# New more challenging and interesting calculus puzzle

• Jan 31st 2011, 11:29 AM
wonderboy1953
New more challenging and interesting calculus puzzle
You're not going to believe the answer to this integral:

$\int u^2\sqrt{(a + bu)} du$ where a and b are constants
• Jan 31st 2011, 11:33 AM
Ackbeet
You can just do v = a + bu.
• Jan 31st 2011, 11:41 AM
wonderboy1953
Quote:

Originally Posted by Ackbeet
You can just do v = a + bu.

Can you go on to solve it? (I'm sure you can, see what answer pops up)
• Jan 31st 2011, 11:49 AM
Ackbeet
Um, sure. Here goes. I get:

Spoiler:

$\displaystyle\int u^{2}\sqrt{a+bu}\,du=\frac{1}{b^{3}}(a+bu)^{3/2}\left(\frac{2(a+bu)^{2}}{7}-\frac{4a(a+bu)}{5}+\frac{2a^{2}}{3}\right)+C.$
• Jan 31st 2011, 11:53 AM
wonderboy1953
Quote:

Originally Posted by Ackbeet
Um, sure. Here goes. I get:

Spoiler:

$\displaystyle\int u^{2}\sqrt{a+bu}\,du=\frac{1}{b^{3}}(a+bu)^{3/2}\left(\frac{2(a+bu)^{2}}{7}-\frac{4a(a+bu)}{5}+\frac{2a^{2}}{3}\right)+C.$

I'll let you know tomorrow as Howard Anton's calculus book seems to give a different answer.
• Jan 31st 2011, 12:02 PM
Ackbeet
This answer is entirely equivalent to the following:

Spoiler:

$\displaystyle\sqrt{a+bu}\left(\frac{16a^{3}}{105 b^{3}}-\frac{8a^{2}u}{105 b^{2}}+\frac{2au^{2}}{35b}+\frac{2u^{3}}{7}\right) +C.$

Just multiply my answer out to get the other. Is this what your book has?
• Feb 1st 2011, 01:23 PM
wonderboy1953
Quote:

Originally Posted by Ackbeet
This answer is entirely equivalent to the following:

Spoiler:

$\displaystyle\sqrt{a+bu}\left(\frac{16a^{3}}{105 b^{3}}-\frac{8a^{2}u}{105 b^{2}}+\frac{2au^{2}}{35b}+\frac{2u^{3}}{7}\right) +C.$

Just multiply my answer out to get the other. Is this what your book has?

The book has $\dfrac{2}{105b^3}(15b^2u^2 - 12abu +8a^2)(a + bu)^{3/2} + C$ which looks similar to your answer and is probably the same (after algebra verification).

It's interesting to note that such a simple looking integral leads to a monstrous result.