intergration problem

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intergration problem

Calculus1
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my firsr step is antderv right
i have an idea (although i totally suck at these)

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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?
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
how to do that
set them equal and solve
\[y=x-1,y^2=2x+6\] 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 )
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
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
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 \(x\)
first one is \[x=y+1\] second is \[x=\frac{y^2-6}{2}\]
so far so good?
okey
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 \[\int_{-2}^5y+1-(\frac{y^2-6}{2})dy\]
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
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
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
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
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?
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
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
\[\frac{\sqrt{x}+1}{x}=\frac{\sqrt{x}}{x}+\frac{1}{x}\] right?
yes
we still have not taken the anti derivative yet the anti derivative of \(\frac{1}{x}\) is what you said \(\ln(x)\)
but in order to take the anti derivative of \[\frac{\sqrt{x}}{x}\] we need to write it as x to some power
ohh k
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
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
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
\[F(x)=2\sqrt{x}+\ln(x)\]
\[F(4)=2\sqrt4 +\ln(4)=4+\ln(4)\]
\[F(1)=2\sqrt 1+\ln(1)=2\]
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
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
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
\[F(4)=\frac{4^3}{3}-\frac{4^2}{2}+2\times 4\] \[F(2)=\frac{2^3}{3}+\frac{2^2}{4}+2\times 2\]need a calculator or sommat
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
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 \(x=4\)
so all the work we did had nothing to do with drawing the graph?
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
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
why bother, you know what \(x^2+1\) looks like right? all you have to compute here is \[\int_0^4(x^2+1)dx\]
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"
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
|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 is the area between two curves you have to subtract the lower one from the upper one BEFORE you integrate
so in your example it was only one curve \(y=x^2+1\) from \(0\) to \(4\) you do \[\int_0^2 (x^2+1)dx\]
ohh u mean we have to simplify then anti but in area under their is no simplfying to do just straight
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

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