Question

need help with this integral

Answer 1

\[\int\limits_{0}^{1}lnxdx \]

Answer 2

pretty sure i need to us l'hopitals rule?just not sure

Answer 3

parts

Answer 4

yep.
btw, I think l'hopitals rule is for limits isn't it?

Answer 5

I'm with those above me. Parts is the way to go.

Answer 6

there will be a limit here...it is an improper integral

Answer 7

I thought there was a limit?Ugh im getting lost

Answer 8

ah..

Answer 9

so how would i use the limit in this?

Answer 10

For parts, let \(u=\ln(x)\) and \(dv=dx\)

Answer 11

\[\int\limits_{0}^{1}\ln(x)dx=\lim_{t\to 0^{+}}\int\limits_{t}^{1}\ln(x)dx\]

Answer 12

yes, notice your lower limit is zero. lnx is not defined there so you'll have to let your lower limit = a and let a approach 0

Answer 13

So I can not just put 0 in at some point?Im not sure why i would use parts here..

Answer 14

if i were you i would memorize this one

Answer 15

math teachers love putting it on test and it would save you a lot of time to just remember that
\[\int\ln(x)dx=x\ln(x)-x\]

Answer 16

check is easy, take the derivative and see that it works

Answer 17

okay so if i get xln(x)-x do I just evaluate from 0 to 1? Sorry never done one like this yet... So wouldnt you get 0-1?=-1?

Answer 18

Does that make sense?or did i skip a step

Answer 19

well you do get -1 but not so fast

Answer 20

because
\[\ln(0)\] is undefined so you have to take
\[\lim_{t\to 0^+}t\ln(t)\]

Answer 21

for this you can use l'hopital as the original post suggests

Answer 22

form is
\[0\times -\infty\]

Answer 23

Im just getting thrown off because of the 1 also...I understand we need to use the limit because of the 0 but just not sure about how to us the 1

Answer 24

the 1 is straight forward
\[1\times \ln(1)-1=1\times 0-1=-1\] the second part is the one that requires work, taking the limit

Answer 25

standard trick for
\[\lim_{t\to 0^+}t\ln(t)\] is to rewrite as
\[\lim_{t\to 0}\frac{\ln(t)}{\frac{1}{t}}\] and use l\hopital

Answer 26

im looking up l'hopitals rule again right now to make sure i remember it right...but i think id get it

Answer 27

so i take the derivatve of ln(t) and the one of 1/t? and that is l'hopitals rule right?

Answer 28

take the derivative separately top and bottom, simplify using algebra and replace t by 0, you will get limit is 0 in one step

Answer 29

right

Answer 30

oh that works. Thanks a lot!

Answer 31

yw