intergration problem

- anonymous

intergration problem

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

##### 1 Attachment

- anonymous

my firsr step is antderv right

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

oh and to answer your question, no, the first step is not to find the anti derivative, the first step is to find the thing you need to integrate, both the integrand and the limits of integration
THEN you can take the anti - derivative to compute the integral

- anonymous

how to do that

- anonymous

set them equal and solve

- anonymous

\[y=x-1,y^2=2x+6\] replace the \(y\) in the second equation by \(x-1\) and solve
\[(x-1)^2=2x+6\] turns out you get an easy quadratic equation to solve (it factor, get integer solutions )

- anonymous

that is if you want to do it by hand
it if was me, i would cut to the chase
http://www.wolframalpha.com/input/?i=%28x-1%29^2%3D2x%2B6

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

o wow yea yea

- anonymous

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

- anonymous

ok then

- anonymous

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

- anonymous

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

- anonymous

so if you want to integrate wrt y, you need to write each of these as a function of \(y\) i.e solve each for \(x\)

- anonymous

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

- anonymous

so far so good?

- anonymous

okey

- anonymous

now if we turn it sideways we see the line is above the parabola
the limits of integration in terms of y means we use the y values
\[\int_{-2}^5y+1-(\frac{y^2-6}{2})dy\]

- anonymous

do some algebra first maybe, then compute that integral

- anonymous

i have to take the antiderv right

- anonymous

i would do the algebra first
then take anti derivatives

- anonymous

oh u mean simplify

- anonymous

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

- xapproachesinfinity

seems good :)

- anonymous

thanks !

- anonymous

now take the anto right

- anonymous

NOW take anti derivatives

- anonymous

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

- anonymous

okey tge y2/2 is confusing

- anonymous

would it be y3/6?

- anonymous

no

- xapproachesinfinity

yes! just don't forgot - sign

- anonymous

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

- anonymous

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

- anonymous

oh yea sorry

- anonymous

the rest should be routine
anti derivative is
\[-\frac{y^2}{6}+\frac{y^2}{2}+4y\]
plug in 5, plug in -2 and subtract
if you do it right, you will get 18i think

- anonymous

so i get 12.5 after simliy

- anonymous

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

- anonymous

i think ur right one my firends got 18 two

- anonymous

me, i like to know what the answer is before i begin
http://www.wolframalpha.com/input/?i=+area+between+the+curves+y%3Dx+-+1%2C+y^2%3D2x%2B6

- anonymous

ok so i have more intergral problems but just much eaiser i am just stuck since i am new to all this so could u help me more?

- anonymous

ok sure why not

- anonymous

if i can
i kind of suck at these though

- anonymous

ok give me 5 minutes i am finishing somthing

- anonymous

kk

- anonymous

lets see

- anonymous

|dw:1433296429808:dw|

- anonymous

i did everything but getting the answer wrong

- anonymous

here is my work just before the solution

- anonymous

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

- anonymous

looks like you have an extra term there

- anonymous

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

- anonymous

not sure where the x came from

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

you got that?

- anonymous

yes

- anonymous

wait i thiught it was +1 when taking anti

- anonymous

ok now take the anti derivative

- anonymous

yes we didn't do that yet

- anonymous

then why did u -1?

- anonymous

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

- anonymous

ok lets go slow

- anonymous

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

- anonymous

yes

- anonymous

then BEFORE we find the anti derivative we are going to divide each term in the numerator by x
that is not integrating, that is dividing

- anonymous

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

- anonymous

yes

- anonymous

we still have not taken the anti derivative yet
the anti derivative of \(\frac{1}{x}\) is what you said \(\ln(x)\)

- anonymous

but in order to take the anti derivative of
\[\frac{\sqrt{x}}{x}\] we need to write it as x to some power

- anonymous

ohh k

- anonymous

then we can use the power rule (backwards) for finding it
as a power it is
\[\frac{\sqrt{x}}{x}=x^{-\frac{1}{2}}\] that is not finding the anti derivative, that is just writing it in exponential form

- anonymous

yea i think that my only mistake

- anonymous

THEN you can add one to the exponent etc

- anonymous

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

- anonymous

questions or clear?

- anonymous

k i am going to finish this uo

- anonymous

kk

- anonymous

i am getting 6.38 , u ?

- anonymous

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

- anonymous

@satellite73 u their ?

- anonymous

plug in 4, plug in 1 subtract

- anonymous

if the book as \(\ln(4)\) as part of the answer, do not use a calculator
you may have got the same answer but as a decimal

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

i did 1^1/2 i got 1

- anonymous

did you forget the 2 out front?

- anonymous

2-1=1

- anonymous

omg yea

- anonymous

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

- anonymous

damn silly mistakes errrr

- anonymous

lol there are lots of ways to make a mistake!

- anonymous

serioulsy the only reasons i lose mrks in calc

- anonymous

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

- anonymous

good!

- anonymous

thats the number 2 reason for doing homework

- anonymous

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

- anonymous

go ahead i will look

- anonymous

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

- anonymous

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

- anonymous

i mean x+1

- anonymous

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

- anonymous

no not in this case

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

simplify then anti ?

- anonymous

no first algebra, then anti derivative

- anonymous

so its x2 -x +2

- anonymous

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

- anonymous

yes looks good

- anonymous

you see the difference in that and the previous one is that you were asked for the area between the curves
then you had to find where they intersect
in this case you are given the limits of integration

- anonymous

ok

- anonymous

then what do we do

- anonymous

plug in 4, plug in 2, subtract

- anonymous

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

- anonymous

what do u get , i get 12.05

- anonymous

idk i didn't do it

- anonymous

ok what do now

- anonymous

i will check your answer

- anonymous

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

- anonymous

oh ur right maybe i did wrong sub

- anonymous

\[F(4)=\frac{4^3}{3}-\frac{4^2}{2}+2\times 4\]
\[F(2)=\frac{2^3}{3}+\frac{2^2}{4}+2\times 2\]need a calculator or sommat

- anonymous

i think i need to stop subbing EVERYRthing into calc

- anonymous

ok so what now

- anonymous

how can i draw the graph from this info

- anonymous

now you are done
go have a beer

- anonymous

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

- anonymous

both are real easy to graph

- anonymous

sooo??|dw:1433301650060:dw|

- anonymous

|dw:1433301637176:dw|

- anonymous

yeah

- anonymous

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

- anonymous

why is that area shaded

- anonymous

because that is what you were asked for the area bounded by the cuve \(y=x^2+3\) the line \(y=x+1\) between the vertical lines \(x=2\) and \(x=4\)

- anonymous

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

- anonymous

we found the area of that region, but no, finding the area is not the same as graphing the region, they are two different things

- anonymous

how long r u gonna stay here?

- anonymous

lol not much longer i hope
you got more?

- anonymous

i wanna do more probs , i might get stuck again

- anonymous

their is a question

- anonymous

find the area under the given curve from a to b

- anonymous

y=x2 +1 from 0 to 4

- anonymous

that is real real easy

- anonymous

set up the integral and do it

- anonymous

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

- anonymous

why bother, you know what \(x^2+1\) looks like right?
all you have to compute here is
\[\int_0^4(x^2+1)dx\]

- anonymous

idk the answer on the back has a graph

- anonymous

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

- anonymous

probs

- anonymous

that integral is real real easy right?

- anonymous

yea

- anonymous

lots easier than the first two for sure

- anonymous

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

- anonymous

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

- anonymous

under the curve means below the curve and above the x axis
between the curve means between the curves
the first one you just use the limits of integration given you
the second you need the upper curve minus the lower curve

- anonymous

|dw:1433302834136:dw|

- anonymous

didnt we do upper minus bottom in both of them

- anonymous

not for the last one no

- anonymous

so the last one i asked was area under the curve

- anonymous

i toke the anti of x2 + 1

- anonymous

i got x3/3 + x

- anonymous

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

- anonymous

yes

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

slow

- anonymous

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

- anonymous

but if it is the area between two curves you have to subtract the lower one from the upper one BEFORE you integrate

- anonymous

so in your example it was only one curve \(y=x^2+1\) from \(0\) to \(4\) you do
\[\int_0^2 (x^2+1)dx\]

- anonymous

ohh u mean we have to simplify then anti but in area under their is no simplfying to do just straight

- anonymous

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

- anonymous

yes right, what you said

- anonymous

now i have a question

- anonymous

do you really love homework?

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.