1. ## Quick Problems

I'm tutoring someone in Calc I over the summer and she came to me with the following two problems on a pre-test. It's been awhile since I've done u-substition, and I think I've got it right, but don't want to set her on the wrong track. The problems:

1) (integral of) ((ln(rt(x))) / x )dx

2) (integral from -8 to 0) (1 / (rt(1-x)))dx

I didn't get much out of the first, but the second, I found it to be -2 + 2(rt(-7)). Would someone be able to verify this result, as well as assist with the first problem?

2. Originally Posted by Physics-is-Phun
1) (integral of) ((ln(rt(x))) / x )dx

2) (integral from -8 to 0) (1 / (rt(1-x)))dx
Does 'rt' mean root? If so $\displaystyle \ln(\sqrt{x})=\frac{\ln(x)}{2}$.

BTW: Typing $$\sqrt{x}$$ gives $\displaystyle \sqrt{x}$

3. Originally Posted by Physics-is-Phun
I'm tutoring someone in Calc I over the summer and she came to me with the following two problems on a pre-test. It's been awhile since I've done u-substition, and I think I've got it right, but don't want to set her on the wrong track. The problems:

1) (integral of) ((ln(rt(x))) / x )dx

2) (integral from -8 to 0) (1 / (rt(1-x)))dx

I didn't get much out of the first, but the second, I found it to be -2 + 2(rt(-7)). Would someone be able to verify this result, as well as assist with the first problem?

1) $\displaystyle \ln{\sqrt{x}} = \frac{1}{2}\ln{x}$

$\displaystyle u = \ln{x}$ ... $\displaystyle du = \frac{1}{x} \, dx$

2) $\displaystyle u = 1-x$

$\displaystyle du = -dx$

$\displaystyle -\int_{-8}^1 \frac{1}{\sqrt{1-x}} \, (-dx) = -\int_9^1 \frac{du}{\sqrt{u}} = \int_1^9 \frac{du}{\sqrt{u}}$

you should get 4 for this integral

4. Ah- thank you for the correction on the second problem. That was rather short-sighted of me. @_@

As for the first problem, I got that far- but I couldn't remember how to proceed with any degree of certainty. Unfortunately, she doesn't have the answer, as it's a "pre-test" problem. What would be the next step moving forward from "u" and "du"?

Aside- thanks for the bracket thingy. I forgot that you could do stuff like that on forums.

5. Originally Posted by Physics-is-Phun
As for the first problem, I got that far- but I couldn't remember how to proceed with any degree of certainty. Unfortunately, she doesn't have the answer, as it's a "pre-test" problem. What would be the next step moving forward from "u" and "du"?
Have you rewritten the integrand in terms of u and du only with no x's or dx?

6. That was actually my difficulty- because the problem has $\displaystyle 1/x$, I wasn't sure if $\displaystyle 1/\sqrt{x}$ could be pulled outside the integral (or something to that effect), which would then leave the proper dx substitution of $\displaystyle (1/\sqrt{x})dx$.

Like I said, I got as far as figuring out what the substitution was supposed to be ($\displaystyle u = ln(\sqrt{x})$ , $\displaystyle du = (1/\sqrt{x})dx$). As far as actually implementing the substitution, it's been awhile since I've done u-sub (much to my chagrin), and I want to get this right for my student- especially as it's a pre-test problem. If I had the next step, I could probably take the problem from there- it's just the next step that's killing me.

7. Originally Posted by Physics-is-Phun
Like I said, I got as far as figuring out what the substitution was supposed to be ($\displaystyle u = ln(\sqrt{x})$ , $\displaystyle du = (1/\sqrt{x})dx$).
If
$\displaystyle u = ln(\sqrt{x})$

then

$\displaystyle du=\frac{1}{2x}dx$

8. Ah... that would be where I first went wrong. That makes the problem so much simpler!

Thank you! I knew I was missing something goofy.