evaluate the integral

Question

johnnydicamillo · Answer

I know this is not hard at all, my algebra is not all the good. $$\int\limits_{1}^{9} \frac{ \sqrt{u}-2u^2 }{ u }$$

johnnydicamillo · Answer

so  I know you basically have to find the anti derivative. then plug in 9 then 1 and subtract them

xapproachesinfinity · Answer

turn square root into power (u)^1/2
and split the fraction

xapproachesinfinity · Answer

$$\large \int_{1}^{9}\frac{u^{\frac{1}{2}}-2u^2}{u}du=\int_{1}^{9}(\frac{u^{\frac{1}{2}}}{u}-2\frac{u^2}{u})du$$

xapproachesinfinity · Answer

can you take it from there?

johnnydicamillo · Answer

let me try

xapproachesinfinity · Answer

okay

xapproachesinfinity · Answer

let me ask you what is $$\frac{a^n}{a^m}=?$$

johnnydicamillo · Answer

$$\frac{ \frac{ u ^{\frac{ 3 }{ 2 }}}{ \frac{ 3 }{ 2 } } }{ u } - 2\frac{ \frac{ u^3 }{ 3}}{ u }$$

johnnydicamillo · Answer

a^n-m?

xapproachesinfinity · Answer

oh no 
first simplify u^1/2 /u and u^2/u
before doing anti derivative

xapproachesinfinity · Answer

yes use that to simplify

xapproachesinfinity · Answer

a=u here n=1/2 m=1

xapproachesinfinity · Answer

and the same thing fro u^2/u=u you know u.u/u left with only one u

xapproachesinfinity · Answer

so work on u^1/2 /u =?

johnnydicamillo · Answer

wait, hold on. U^-1/2

xapproachesinfinity · Answer

you need to really work on your algebra

xapproachesinfinity · Answer

yes correct

johnnydicamillo · Answer

hahaaha

xapproachesinfinity · Answer

yeah seriously this is needs somewhat good algebra 
just do some reviews and stuff

johnnydicamillo · Answer

wonder how I made it to Calc 2

xapproachesinfinity · Answer

okay so we have $$\large \int_{1}^{9}(u^{\frac{-1}{2}}-2u )du$$

xapproachesinfinity · Answer

hehe oh this is calc2 wow i'm surprised, i'm in calc 2 as well hehe

xapproachesinfinity · Answer

now use the power rule for anti derivatives

johnnydicamillo · Answer

wait isn't it $$-2u^2/u$$

xapproachesinfinity · Answer

well you must be good ate any rate you just need to do more hard work :)

xapproachesinfinity · Answer

yes! it is simplified to -2u i explain how above 
u^2/u=u

johnnydicamillo · Answer

right right

xapproachesinfinity · Answer

you should at least know that one lol

johnnydicamillo · Answer

okay so now I integrate?

xapproachesinfinity · Answer

so can you carry it now

xapproachesinfinity · Answer

yes

johnnydicamillo · Answer

$$u ^{\frac{ 1 }{ 2}} - u^2$$

xapproachesinfinity · Answer

hmm try again 
if you do derivative of that you get 1/2 u^{-1/2}-2u

xapproachesinfinity · Answer

we don't have that 1/2 there
so we have to make up for it

xapproachesinfinity · Answer

the other part you got it right it u^2

johnnydicamillo · Answer

wait a minute it is $$\frac{ 1 }{ 2 } u ^{\frac{ 1 }{ 2 }} - u^2$$

xapproachesinfinity · Answer

no no

johnnydicamillo · Answer

because -.5 + 1 is .5?

xapproachesinfinity · Answer

you are making it 1/4 instead hehe

xapproachesinfinity · Answer

to get rid of 1/2 you multiply by 2

xapproachesinfinity · Answer

it's not about that 
its the near u^1/2
that's what you got wrong

xapproachesinfinity · Answer

it should be 2u^1/2-u^2 okay

johnnydicamillo · Answer

gotcha

xapproachesinfinity · Answer

now evaluate $$[.......]_{1}^{9}$$

xapproachesinfinity · Answer

you haven't forgotten the limits do you? haha

johnnydicamillo · Answer

-76

xapproachesinfinity · Answer

yes sound good!

johnnydicamillo · Answer

thanks for you help

xapproachesinfinity · Answer

welcome!

xapproachesinfinity · Answer

review some algebra
the calc 2 has a lot of work needs to be done 
i take it you just started the course

johnnydicamillo · Answer

yep, I will review asap. midterm in 3 days, yikes!

johnnydicamillo · Answer

the quarter system is super fast

xapproachesinfinity · Answer

really! that's super fast ^^
good luck in your midterm

xapproachesinfinity · Answer

my calc2 class hasn't started yet

johnnydicamillo · Answer

oh wow, well thanks

xapproachesinfinity · Answer

welcome!