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my firsr step is antderv right

i have an idea (although i totally suck at these)

it will be easier to integrate in terms of \(y\) not \(x\) so lets solve each of these for \(x\)

the first one is \(x=y+1\) and the second is \(x=\frac{y^2-6}{2}\)

we also need to find the limits of integration
did you find where they intersect?

how to do that

set them equal and solve

k so i get x^2-4x+8

you get \(x=-1\) or \(x=5\)
the points where they intersect are \((-1,-2)\) and \((5,4)\)

hmm no i think if you do it by hand you get
\[x^2-2x+1=2x+6\\
x^2-4x-5=0\]

o wow yea yea

factors as\[(x-5)(x+1)=0\\
x=-1,x=5\]

ok then

now this is a parabola that opens sideways
so it will be easier to integrate wrt y not x

here is a picture
http://www.wolframalpha.com/input/?i=+y%3Dx+-+1%2C+y^2%3D2x%2B6

first one is
\[x=y+1\] second is
\[x=\frac{y^2-6}{2}\]

so far so good?

okey

do some algebra first maybe, then compute that integral

i have to take the antiderv right

i would do the algebra first
then take anti derivatives

oh u mean simplify

oops
\[\int_{-2}^5\left(\frac{-y^2}{2}+y+4\right)dy\]

seems good :)

thanks !

now take the anto right

NOW take anti derivatives

yeah i know you are dying to do it, do it now

okey tge y2/2 is confusing

would it be y3/6?

no

yes! just don't forgot - sign

\[-\frac{y^2}{6}\] for the first term

ok as @xapproachesinfinity said "yes" only "no"

oh yea sorry

so i get 12.5 after simliy

damn typo \[-\frac{y^3}{6}+\frac{y^2}{2}+4y\]

i think ur right one my firends got 18 two

ok sure why not

if i can
i kind of suck at these though

ok give me 5 minutes i am finishing somthing

kk

lets see

|dw:1433296429808:dw|

i did everything but getting the answer wrong

here is my work just before the solution

2/3x^3/2 + x +lnx , am i right

looks like you have an extra term there

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

not sure where the x came from

how is squrtx , x ^-1/2? half should be positve right

also note that
\[\frac{\sqrt{x}}{x}=\frac{1}{\sqrt{x}}\] not \(\sqrt{x}\)

or if you like
\[\frac{\sqrt{x}}{x}=\frac{x^{\frac{1}{2}}}{x}=x^{\frac{1}{2}-1}=x^{-\frac{1}{2}}\]

you got that?

yes

wait i thiught it was +1 when taking anti

ok now take the anti derivative

yes we didn't do that yet

then why did u -1?

as you put it, we 'simplified' first to get
\[x^{-\frac{1}{2}}+\frac{1}{x}\] now go from there

ok lets go slow

first off you are given
\[\frac{\sqrt{x}+1}{x}\] right?

yes

\[\frac{\sqrt{x}+1}{x}=\frac{\sqrt{x}}{x}+\frac{1}{x}\] right?

yes

ohh k

yea i think that my only mistake

THEN you can add one to the exponent etc

you should get
\[2x^{\frac{1}{2}}\] or if you prefer
\[2\sqrt{x}\]

questions or clear?

k i am going to finish this uo

kk

i am getting 6.38 , u ?

@satellite73 the answer on my book is like 2+ln4

@satellite73 u their ?

plug in 4, plug in 1 subtract

\[F(x)=2\sqrt{x}+\ln(x)\]

\[F(4)=2\sqrt4 +\ln(4)=4+\ln(4)\]

\[F(1)=2\sqrt 1+\ln(1)=2\]

i did 1^1/2 i got 1

did you forget the 2 out front?

2-1=1

omg yea

\[\int x^{-\frac{1}{2}}dx=2\sqrt{x}\]

damn silly mistakes errrr

lol there are lots of ways to make a mistake!

serioulsy the only reasons i lose mrks in calc

but i am making miskates here so i wont makw them on the test

good!

thats the number 2 reason for doing homework

do u know questions realted to area under graph where u do the same thing just now draw

go ahead i will look

"fnid teh aera fo teh reiogn beteewn the gienv curesv, setch of the reiogn

y=x^2 +3 and y= x+a from 2 to 4

i mean x+1

what i did was made them equal to each other then anti

no not in this case

you are given the limits of integration, you don't have to find them

the only thing you need to know is that the parabola lies above the line

so
\[\int_2^2\left(x^2+3-(x+1)\right)dx\]

simplify then anti ?

no first algebra, then anti derivative

so its x2 -x +2

so after antoi i get the wuation 1/3x3 - 1/2x2 +2x , am i right

yes looks good

ok

then what do we do

plug in 4, plug in 2, subtract

that is what you always do
\[F(x)=\frac{x^3}{3}-\frac{x^2}{2}+2x\] compute
\[F(4)-F(2)\]

what do u get , i get 12.05

idk i didn't do it

ok what do now

i will check your answer

i got \(\frac{50}{3}\) hmmm

oh ur right maybe i did wrong sub

i think i need to stop subbing EVERYRthing into calc

ok so what now

how can i draw the graph from this info

now you are done
go have a beer

if you have to graph, graph the line and the parabola, that is all

both are real easy to graph

sooo??|dw:1433301650060:dw|

|dw:1433301637176:dw|

yeah

on the back of my book its like |dw:1433301723404:dw|

why is that area shaded

so all the work we did had nothing to do with drawing the graph?

how long r u gonna stay here?

lol not much longer i hope
you got more?

i wanna do more probs , i might get stuck again

their is a question

find the area under the given curve from a to b

y=x2 +1 from 0 to 4

that is real real easy

set up the integral and do it

we can graph it even before we solve it right ? and whats the difference

idk the answer on the back has a graph

that is because... they have a graph in the back
the graphics jack up the price of the book

probs

that integral is real real easy right?

yea

lots easier than the first two for sure

their are different chapters on my homwork but all the same thing

one says "area under curve " and the other "area between curves"

|dw:1433302834136:dw|

didnt we do upper minus bottom in both of them

not for the last one no

so the last one i asked was area under the curve

i toke the anti of x2 + 1

i got x3/3 + x

then i did F(4)-F(0)

yes

final soltuion 76/3 which is correct is that top minus bottom ?

\[F(0)=0\] in this case it should be easier than most

\[\frac{4^3}{3}+4\] is all

u said area under the curve is NOT top minus bottom but thats what i just did ?

slow

it is always
\[F(b)-F(a)\]

but if it was between to curves, you have to subtract first, then integrate

yes right, what you said

now i have a question

do you really love homework?

according to os you posted this question two hours ago! so i guess you must