Question

evaluate the integral

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

can you take it from there?

Answer 6

let me try

Answer 7

okay

Answer 8

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

Answer 9

\[\frac{ \frac{ u ^{\frac{ 3 }{ 2 }}}{ \frac{ 3 }{ 2 } } }{ u } - 2\frac{ \frac{ u^3 }{ 3}}{ u }\]

Answer 10

a^n-m?

Answer 11

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

Answer 12

yes use that to simplify

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

wait, hold on. U^-1/2

Answer 17

you need to really work on your algebra

Answer 18

yes correct

Answer 19

hahaaha

Answer 20

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

Answer 21

wonder how I made it to Calc 2

Answer 22

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

Answer 23

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

Answer 24

now use the power rule for anti derivatives

Answer 25

wait isn't it \[-2u^2/u\]

Answer 26

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

Answer 27

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

Answer 28

right right

Answer 29

you should at least know that one lol

Answer 30

okay so now I integrate?

Answer 31

so can you carry it now

Answer 32

yes

Answer 33

\[u ^{\frac{ 1 }{ 2}} - u^2\]

Answer 34

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

Answer 35

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

Answer 36

the other part you got it right it u^2

Answer 37

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

Answer 38

no no

Answer 39

because -.5 + 1 is .5?

Answer 40

you are making it 1/4 instead hehe

Answer 41

to get rid of 1/2 you multiply by 2

Answer 42

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

Answer 43

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

Answer 44

gotcha

Answer 45

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

Answer 46

you haven't forgotten the limits do you? haha

Answer 47

-76

Answer 48

yes sound good!

Answer 49

thanks for you help

Answer 50

welcome!

Answer 51

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

Answer 52

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

Answer 53

the quarter system is super fast

Answer 54

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

Answer 55

my calc2 class hasn't started yet

Answer 56

oh wow, well thanks

Answer 57

welcome!