anonymous
  • anonymous
Area under a curve stuff. Find the area bounded by the curves y^2 = 2x + 6 and x = y + 1. Your work must include an integral in one variable.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
so far i've found that the curves in terms of x are y = sqrt(2x + 6) and y = x-1
anonymous
  • anonymous
@Hero do you know how to do these ?
anonymous
  • anonymous
oh okay thanks for stopping by either way :)

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anonymous
  • anonymous
Graph it and find the bounded region in order to set up the integral.
anonymous
  • anonymous
The way you got the y by itself is correct.
anonymous
  • anonymous
yep i already graphed it
anonymous
  • anonymous
which function is on top of the bounded region?
anonymous
  • anonymous
um the y = sqrt(2x + 6) one right?
anonymous
  • anonymous
Does the problem also say bounded by x=0?
anonymous
  • anonymous
no copyed and pasted exactly what was given to me .
anonymous
  • anonymous
i'd assume it is a given though right?
anonymous
  • anonymous
It should be or the answer is infinite from just looking at the graph. So the bottom function is y=x-1 \[\int\limits_{0}^{5}\sqrt{2x+6}-(x-1)dx\]
anonymous
  • anonymous
since x-1 is cut off by the x=0, the square root of (2x+6) remains you add the first integral I gave you to this \[\int\limits_{-3}^{0}\sqrt{2x+6}\]
anonymous
  • anonymous
Know how I got that?
anonymous
  • anonymous
yes , why dont you just use the \[\int\limits_{-3}^{5}\sqrt{2x+6}-(x-1)\]
anonymous
  • anonymous
Because the x=0 cuts the x-1 out. You can see that the bounded region does not touch the x-1 from x=-3 to x=0 Do you see the graph?
anonymous
  • anonymous
yes i do |dw:1439509331610:dw| is this what that integral would find?
anonymous
  • anonymous
i mean the integral of the top function - the bottom function by the way
anonymous
  • anonymous
|dw:1439509369012:dw| Only here, not the bottom. The bottom left part you drew is not bounded with x=0
anonymous
  • anonymous
lol
anonymous
  • anonymous
So you got to separate it.
anonymous
  • anonymous
\[y^2=2x+6\] y=x-1 \[\left( x-1 \right)^2=2x+6\] \[x^2-2x+1-2x-6=0\] \[x^2-4x-5=0\] \[x^2-5x+x-5=0\] \[x \left( x-5 \right)+1\left( x-5 \right)=0\] \[\left( x-5 \right)\left( x+1 \right)=0\] x=-1,5
anonymous
  • anonymous
im sorry your picture is jiberish XD
anonymous
  • anonymous
Go to desmos graphing calculator online.
anonymous
  • anonymous
https://www.desmos.com/calculator
anonymous
  • anonymous
i have a graphing calculator, iv graphed it
anonymous
  • anonymous
My picture is same as yours, except dont shade the bottom since it is not enclosed by x=0, y=x-1 and y=(x^2+6)^(1/2)
anonymous
  • anonymous
yes i understand what your trying to say
anonymous
  • anonymous
Do you know how I got the 2 integrals and why I added them?
anonymous
  • anonymous
so your saying that this |dw:1439509856029:dw| equals this \[(\int\limits_{0}^{5}\sqrt{2x+6} - (x-1) dx) + \]\[\int\limits_{-3}^{0}\sqrt{2x+6}\]
anonymous
  • anonymous
Actually I was wrong
anonymous
  • anonymous
-_-
anonymous
  • anonymous
It should be \[\int\limits_{1}^{5}\sqrt{2x+6}-(x-1)dx+\int\limits_{0}^{1}\sqrt{2x+6}-(x-1)dx+\int\limits_{-3}^{0}\sqrt{x^2+6}dx\]
anonymous
  • anonymous
Do not forget dx at the end. I forgot too.
anonymous
  • anonymous
Oh wait I am wrong again. Rusty at this. Now I am 100% sure.
anonymous
  • anonymous
\[\int\limits_{1}^{5}\sqrt{2x+6}-(x-1)dx+\int\limits_{-3}^{1}\sqrt{2x+6}\]
anonymous
  • anonymous
|dw:1439520428984:dw| \[\left\{ \int\limits_{-3}^{5}\sqrt{2x+6}dx-\int\limits_{-3}^{-1}\sqrt{2x+6}dx-\int\limits_{-1}^{1}\left( x-1 \right)dx \right\}-\int\limits_{1}^{5 }\left( x-1 \right)dx\] i may be wrong somewhere.
anonymous
  • anonymous
i seriously think its just the top function minus the bottom functin integral
anonymous
  • anonymous
okay actually im not so sure anymore , we need a fourth opinion
anonymous
  • anonymous
actually that's pretty smart lets do that and match it to our other answers
anonymous
  • anonymous
I 5.79+7.54=13.33 That is what I got.
anonymous
  • anonymous
See if anyone else get the same as me even with different graphs and methods.
anonymous
  • anonymous
\[\int\limits_{-3}^{5}\sqrt{2x+6}\] \[put~\sqrt{2x+6}=t,2x+6=t^2,2dx=2t~dt,dx=t~dt\] when x=-3,t=0 when x=5,t=4 (t is positive) \[\int\limits_{0}^{4}t^2~dt=\frac{ t^3 }{ 3 }|0\rightarrow 4\] \[=\frac{ 1 }{ 3 }\left( 4^3-0^3 \right)=\frac{ 64 }{ 3 }\]
anonymous
  • anonymous
similarly solve other integrals
anonymous
  • anonymous
looks smaller than 21 and 1/3 since the bottom shaded graph is negative
anonymous
  • anonymous
you can go on desmo graphing calculator and count the squares.
anonymous
  • anonymous
The way without moving the y over produces a different graph which makes it harder to solve.
phi
  • phi
it looks easier to integrate over dy the right bound is x= y+1 and the left bound is y^2/2 - 3 thus \[ \int_{-2}^4 y+1-\left( \frac{1}{2}y^2 -3\right) \ dy \]
anonymous
  • anonymous
I got area 18 for yours.
anonymous
  • anonymous
okay i also got 18 also, to be honest it seems like this would be the anser just because its rounder
anonymous
  • anonymous
Not a multiple choice right?
anonymous
  • anonymous
nope, open ended
IrishBoy123
  • IrishBoy123
18 http://www.wolframalpha.com/input/?i=%5Cint_%7B-2%7D%5E%7B4%7D+%5Cint_%7B%5Cfrac%7By%5E2+-+6%7D%7B2%7D%7D%5E%7By+%2B+1%7D+dx+dy
phi
  • phi
we can do it integrating over dx and you can get 18 it is messier.
anonymous
  • anonymous
no need this method will do , thanks again @phi
anonymous
  • anonymous
What is the graph for x?
IrishBoy123
  • IrishBoy123
https://gyazo.com/4704acb0c39722d70976e884310e21fa you have to deal with the -ve signs if you do a "normal" integration, which means that you have to move both curves over and up, and that really messes up the algebra none of the solutions posted that don't involve a double integration look convincing to me.
phi
  • phi
like this
1 Attachment
phi
  • phi
each small rectangle will have area width* height height is dy width is (for a positive number) is right side minus left side.
phi
  • phi
thus to set up the integral you write \[ (x_r -x_l) \ dy \] in terms of y and sum up over dy (imagine lots of rectangles stacked on top of each other)
anonymous
  • anonymous
Still confused because everything under the x coordinate should be negative In the gyazo every 16 square is one. It seems to be less, unless you should the negative below x coordinate is positive?
anonymous
  • anonymous
Even if you flip it, it still seems less.
anonymous
  • anonymous
The area under a curve between two points can be found by doing a definite integral between the two points. To find the area under the curve y = f(x) between x = a and x = b, integrate y = f(x) between the limits of a and b. Areas under the x-axis will come out negative and areas above the x-axis will be positive. This is what it said when I searched "can area of a curve be negative.
IrishBoy123
  • IrishBoy123
hi @OOOPS i think you are quite right: @Phi's single dy integral is simply a better version of the double integral i stuffed through the engine. took me a while but my bad.
anonymous
  • anonymous
\[\int\limits_{0}^{4}y+1-(\frac{ 1 }{ 2}y^2-3)dy - \int\limits_{-2}^{0}y+1-(\frac{ 1 }{ 2}y^2-3)dy +\] =\[\frac{ 40 }{ 3 }-\frac{ 14 }{ 3 }=\frac{ 26 }{ }\]
anonymous
  • anonymous
26/3 is the real answer
anonymous
  • anonymous
because the (-3,0) graph goes up again.
anonymous
  • anonymous
it increased.
anonymous
  • anonymous
@phi think you made a mistake.
anonymous
  • anonymous
the parabola increased increased at y=0
anonymous
  • anonymous
In a different graph such as x=f(y), I wonder if they do not care about negative values?
anonymous
  • anonymous
If \[\int\limits_{\pi/2}^{-\pi/2}sinxdx=0\] cancels and makes it zero. Wonder why this does not?
anonymous
  • anonymous
oops other way around in the upper and lower limits.
IrishBoy123
  • IrishBoy123
that's the balls-up i made initially here, it is not about whether f(x) is above or below the axis in ∫f(x), it is about the difference between f(x) and g(x) in ∫ [f(x) - g(x)]
anonymous
  • anonymous
okay i went to eat what'd i miss?
anonymous
  • anonymous
If it says find the area of sinx bounded by y=0 from -pi/2 to pi/2 \[\int\limits_{-\pi/2}^{\pi/2}\sin(x)-0dx=0\] It would still be zero. y=0 is a function ( a constant one)
anonymous
  • anonymous
or a better one Find the area bounded by y=-0.50 and y=sin(x) |dw:1439517020584:dw| \[\int\limits_{-\pi/2}^{\pi/2}-0.50-\sin(x)dx=-\frac{ 157 }{ 100}\] -0.50 -sin(x) because -0.50 is on top of sinx when bounded
IrishBoy123
  • IrishBoy123
\(\large f(x) = \sin x, \ g(x) = -C\) \(\large \int_{-\pi /2}^{\pi /2} f(x) - g(x) \ dx\) \(\large = \int_{-\pi /2}^{\pi /2} \sin x + \ C \ dx\) \(\large = [ -\cos x + \ Cx ]_{-\pi /2}^{\pi /2}\) \(\large = [Cx ]_{-\pi /2}^{\pi /2}\) \(\large = C\pi\)
IrishBoy123
  • IrishBoy123
\(C = 0 \implies I = 0\)
anonymous
  • anonymous
I said -0.5 not 0.5
anonymous
  • anonymous
You do not know what bounded means right?
anonymous
  • anonymous
you did not say bounded x=0
anonymous
  • anonymous
|dw:1439518158177:dw|
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
@IrishBoy123 It is g(x)-f(x) not f(x)-g(x) -0.5 is on top of sin(x) when bounded
anonymous
  • anonymous
cannot open it.
anonymous
  • anonymous
phi
  • phi
For what it is worth
1 Attachment

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