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.

- anonymous

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- anonymous

so far i've found that the curves in terms of x are
y = sqrt(2x + 6)
and
y = x-1

- anonymous

@Hero do you know how to do these ?

- anonymous

oh okay thanks for stopping by either way :)

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## More answers

- anonymous

Graph it and find the bounded region in order to set up the integral.

- anonymous

The way you got the y by itself is correct.

- anonymous

yep i already graphed it

- anonymous

which function is on top of the bounded region?

- anonymous

um the y = sqrt(2x + 6) one right?

- anonymous

Does the problem also say bounded by x=0?

- anonymous

no copyed and pasted exactly what was given to me .

- anonymous

i'd assume it is a given though right?

- 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

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

Know how I got that?

- anonymous

yes , why dont you just use the \[\int\limits_{-3}^{5}\sqrt{2x+6}-(x-1)\]

- 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

yes i do |dw:1439509331610:dw|
is this what that integral would find?

- anonymous

i mean the integral of the top function - the bottom function by the way

- anonymous

|dw:1439509369012:dw|
Only here, not the bottom.
The bottom left part you drew is not bounded with x=0

- anonymous

lol

- anonymous

So you got to separate it.

- 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

im sorry your picture is jiberish XD

- anonymous

Go to desmos graphing calculator online.

- anonymous

https://www.desmos.com/calculator

- anonymous

i have a graphing calculator, iv graphed it

- 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

yes i understand what your trying to say

- anonymous

Do you know how I got the 2 integrals and why I added them?

- 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

Actually I was wrong

- 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

Do not forget dx at the end.
I forgot too.

- anonymous

Oh wait I am wrong again. Rusty at this. Now I am 100% sure.

- anonymous

\[\int\limits_{1}^{5}\sqrt{2x+6}-(x-1)dx+\int\limits_{-3}^{1}\sqrt{2x+6}\]

- 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

i seriously think its just the top function minus the bottom functin integral

- anonymous

okay actually im not so sure anymore , we need a fourth opinion

- anonymous

actually that's pretty smart lets do that and match it to our other answers

- anonymous

I 5.79+7.54=13.33
That is what I got.

- anonymous

See if anyone else get the same as me even with different graphs and methods.

- 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

similarly solve other integrals

- anonymous

looks smaller than 21 and 1/3 since the bottom shaded graph is negative

- anonymous

you can go on desmo graphing calculator and count the squares.

- anonymous

The way without moving the y over produces a different graph which makes it harder to solve.

- 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

I got area 18 for yours.

- anonymous

okay i also got 18 also, to be honest it seems like this would be the anser just because its rounder

- anonymous

Not a multiple choice right?

- anonymous

nope, open ended

- 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

we can do it integrating over dx and you can get 18
it is messier.

- anonymous

no need this method will do , thanks again @phi

- anonymous

What is the graph for x?

- 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

like this

##### 1 Attachment

- 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

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

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

Even if you flip it, it still seems less.

- 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

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

\[\int\limits_{0}^{4}y+1-(\frac{ 1 }{ 2}y^2-3)dy - \int\limits_{-2}^{0}y+1-(\frac{ 1 }{ 2}y^2-3)dy +\]
=\[\frac{ 40 }{ 3 }-\frac{ 14 }{ 3 }=\frac{ 26 }{ }\]

- anonymous

26/3 is the real answer

- anonymous

because the (-3,0) graph goes up again.

- anonymous

it increased.

- anonymous

@phi
think you made a mistake.

- anonymous

the parabola increased increased at y=0

- anonymous

In a different graph such as x=f(y), I wonder if they do not care about negative values?

- anonymous

If \[\int\limits_{\pi/2}^{-\pi/2}sinxdx=0\]
cancels and makes it zero.
Wonder why this does not?

- anonymous

oops other way around in the upper and lower limits.

- IrishBoy123

that's the balls-up i made initially
here, it is not about whether f(x) is above or below the axis in âˆ«f(x), it is about the difference between f(x) and g(x) in âˆ«Â [f(x) - g(x)]

- anonymous

okay i went to eat what'd i miss?

- 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

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

\(\large f(x) = \sin x, \ g(x) = -C\)
\(\large \int_{-\pi /2}^{\pi /2} f(x) - g(x) \ dx\)
\(\large = \int_{-\pi /2}^{\pi /2} \sin x + \ C \ dx\)
\(\large = [ -\cos x + \ Cx ]_{-\pi /2}^{\pi /2}\)
\(\large = [Cx ]_{-\pi /2}^{\pi /2}\)
\(\large = C\pi\)

- IrishBoy123

\(C = 0 \implies I = 0\)

- anonymous

I said -0.5 not 0.5

- anonymous

You do not know what bounded means right?

- anonymous

you did not say bounded x=0

- anonymous

|dw:1439518158177:dw|

- anonymous

##### 1 Attachment

- anonymous

@IrishBoy123
It is g(x)-f(x)
not f(x)-g(x)
-0.5 is on top of sin(x) when bounded

- anonymous

cannot open it.

- anonymous

##### 1 Attachment

- phi

For what it is worth

##### 1 Attachment

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