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anonymous
 one year ago
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
 one year ago
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.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so far i've found that the curves in terms of x are y = sqrt(2x + 6) and y = x1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Hero do you know how to do these ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh okay thanks for stopping by either way :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Graph it and find the bounded region in order to set up the integral.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The way you got the y by itself is correct.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yep i already graphed it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which function is on top of the bounded region?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0um the y = sqrt(2x + 6) one right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Does the problem also say bounded by x=0?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no copyed and pasted exactly what was given to me .

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i'd assume it is a given though right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It should be or the answer is infinite from just looking at the graph. So the bottom function is y=x1 \[\int\limits_{0}^{5}\sqrt{2x+6}(x1)dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0since x1 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
 one year ago
Best ResponseYou've already chosen the best response.0Know how I got that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes , why dont you just use the \[\int\limits_{3}^{5}\sqrt{2x+6}(x1)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Because the x=0 cuts the x1 out. You can see that the bounded region does not touch the x1 from x=3 to x=0 Do you see the graph?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes i do dw:1439509331610:dw is this what that integral would find?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i mean the integral of the top function  the bottom function by the way

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439509369012:dw Only here, not the bottom. The bottom left part you drew is not bounded with x=0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So you got to separate it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[y^2=2x+6\] y=x1 \[\left( x1 \right)^2=2x+6\] \[x^22x+12x6=0\] \[x^24x5=0\] \[x^25x+x5=0\] \[x \left( x5 \right)+1\left( x5 \right)=0\] \[\left( x5 \right)\left( x+1 \right)=0\] x=1,5

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im sorry your picture is jiberish XD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Go to desmos graphing calculator online.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i have a graphing calculator, iv graphed it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0My picture is same as yours, except dont shade the bottom since it is not enclosed by x=0, y=x1 and y=(x^2+6)^(1/2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes i understand what your trying to say

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do you know how I got the 2 integrals and why I added them?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so your saying that this dw:1439509856029:dw equals this \[(\int\limits_{0}^{5}\sqrt{2x+6}  (x1) dx) + \]\[\int\limits_{3}^{0}\sqrt{2x+6}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Actually I was wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It should be \[\int\limits_{1}^{5}\sqrt{2x+6}(x1)dx+\int\limits_{0}^{1}\sqrt{2x+6}(x1)dx+\int\limits_{3}^{0}\sqrt{x^2+6}dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do not forget dx at the end. I forgot too.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh wait I am wrong again. Rusty at this. Now I am 100% sure.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{1}^{5}\sqrt{2x+6}(x1)dx+\int\limits_{3}^{1}\sqrt{2x+6}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439520428984:dw \[\left\{ \int\limits_{3}^{5}\sqrt{2x+6}dx\int\limits_{3}^{1}\sqrt{2x+6}dx\int\limits_{1}^{1}\left( x1 \right)dx \right\}\int\limits_{1}^{5 }\left( x1 \right)dx\] i may be wrong somewhere.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i seriously think its just the top function minus the bottom functin integral

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay actually im not so sure anymore , we need a fourth opinion

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0actually that's pretty smart lets do that and match it to our other answers

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 5.79+7.54=13.33 That is what I got.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0See if anyone else get the same as me even with different graphs and methods.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{3}^{5}\sqrt{2x+6}\] \[put~\sqrt{2x+6}=t,2x+6=t^2,2dx=2t~dt,dx=t~dt\] when x=3,t=0 when x=5,t=4 (t is positive) \[\int\limits_{0}^{4}t^2~dt=\frac{ t^3 }{ 3 }0\rightarrow 4\] \[=\frac{ 1 }{ 3 }\left( 4^30^3 \right)=\frac{ 64 }{ 3 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0similarly solve other integrals

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0looks smaller than 21 and 1/3 since the bottom shaded graph is negative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you can go on desmo graphing calculator and count the squares.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The way without moving the y over produces a different graph which makes it harder to solve.

phi
 one year ago
Best ResponseYou've already chosen the best response.2it 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
 one year ago
Best ResponseYou've already chosen the best response.0I got area 18 for yours.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay i also got 18 also, to be honest it seems like this would be the anser just because its rounder

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Not a multiple choice right?

phi
 one year ago
Best ResponseYou've already chosen the best response.2we can do it integrating over dx and you can get 18 it is messier.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no need this method will do , thanks again @phi

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What is the graph for x?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0https://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
 one year ago
Best ResponseYou've already chosen the best response.2each small rectangle will have area width* height height is dy width is (for a positive number) is right side minus left side.

phi
 one year ago
Best ResponseYou've already chosen the best response.2thus 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
 one year ago
Best ResponseYou've already chosen the best response.0Still 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
 one year ago
Best ResponseYou've already chosen the best response.0Even if you flip it, it still seems less.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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 xaxis will come out negative and areas above the xaxis will be positive. This is what it said when I searched "can area of a curve be negative.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0hi @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
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{4}y+1(\frac{ 1 }{ 2}y^23)dy  \int\limits_{2}^{0}y+1(\frac{ 1 }{ 2}y^23)dy +\] =\[\frac{ 40 }{ 3 }\frac{ 14 }{ 3 }=\frac{ 26 }{ }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.026/3 is the real answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because the (3,0) graph goes up again.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@phi think you made a mistake.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the parabola increased increased at y=0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In a different graph such as x=f(y), I wonder if they do not care about negative values?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If \[\int\limits_{\pi/2}^{\pi/2}sinxdx=0\] cancels and makes it zero. Wonder why this does not?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops other way around in the upper and lower limits.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0that's the ballsup 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
 one year ago
Best ResponseYou've already chosen the best response.0okay i went to eat what'd i miss?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If 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
 one year ago
Best ResponseYou've already chosen the best response.0or 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
 one year ago
Best ResponseYou've already chosen the best response.0\(\large f(x) = \sin x, \ g(x) = C\) \(\large \int_{\pi /2}^{\pi /2} f(x)  g(x) \ dx\) \(\large = \int_{\pi /2}^{\pi /2} \sin x + \ C \ dx\) \(\large = [ \cos x + \ Cx ]_{\pi /2}^{\pi /2}\) \(\large = [Cx ]_{\pi /2}^{\pi /2}\) \(\large = C\pi\)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0\(C = 0 \implies I = 0\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You do not know what bounded means right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you did not say bounded x=0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439518158177:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@IrishBoy123 It is g(x)f(x) not f(x)g(x) 0.5 is on top of sin(x) when bounded
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