# Thread: Integral with Sq Root of u on bottom question

1. ## 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

2. You may wish to note this:

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

Looking easier, yet?

3. $\displaystyle \int \frac{u-2}{u^{1/2}} du = \int u^{1/2}du - 2\int u^{-1/2}du.$

Should be enough for you to find the answer.

4. 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.

Thanks to both of you again

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 $\displaystyle \frac{u}{u^{1/2}}=u^{1-1/2}=u^{1/2}$ by the basic rules of exponents. So $\displaystyle \frac{u-2}{u^{1/2}}=u^{1/2}-2u^{-1/2}.$

The integral portion follows from the basic rule

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

Make sense?

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

The integral portion follows from the basic rule

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

Make sense?
It all clicked after reading that! Thank you so much

7. Technical note: Is $\displaystyle \frac{u}{\sqrt{u}} = \sqrt{u}$?

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)?

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

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 $\displaystyle \frac{x^2- 4}{x- 2}= x+ 2$ as long as $\displaystyle x\ne 2$.

If the lower limit of the integral were 0, that is, if it were $\displaystyle \int_0^4 \frac{u}{\sqrt{u}}du$, 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 $\displaystyle \int_0^4 \frac{u}{\sqrt{u}}du$ with $\displaystyle \lim_{\epsilon\rightarrow 0}^4 u^{1/2} du$. We can do that because u is not 0 there. That gives $\displaystyle \left[\frac{2}{3}u^{3/2}\right]_\epsilon^4= \frac{2}{3}\left(8- \epsilon^{3/2}\right)$. And that goes to $\displaystyle \frac{16}{3}$ as $\displaystyle \epsilon$ goes to 0, the same as if we had must integrated $\displaystyle \int_0^4 u^{1/2} du$

9. 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.