Integration by parts

**spazzed** · Jul 4th 2011, 07:53 AM

Hi all, I need some help. I'm taking system dynamics and control over the summer, but it's been over 3 years since my last calculus course, and some of these integrals are killing me.

$\int_1^{2.5} \frac{1}{0.05-0.02\sqrt{H}}\,dH$

is evaluated to be

$-250ln(0.05-0.02\sqrt{H})-100\sqrt{H}+250$ before plugging integration limits, but I can't even get here.

I've tried substituting the lower half with X, then going from there which gets me a partial answer, and I've tried integrating by parts. I'm so out of mental shape it hurts.

Happy 4th

**TKHunny** · Jul 4th 2011, 08:11 AM

Parts will get you there. How did you do it?

You may wish to simplify your life a little, unless you are married to significant figures, and rewrite as $\frac{100}{5-2\sqrt{H}}$ .

**spazzed** · Jul 4th 2011, 08:35 AM

$u = \frac {100} {5-2\sqrt{H}}$
$dv = dH \, v = H$
$du = 100 ln(5-2\Sqrt{H})\frac {-2}{\sqrt{H}}\,dH$

$\frac {100H} {5-2\sqrt{H}} + 200 \int_1^{2.5} \sqrt{H}ln(5-2\sqrt{H})\, dH$

Have I made a mistake anywhere? If no, how do I go about evaluating the vdu integral?

**TKHunny** · Jul 4th 2011, 09:42 AM

Well done. You might have resisted trying to simplify $\frac{H}{\sqrt{H}}$ . You'll need that back.

Try a substitution: $u = 5 - 2\sqrt{H}$ .

**spazzed** · Jul 4th 2011, 11:01 AM

$u = 5 - 2\sqrt{H}$
$du = \frac{-2}{\sqrt{H}}\,dH$
$\frac{100H}{5-2\sqrt{H}} + 200 \int_1^{2.5} H ln(u) du$

where does the leftover H term go? do you just consider it another constant now? and how do I have to alter the limits of integration for this again?

I would really appreciate it if you could just give me a step by step. This would be like a quick review, then. I took calc 1 in 2004, I haven't been in school for 2.5 years, and I'm jumping back to finish my last year playing catchup in higher engineering classes, and I learn best by seeing a solution then picking it apart.

**TKHunny** · Jul 4th 2011, 11:10 AM

Use your definition of 'u' and solve for H. Good work. That's kind of a mess. Of course, there's still some more work to go.

Note: There may be other (and possibly MUCH EASIER) ways.

**tom@ballooncalculus** · Jul 4th 2011, 12:06 PM

I don't think parts is working for you there - getting your u's and du's muddled. I would of course recommend lazy integration by parts...

http://www.ballooncalculus.org/examp...nce.html#parts

... but not in this case. A simple u-sub would be H = u^2.

Just in case a picture helps...

i.e. replace H with something... as we said, u^2...

... where (key in spoiler) ...

Spoiler:

So,

$F(u^2) = \sqrt{u^2} + \frac{5}{2} \ln |5 - 2\sqrt{u^2}| + c$

i.e...

$I = \sqrt{H} + \frac{5}{2} \ln |5 - 2\sqrt{H}| + c$

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