anonymous
  • anonymous
intergration problem
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
anonymous
  • anonymous
my firsr step is antderv right
anonymous
  • anonymous
i have an idea (although i totally suck at these)

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anonymous
  • anonymous
it will be easier to integrate in terms of \(y\) not \(x\) so lets solve each of these for \(x\)
anonymous
  • anonymous
the first one is \(x=y+1\) and the second is \(x=\frac{y^2-6}{2}\)
anonymous
  • anonymous
we also need to find the limits of integration did you find where they intersect?
anonymous
  • 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
  • anonymous
how to do that
anonymous
  • anonymous
set them equal and solve
anonymous
  • 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
  • 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
  • anonymous
k so i get x^2-4x+8
anonymous
  • anonymous
you get \(x=-1\) or \(x=5\) the points where they intersect are \((-1,-2)\) and \((5,4)\)
anonymous
  • 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
  • anonymous
o wow yea yea
anonymous
  • anonymous
factors as\[(x-5)(x+1)=0\\ x=-1,x=5\]
anonymous
  • anonymous
ok then
anonymous
  • anonymous
now this is a parabola that opens sideways so it will be easier to integrate wrt y not x
anonymous
  • anonymous
here is a picture http://www.wolframalpha.com/input/?i=+y%3Dx+-+1%2C+y^2%3D2x%2B6
anonymous
  • 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
  • anonymous
first one is \[x=y+1\] second is \[x=\frac{y^2-6}{2}\]
anonymous
  • anonymous
so far so good?
anonymous
  • anonymous
okey
anonymous
  • 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
  • anonymous
do some algebra first maybe, then compute that integral
anonymous
  • anonymous
i have to take the antiderv right
anonymous
  • anonymous
i would do the algebra first then take anti derivatives
anonymous
  • anonymous
oh u mean simplify
anonymous
  • anonymous
oops \[\int_{-2}^5\left(\frac{-y^2}{2}+y+4\right)dy\]
xapproachesinfinity
  • xapproachesinfinity
seems good :)
anonymous
  • anonymous
thanks !
anonymous
  • anonymous
now take the anto right
anonymous
  • anonymous
NOW take anti derivatives
anonymous
  • anonymous
yeah i know you are dying to do it, do it now
anonymous
  • anonymous
okey tge y2/2 is confusing
anonymous
  • anonymous
would it be y3/6?
anonymous
  • anonymous
no
xapproachesinfinity
  • xapproachesinfinity
yes! just don't forgot - sign
anonymous
  • anonymous
\[-\frac{y^2}{6}\] for the first term
anonymous
  • anonymous
ok as @xapproachesinfinity said "yes" only "no"
anonymous
  • anonymous
oh yea sorry
anonymous
  • 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
  • anonymous
so i get 12.5 after simliy
anonymous
  • anonymous
damn typo \[-\frac{y^3}{6}+\frac{y^2}{2}+4y\]
anonymous
  • anonymous
i think ur right one my firends got 18 two
anonymous
  • 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
  • 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
  • anonymous
ok sure why not
anonymous
  • anonymous
if i can i kind of suck at these though
anonymous
  • anonymous
ok give me 5 minutes i am finishing somthing
anonymous
  • anonymous
kk
anonymous
  • anonymous
lets see
anonymous
  • anonymous
|dw:1433296429808:dw|
anonymous
  • anonymous
i did everything but getting the answer wrong
anonymous
  • anonymous
here is my work just before the solution
anonymous
  • anonymous
2/3x^3/2 + x +lnx , am i right
anonymous
  • anonymous
looks like you have an extra term there
anonymous
  • anonymous
\[\frac{\sqrt{x}+1}{x}=x^{-\frac{1}{2}}+\frac{1}{x}\]
anonymous
  • anonymous
not sure where the x came from
anonymous
  • anonymous
how is squrtx , x ^-1/2? half should be positve right
anonymous
  • anonymous
also note that \[\frac{\sqrt{x}}{x}=\frac{1}{\sqrt{x}}\] not \(\sqrt{x}\)
anonymous
  • 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
  • anonymous
you got that?
anonymous
  • anonymous
yes
anonymous
  • anonymous
wait i thiught it was +1 when taking anti
anonymous
  • anonymous
ok now take the anti derivative
anonymous
  • anonymous
yes we didn't do that yet
anonymous
  • anonymous
then why did u -1?
anonymous
  • anonymous
as you put it, we 'simplified' first to get \[x^{-\frac{1}{2}}+\frac{1}{x}\] now go from there
anonymous
  • anonymous
ok lets go slow
anonymous
  • anonymous
first off you are given \[\frac{\sqrt{x}+1}{x}\] right?
anonymous
  • anonymous
yes
anonymous
  • 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
  • anonymous
\[\frac{\sqrt{x}+1}{x}=\frac{\sqrt{x}}{x}+\frac{1}{x}\] right?
anonymous
  • anonymous
yes
anonymous
  • anonymous
we still have not taken the anti derivative yet the anti derivative of \(\frac{1}{x}\) is what you said \(\ln(x)\)
anonymous
  • 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
  • anonymous
ohh k
anonymous
  • 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
  • anonymous
yea i think that my only mistake
anonymous
  • anonymous
THEN you can add one to the exponent etc
anonymous
  • anonymous
you should get \[2x^{\frac{1}{2}}\] or if you prefer \[2\sqrt{x}\]
anonymous
  • anonymous
questions or clear?
anonymous
  • anonymous
k i am going to finish this uo
anonymous
  • anonymous
kk
anonymous
  • anonymous
i am getting 6.38 , u ?
anonymous
  • anonymous
@satellite73 the answer on my book is like 2+ln4
anonymous
  • anonymous
@satellite73 u their ?
anonymous
  • anonymous
plug in 4, plug in 1 subtract
anonymous
  • 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
  • anonymous
\[F(x)=2\sqrt{x}+\ln(x)\]
anonymous
  • anonymous
\[F(4)=2\sqrt4 +\ln(4)=4+\ln(4)\]
anonymous
  • anonymous
\[F(1)=2\sqrt 1+\ln(1)=2\]
anonymous
  • anonymous
i did 1^1/2 i got 1
anonymous
  • anonymous
did you forget the 2 out front?
anonymous
  • anonymous
2-1=1
anonymous
  • anonymous
omg yea
anonymous
  • anonymous
\[\int x^{-\frac{1}{2}}dx=2\sqrt{x}\]
anonymous
  • anonymous
damn silly mistakes errrr
anonymous
  • anonymous
lol there are lots of ways to make a mistake!
anonymous
  • anonymous
serioulsy the only reasons i lose mrks in calc
anonymous
  • anonymous
but i am making miskates here so i wont makw them on the test
anonymous
  • anonymous
good!
anonymous
  • anonymous
thats the number 2 reason for doing homework
anonymous
  • anonymous
do u know questions realted to area under graph where u do the same thing just now draw
anonymous
  • anonymous
go ahead i will look
anonymous
  • anonymous
"fnid teh aera fo teh reiogn beteewn the gienv curesv, setch of the reiogn
anonymous
  • anonymous
y=x^2 +3 and y= x+a from 2 to 4
anonymous
  • anonymous
i mean x+1
anonymous
  • anonymous
what i did was made them equal to each other then anti
anonymous
  • anonymous
no not in this case
anonymous
  • anonymous
you are given the limits of integration, you don't have to find them
anonymous
  • anonymous
the only thing you need to know is that the parabola lies above the line
anonymous
  • anonymous
so \[\int_2^2\left(x^2+3-(x+1)\right)dx\]
anonymous
  • anonymous
simplify then anti ?
anonymous
  • anonymous
no first algebra, then anti derivative
anonymous
  • anonymous
so its x2 -x +2
anonymous
  • anonymous
so after antoi i get the wuation 1/3x3 - 1/2x2 +2x , am i right
anonymous
  • anonymous
yes looks good
anonymous
  • 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
  • anonymous
ok
anonymous
  • anonymous
then what do we do
anonymous
  • anonymous
plug in 4, plug in 2, subtract
anonymous
  • anonymous
that is what you always do \[F(x)=\frac{x^3}{3}-\frac{x^2}{2}+2x\] compute \[F(4)-F(2)\]
anonymous
  • anonymous
what do u get , i get 12.05
anonymous
  • anonymous
idk i didn't do it
anonymous
  • anonymous
ok what do now
anonymous
  • anonymous
i will check your answer
anonymous
  • anonymous
i got \(\frac{50}{3}\) hmmm
anonymous
  • anonymous
oh ur right maybe i did wrong sub
anonymous
  • 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
  • anonymous
i think i need to stop subbing EVERYRthing into calc
anonymous
  • anonymous
ok so what now
anonymous
  • anonymous
how can i draw the graph from this info
anonymous
  • anonymous
now you are done go have a beer
anonymous
  • anonymous
if you have to graph, graph the line and the parabola, that is all
anonymous
  • anonymous
both are real easy to graph
anonymous
  • anonymous
sooo??|dw:1433301650060:dw|
anonymous
  • anonymous
|dw:1433301637176:dw|
anonymous
  • anonymous
yeah
anonymous
  • anonymous
on the back of my book its like |dw:1433301723404:dw|
anonymous
  • anonymous
why is that area shaded
anonymous
  • 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
  • anonymous
so all the work we did had nothing to do with drawing the graph?
anonymous
  • 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
  • anonymous
how long r u gonna stay here?
anonymous
  • anonymous
lol not much longer i hope you got more?
anonymous
  • anonymous
i wanna do more probs , i might get stuck again
anonymous
  • anonymous
their is a question
anonymous
  • anonymous
find the area under the given curve from a to b
anonymous
  • anonymous
y=x2 +1 from 0 to 4
anonymous
  • anonymous
that is real real easy
anonymous
  • anonymous
set up the integral and do it
anonymous
  • anonymous
we can graph it even before we solve it right ? and whats the difference
anonymous
  • 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
  • anonymous
idk the answer on the back has a graph
anonymous
  • anonymous
that is because... they have a graph in the back the graphics jack up the price of the book
anonymous
  • anonymous
probs
anonymous
  • anonymous
that integral is real real easy right?
anonymous
  • anonymous
yea
anonymous
  • anonymous
lots easier than the first two for sure
anonymous
  • anonymous
their are different chapters on my homwork but all the same thing
anonymous
  • anonymous
one says "area under curve " and the other "area between curves"
anonymous
  • 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
  • anonymous
|dw:1433302834136:dw|
anonymous
  • anonymous
didnt we do upper minus bottom in both of them
anonymous
  • anonymous
not for the last one no
anonymous
  • anonymous
so the last one i asked was area under the curve
anonymous
  • anonymous
i toke the anti of x2 + 1
anonymous
  • anonymous
i got x3/3 + x
anonymous
  • anonymous
then i did F(4)-F(0)
anonymous
  • anonymous
yes
anonymous
  • anonymous
final soltuion 76/3 which is correct is that top minus bottom ?
anonymous
  • anonymous
\[F(0)=0\] in this case it should be easier than most
anonymous
  • anonymous
\[\frac{4^3}{3}+4\] is all
anonymous
  • anonymous
u said area under the curve is NOT top minus bottom but thats what i just did ?
anonymous
  • anonymous
slow
anonymous
  • anonymous
it is always \[F(b)-F(a)\]
anonymous
  • 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
  • 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
  • anonymous
ohh u mean we have to simplify then anti but in area under their is no simplfying to do just straight
anonymous
  • anonymous
but if it was between to curves, you have to subtract first, then integrate
anonymous
  • anonymous
yes right, what you said
anonymous
  • anonymous
now i have a question
anonymous
  • anonymous
do you really love homework?
anonymous
  • anonymous
according to os you posted this question two hours ago! so i guess you must

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