Question

integrate 1dx/xsqrt(lnx) from e to e^x I get 2u^(1/2) =2(lnx)^(1/2) but I am not sure how to do the rest.

Answer 1

2(lnx)^(1/2)

Mmm ok ya I think your integration is correct.
So you're confused on what to do with the boundaries I guess? :o

So when we plug in the upper bound:

2(ln e^x)^(1/2)

We apply a rule of logs `log(a^b) = b log(a)`

2(x*ln e)^(1/2)

Answer 2

Do you understand what the ln e will simplify to?

Answer 3

oh ok I was stuck with the e^x bound....ln e should be 1 right?

Answer 4

We're currently working on the e^x bound. Do you understand how I used the log rule to move the x OUT of the log?

Answer 5

Yes 1.

Answer 6

Its been awhile but I vaguely remember the rule now.

Answer 7

once the X is out...if you plug in e^x aren't you just going to get stuck in an infinite loop of plugging in e^x's

Answer 8

Oh interesting.
So you're seeing a new x and thinking 'Oh I should plug my bound in'.

No we only want to plug the boundary in once.
We're not evaluating the expression at e^x and then plugging e^x into that, and then pulgging e^x into that.
That wouldn't really make sense :o

We plugged e^x in for x, giving us:
2(ln e^x)^(1/2)

Which simplified to:

2x^(1/2)

Answer 9

That's our expression evaluated at the upper bound, simplified.

Answer 10

oh great thanks....I think my ability to reason after hours of calc and physics has become warped!

Answer 11

hehe