Quick Problems

**Physics-is-Phun** · August 5th 2009, 01:56 PM

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?

Thanks in advance!

**Plato** · August 5th 2009, 02:09 PM

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 $\ln(\sqrt{x})=\frac{\ln(x)}{2}$ .

BTW: Typing [tex]\sqrt{x}[/tex] gives $\sqrt{x}$

**skeeter** · August 5th 2009, 02:13 PM

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?

Thanks in advance!

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

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

2) $u = 1-x$

$du = -dx$

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

you should get 4 for this integral

**Physics-is-Phun** · August 5th 2009, 07:14 PM

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.

**slider142** · August 6th 2009, 02:40 AM

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?

**Physics-is-Phun** · August 6th 2009, 04:38 AM

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

Like I said, I got as far as figuring out what the substitution was supposed to be ( $u = ln(\sqrt{x})$ , $du = (1/\sqrt{x})dx$ ). 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.

**malaygoel** · August 6th 2009, 04:41 AM

Originally Posted by Physics-is-Phun

Like I said, I got as far as figuring out what the substitution was supposed to be ( $u = ln(\sqrt{x})$ , $du = (1/\sqrt{x})dx$ ).

If
$u = ln(\sqrt{x})$

then

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

**Physics-is-Phun** · August 6th 2009, 04:48 AM

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.

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