# Integral with Sq Root of u on bottom question

• Jul 8th 2009, 07:46 PM
Integral with Sq Root of u on bottom question
Going through my book I've found this problem which I can't seem to find the answer to.

Directions: Evaluate the definite integral of the algebraic function.
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Problem: Inegral symbol (the bottom of the symbol there is a 1 and at the top a 4)

u-2/ u^(1/2) .

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Can anyone help me in answering this? This probably seems like a simple and dumb question, but it seems that my algebra may be a bit weak.

Thanks
• Jul 8th 2009, 07:56 PM
TKHunny
You may wish to note this:

$
\frac{u-2}{\sqrt{u}}\;=\;\sqrt{u} - \frac{2}{\sqrt{u}}\;=\; u^{1/2} - 2u^{-1/2}
$

Looking easier, yet?
• Jul 8th 2009, 08:01 PM
AlephZero
$\int \frac{u-2}{u^{1/2}} du = \int u^{1/2}du - 2\int u^{-1/2}du.$

Should be enough for you to find the answer.
• Jul 8th 2009, 08:25 PM
I can see what both of you did, and thank you. My last question would be what exactly do I do to bring the u^(1/2) to the numerator?

AlephZero: How exactly did you get to the second part ? I'm looking for small steps each time so I can diretly see whats going on. I can't seem to grasp it still. (Headbang)

Thanks to both of you again :)
• Jul 8th 2009, 08:40 PM
AlephZero
Quote:

My last question would be what exactly do I do to bring the u^(1/2) to the numerator?

AlephZero: How exactly did you get to the second part ? I'm looking for small steps each time so I can diretly see whats going on. I can't seem to grasp it still.

I'm not exactly clear on what you mean by "bringing it to the numerator," but $\frac{u}{u^{1/2}}=u^{1-1/2}=u^{1/2}$ by the basic rules of exponents. So $\frac{u-2}{u^{1/2}}=u^{1/2}-2u^{-1/2}.$

The integral portion follows from the basic rule

$\int [f_1(x)+f_2(x)] dx = \int f_1(x) dx + \int f_2(x) dx.$

Make sense?
• Jul 8th 2009, 08:47 PM
Quote:

Originally Posted by AlephZero
I'm not exactly clear on what you mean by "bringing it to the numerator," but $\frac{u}{u^{1/2}}=u^{1-1/2}=u^{1/2}$ by the basic rules of exponents. So $\frac{u-2}{u^{1/2}}=u^{1/2}-2u^{-1/2}.$

The integral portion follows from the basic rule

$\int [f_1(x)+f_2(x)] dx = \int f_1(x) dx + \int f_2(x) dx.$

Make sense?

It all clicked after reading that! Thank you so much :D
• Jul 10th 2009, 06:36 PM
TKHunny
Technical note: Is $\frac{u}{\sqrt{u}} = \sqrt{u}$?

Really? Truly? Absolutely in every way?

It's not particularly important for this problem, since the lower limit is one (1). Would it make a difference if the lower limit were zero (0)?
• Jul 11th 2009, 10:38 AM
HallsofIvy
Quote:

Originally Posted by TKHunny
Technical note: Is $\frac{u}{\sqrt{u}} = \sqrt{u}$?

Really? Truly? Absolutely in every way?

It's not particularly important for this problem, since the lower limit is one (1). Would it make a difference if the lower limit were zero (0)?

Well, obviously not if the the denominator were 0. In the same sense that $\frac{x^2- 4}{x- 2}= x+ 2$ as long as $x\ne 2$.

If the lower limit of the integral were 0, that is, if it were $\int_0^4 \frac{u}{\sqrt{u}}du$, the integrand is not defined at u= 0 and so the integral is an improper integral. Your point is a good one but we can do the following:

Replace $\int_0^4 \frac{u}{\sqrt{u}}du$ with $\lim_{\epsilon\rightarrow 0}^4 u^{1/2} du$. We can do that because u is not 0 there. That gives $\left[\frac{2}{3}u^{3/2}\right]_\epsilon^4= \frac{2}{3}\left(8- \epsilon^{3/2}\right)$. And that goes to $\frac{16}{3}$ as $\epsilon$ goes to 0, the same as if we had must integrated $\int_0^4 u^{1/2} du$
• Jul 11th 2009, 05:28 PM
TKHunny
I was just wondering if anyone ever had introduced Shadoowned to Domain considerations. Ploughing through without validity is bad business. One should at least think about it, I think.