Question

Evaluate the Integral.

Answer 1

\[\int\limits_{36}^{49} \frac{ \ln y }{ \sqrt{y} } dy\]

Answer 2

When you differentiate log(y), it turns into some y stuff, ya? 1/y.
No more log after that.
So it makes sense to try Parts again, with log(y) being your u.

Answer 3

Oh okay. Ill try that

Answer 4

I got stuck :/

Answer 5

No no, dv is wrong.

Answer 6

\[\large\rm \int\limits\limits \frac{\ln y}{\sqrt y}dy\quad=\quad \int\limits\limits \ln y\left(\frac{1}{\sqrt y}dy\right)\quad=\quad \int\limits u(dv)\]

Answer 7

Oh i see. so

\[\ln y \times 2(y)^{1/2} - 2 \int\limits_{36}^{49} \frac{ \sqrt{y} }{y }\]

Answer 8

Good, yes.

Answer 9

I am stuck at the integration part

Answer 10

y^(1/2) / y^1
Simplify this using exponent rule.

Answer 11

so...\[\frac{ 2\left( y \right)^{3/2} }{ 3y } ?\]

Answer 12

\[\frac{ \sqrt{y} }{ y }=\frac{ 1 }{ \sqrt{y} }=y ^{-\frac{ 1 }{ 2 }}\]

Answer 13

No, perhaps you made the same mistake as before.

Answer 14

\[\large\rm \frac{\sqrt y}{y}=\frac{1}{\sqrt y}\]This is not the same as \(\large\rm \sqrt y\) though.

Answer 15

Oh you left a y in the denominator :P Yah something weird going on there.
Didn't notice that til now.

Answer 16

\[\large\rm 2\sqrt y \ln y-2\int\limits y^{-1/2}dy\]It's pretty much the same integration as your dv to v, ya?

Answer 17

why is y to the (-1/2) power? Isnt it suppose to be positive? and what happened to the 1/y?

Answer 18

oh did you do \[y^{1/2} - y^{2/2}\]

Answer 19

\[\large\rm \frac{\sqrt y}{y}=\frac{y^{1/2}}{y^1}\]From that point, applying exponent rule,\[\large\rm \frac{y^{1/2}}{y^1}=y^{1/2-1}\]

Answer 20

Oh yea.

Answer 21

So then it would be
\[2\sqrt{y}\ln y - 2 \left[ y^{1/2} \right] 36 \to 49\]

Answer 22

Hmm don't you end up with another 2 after that integration?
Thinking..

Answer 23

\[\large\rm 2\sqrt y \ln y-2\color{orangered}{\int\limits\frac{1}{\sqrt y}dy}\]becomes,\[\large\rm 2\sqrt y \ln y-2\color{orangered}{(2\sqrt y)}\]

Answer 24

Oh yes, i had forgotten to include it

Answer 25

So the answer it \[2\sqrt{y}\ln y - 4\]

Answer 26

?

Answer 27

Nvm, it says that my answer is wrong

Answer 28

The integral has bounds.

Answer 29

Hm?

Answer 30

I plugged in the 49 and 36 and got -2[14-12]= -2[2] = -4

Answer 31

What?
How are you getting these nice simple numbers with natural log involved?

Answer 32

@zepdrix I just wonder if I can let u = sqrt y?

Answer 33

Oh no. lol I meant for the integration part.
So I have\[2\sqrt{y}\ln y - 2 \int\limits_{36}^{49} \frac{ 1 }{ \sqrt{y} }dy\]

Which then becomes...
\[2\sqrt{y} \ln y - 2 (2\sqrt{y}) \]

Then I plugged in the numbers (49 and 36)
\[2\sqrt{y} \ln y - 2 \left[ 2\sqrt{49}-2\sqrt{36} \right]\]

Then I got..
\[2\sqrt{y} \ln y - 2 \left[ 2(7)-2(6) \right]\]
\[2\sqrt{y} \ln y - 2 \left[ 14-12 \right]\]
\[2\sqrt{y} \ln y - 2 \left[ 2 \right]\]
\[2\sqrt{y} \ln y - 4\]

Answer 34

But its wrong :/

Answer 35

Oh woah woah woah... no no.
The bounds still apply to the first part of our IBP.

Answer 36

So what we would like to do is this,\[\large\rm \left(2\sqrt y \ln y-4\sqrt y\right)_{36}^{49}\]Plug these values into the entire thing.

Answer 37

If you like, maybe remember your integration by parts formula like this,\[\large\rm \int\limits_a^b u~dv=uv|_a^b-\int\limits_a^b v~du\]

Answer 38

Oh. okay

Answer 39

\[14\ln 7-12\ln 6 - 4\] ?

Answer 40

It's log(y), not log(sqrty)

Answer 41

You're plugging 49 and 36 into your logs, not 7 and 6, right?

Answer 42

Yea

Answer 43

Oh i see

\[14 \ln 49 - 12 \ln 36 -4\]

Answer 44

Ok that looks much better.
You can apply some log rules if you want to simplify it further,\[\large\rm \ln\left(\frac{49^{14}}{36^{12}}\right)-4\]But where you stopped is probably fine.

Answer 45

SID...
SIDDDDDDDDDDDDDDDDDDD...

Dude bro... keep studying...
boy oh boy D:

Some of your Algebra skills are a bit rusty.
Practice practice practice.

Answer 46

Yea I know haha.
Will do!