx\)


Sh*t don't make sense!

We will start out with U-sub


sub U= 4-x
dU = -1
The book says it turns it this
\(\displaystyle 2\pi \int_0^4 \! (4-U)(U)^{-1/3} du\)

But I think it should be

\(\displaystyle 2\pi - \int_0^4 \! (4-U)(U)^{-1/3} du\)

How exactly are you getting \(\displaystyle 2\pi\) minus your integral? I think you mean \(\displaystyle -2\pi\) times your integral. If that is what your book has, a minus sign is missing. Neither are correct though. Of course:

\(\displaystyle \displaystyle -2\pi\int_0^4(4-u)u^{-1/3}du = 2\pi\int_0^4(u-4)u^{-1/3}du \)

And \(\displaystyle du = -1dx\), not \(\displaystyle -1\).