Question

integral problem

Answer 1

what is it

Answer 2

\[\int\limits_{4}^{9} \frac{ lny }{ \sqrt(y) }\]

Answer 3

ooo srry im not good at that

Answer 4

Set \(u=\sqrt y\), which gives \(u^2=y\) and \(2u\,du=dy\).
\[\int_4^9\frac{\ln y}{\sqrt y}\,dy=\int_2^3\frac{\ln u^2}{u}\,du=2\int_2^3\frac{\ln u}{u}\,du\]

Answer 5

lol

Answer 6

is it okay if i set u=lny

Answer 7

Oops, left out a factor:
\[\int\cdots dy=\int_2^3\frac{\ln u^2}{u}(2u\,du)=4\int_2^3\ln u\,du\]
Sorry for the mixup

Answer 8

i got 2sqrt(Y)*lny - integral of 2Sqrt(y)*1/y dy is that right too?

Answer 9

Did you integrate by parts?

Answer 10

yes

Answer 11

Alright, I imagine you used
\[\begin{matrix}u=\ln y&&&dv=\dfrac{dy}{\sqrt y}\\
du=\dfrac{dy}{y}&&&v=2\sqrt y\end{matrix}\]
which gives
\[\int_4^9\frac{\ln y}{\sqrt y}\,dy=2\left[\sqrt y\ln y\right]_4^9-2\int_4^9\frac{\sqrt y}{y}\,dy\]
Yes, your method works out as well.

Answer 12

okay one more thing do the y's cancel out?

Answer 13

Yes, \(\dfrac{\sqrt y}{y}=\dfrac{1}{\sqrt y}\).

Answer 14

okay thank you

Answer 15

yw

Answer 16

ah

Answer 17

com to my question ya bafoone

Answer 18

@xapproachesinfinity