Question

honestly i havent tried solving this yet...but i do need to see how it is solved as i still do NOT know how to start this :/

\[\Large \int_{0}^{1} \frac{\ln x}{\sqrt x} dx\]

Answer 1

This problem scares me....

Answer 2

i am thinking integration by parts?

Answer 3

sorta all of it *gulp*

Answer 4

btw...for the record i did not get how to solve my previous question -_- if anyone has an idea do check http://openstudy.com/users/lgbasallote#/updates/4fa872e9e4b059b524f46f3a

Answer 5

^i am troubled by the last part during the plugging in of limits

anyway..back tot his one...

Answer 6

@Mimi_x3 ? :(

Answer 7

Integration by parts.

u =lnx dv/sqrt(x)
du=1/x , u=2sqrt(x)

Answer 8

ugh...i gotta eat -__- owl be back in 30 minutes :/

Answer 9

i made a typo.
u=lnx dv=1/sqrt(x)
du=1/x , v = 2sqrt(x)

Answer 10

because i feel bad not completing it so..
\[\int\limits\frac{lnx}{\sqrt{x}} dx => \int\limits lnx*\frac{1}{\sqrt{x}} dx\]
\(u=lnx\) => dv=1/sqrt(x) => v = 2sqrt(x)

\(du= \frac{1}{x}dx\)
\[=>lnx*2\sqrt{x} -\int\limits2\sqrt{x}*\frac{1}{x} dx => lnx*2\sqrt{x}
-2\int\frac{\sqrt{x}}{x}dx\]
you can do it now i think..

Answer 11

my biggest problem in improper integrals is the plugging in NOT the integration -_-

Answer 12

this does not seem like improper integrals..

Answer 13

since the limit is 1 to 0 not infinity to 0

Answer 14

well zero makes it divergent :p

Answer 15

wait yeah sorry i didnt read the question properly; sorry i cant help you. xD

Answer 16

awww :(

Answer 17

tke lnx =t

Answer 18

uh huh?

Answer 19

dt=dx/x
x=exp(t)

Answer 20

exp (t)?

Answer 21

exponential of t
\[e ^{t}\]

Answer 22

uhmm uhmm i think IPB works better...i just wanna know the steps AFTER the integration

Answer 23

whats IPB

Answer 24

integration by parts

Answer 25

IBP sorry :))

Answer 26

the equation becomes
\[\int\limits_{a}^{b} te ^{t/2}dt\]

Answer 27

now it is easy to solve

Answer 28

use IBP

Answer 29

uhmm..like i said...what i need to know is AFTER the integration...when plugging in those limits..that's where i get confused -_-

Answer 30

You use L'hospital's rule.

Answer 31

huh? pls explain further :P

Answer 32

Assuming mimi's integration is correct, the answer should be:
\[\lim_{t \rightarrow 0^+}(\ln{x}\sqrt{x}-4\sqrt{x})|_t^1\\
=-\sqrt{4}-\lim_{t \rightarrow 0^+}(\ln{t}\sqrt{t}-4\sqrt{t})\]
Since \(\lim_{t \rightarrow 0^+}\ln{t}=-\infty\) while \(\lim_{t \rightarrow 0^+}\sqrt{t}=0\), there's an indeterminate form 0*infinity. So you use L'hospital's rule on \(\large \lim_{t \rightarrow 0^+}\frac{\ln{t}}{t^{-1/2}}\).

Answer 33

i see`

Answer 34

\[ \lim_{t \rightarrow 0^+}(\ln{x}\sqrt{x}-4\sqrt{x})|_t^1\
=-4-\lim_{t \rightarrow 0^+}(\ln{t}\sqrt{t}-4\sqrt{t}) \]

Answer 35

Oops. My bad. Stupid typo. =))

Answer 36

i see..and 4 sqrt t is 0 right?

Answer 37

y

Answer 38

4(0) is zero right o.O