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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.

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

  2. anonymous
    • one year ago
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    @Hero do you know how to do these ?

  3. anonymous
    • one year ago
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    oh okay thanks for stopping by either way :)

  4. anonymous
    • one year ago
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    Graph it and find the bounded region in order to set up the integral.

  5. anonymous
    • one year ago
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    The way you got the y by itself is correct.

  6. anonymous
    • one year ago
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    yep i already graphed it

  7. anonymous
    • one year ago
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    which function is on top of the bounded region?

  8. anonymous
    • one year ago
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    um the y = sqrt(2x + 6) one right?

  9. anonymous
    • one year ago
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    Does the problem also say bounded by x=0?

  10. anonymous
    • one year ago
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    no copyed and pasted exactly what was given to me .

  11. anonymous
    • one year ago
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    i'd assume it is a given though right?

  12. anonymous
    • one year ago
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    It should be or the answer is infinite from just looking at the graph. So the bottom function is y=x-1 \[\int\limits_{0}^{5}\sqrt{2x+6}-(x-1)dx\]

  13. anonymous
    • one year ago
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    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 \[\int\limits_{-3}^{0}\sqrt{2x+6}\]

  14. anonymous
    • one year ago
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    Know how I got that?

  15. anonymous
    • one year ago
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    yes , why dont you just use the \[\int\limits_{-3}^{5}\sqrt{2x+6}-(x-1)\]

  16. anonymous
    • one year ago
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    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?

  17. anonymous
    • one year ago
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    yes i do |dw:1439509331610:dw| is this what that integral would find?

  18. anonymous
    • one year ago
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    i mean the integral of the top function - the bottom function by the way

  19. anonymous
    • one year ago
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    |dw:1439509369012:dw| Only here, not the bottom. The bottom left part you drew is not bounded with x=0

  20. anonymous
    • one year ago
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    lol

  21. anonymous
    • one year ago
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    So you got to separate it.

  22. anonymous
    • one year ago
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    \[y^2=2x+6\] 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

  23. anonymous
    • one year ago
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    im sorry your picture is jiberish XD

  24. anonymous
    • one year ago
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    Go to desmos graphing calculator online.

  25. anonymous
    • one year ago
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    https://www.desmos.com/calculator

  26. anonymous
    • one year ago
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    i have a graphing calculator, iv graphed it

  27. anonymous
    • one year ago
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    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)

  28. anonymous
    • one year ago
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    yes i understand what your trying to say

  29. anonymous
    • one year ago
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    Do you know how I got the 2 integrals and why I added them?

  30. anonymous
    • one year ago
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    so your saying that this |dw:1439509856029:dw| equals this \[(\int\limits_{0}^{5}\sqrt{2x+6} - (x-1) dx) + \]\[\int\limits_{-3}^{0}\sqrt{2x+6}\]

  31. anonymous
    • one year ago
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    Actually I was wrong

  32. anonymous
    • one year ago
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    -_-

  33. anonymous
    • one year ago
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    It should be \[\int\limits_{1}^{5}\sqrt{2x+6}-(x-1)dx+\int\limits_{0}^{1}\sqrt{2x+6}-(x-1)dx+\int\limits_{-3}^{0}\sqrt{x^2+6}dx\]

  34. anonymous
    • one year ago
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    Do not forget dx at the end. I forgot too.

  35. anonymous
    • one year ago
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    Oh wait I am wrong again. Rusty at this. Now I am 100% sure.

  36. anonymous
    • one year ago
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    \[\int\limits_{1}^{5}\sqrt{2x+6}-(x-1)dx+\int\limits_{-3}^{1}\sqrt{2x+6}\]

  37. anonymous
    • one year ago
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    |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.

  38. anonymous
    • one year ago
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    i seriously think its just the top function minus the bottom functin integral

  39. anonymous
    • one year ago
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    okay actually im not so sure anymore , we need a fourth opinion

  40. anonymous
    • one year ago
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    actually that's pretty smart lets do that and match it to our other answers

  41. anonymous
    • one year ago
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    I 5.79+7.54=13.33 That is what I got.

  42. anonymous
    • one year ago
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    See if anyone else get the same as me even with different graphs and methods.

  43. anonymous
    • one year ago
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    \[\int\limits_{-3}^{5}\sqrt{2x+6}\] \[put~\sqrt{2x+6}=t,2x+6=t^2,2dx=2t~dt,dx=t~dt\] when x=-3,t=0 when x=5,t=4 (t is positive) \[\int\limits_{0}^{4}t^2~dt=\frac{ t^3 }{ 3 }|0\rightarrow 4\] \[=\frac{ 1 }{ 3 }\left( 4^3-0^3 \right)=\frac{ 64 }{ 3 }\]

  44. anonymous
    • one year ago
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    similarly solve other integrals

  45. anonymous
    • one year ago
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    looks smaller than 21 and 1/3 since the bottom shaded graph is negative

  46. anonymous
    • one year ago
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    you can go on desmo graphing calculator and count the squares.

  47. anonymous
    • one year ago
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    The way without moving the y over produces a different graph which makes it harder to solve.

  48. phi
    • one year ago
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    it looks easier to integrate over dy the right bound is x= y+1 and the left bound is y^2/2 - 3 thus \[ \int_{-2}^4 y+1-\left( \frac{1}{2}y^2 -3\right) \ dy \]

  49. anonymous
    • one year ago
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    I got area 18 for yours.

  50. anonymous
    • one year ago
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    okay i also got 18 also, to be honest it seems like this would be the anser just because its rounder

  51. anonymous
    • one year ago
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    Not a multiple choice right?

  52. anonymous
    • one year ago
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    nope, open ended

  53. phi
    • one year ago
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    we can do it integrating over dx and you can get 18 it is messier.

  54. anonymous
    • one year ago
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    no need this method will do , thanks again @phi

  55. anonymous
    • one year ago
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    What is the graph for x?

  56. IrishBoy123
    • one year ago
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    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.

  57. phi
    • one year ago
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    like this

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  58. phi
    • one year ago
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    each small rectangle will have area width* height height is dy width is (for a positive number) is right side minus left side.

  59. phi
    • one year ago
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    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)

  60. anonymous
    • one year ago
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    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?

  61. anonymous
    • one year ago
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    Even if you flip it, it still seems less.

  62. anonymous
    • one year ago
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    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.

  63. IrishBoy123
    • one year ago
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    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.

  64. anonymous
    • one year ago
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    \[\int\limits_{0}^{4}y+1-(\frac{ 1 }{ 2}y^2-3)dy - \int\limits_{-2}^{0}y+1-(\frac{ 1 }{ 2}y^2-3)dy +\] =\[\frac{ 40 }{ 3 }-\frac{ 14 }{ 3 }=\frac{ 26 }{ }\]

  65. anonymous
    • one year ago
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    26/3 is the real answer

  66. anonymous
    • one year ago
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    because the (-3,0) graph goes up again.

  67. anonymous
    • one year ago
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    it increased.

  68. anonymous
    • one year ago
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    @phi think you made a mistake.

  69. anonymous
    • one year ago
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    the parabola increased increased at y=0

  70. anonymous
    • one year ago
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    In a different graph such as x=f(y), I wonder if they do not care about negative values?

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

  72. anonymous
    • one year ago
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    oops other way around in the upper and lower limits.

  73. IrishBoy123
    • one year ago
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    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)]

  74. anonymous
    • one year ago
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    okay i went to eat what'd i miss?

  75. anonymous
    • one year ago
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    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)

  76. anonymous
    • one year ago
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    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

  77. IrishBoy123
    • one year ago
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    \(\large f(x) = \sin x, \ g(x) = -C\) \(\large \int_{-\pi /2}^{\pi /2} f(x) - g(x) \ dx\) \(\large = \int_{-\pi /2}^{\pi /2} \sin x + \ C \ dx\) \(\large = [ -\cos x + \ Cx ]_{-\pi /2}^{\pi /2}\) \(\large = [Cx ]_{-\pi /2}^{\pi /2}\) \(\large = C\pi\)

  78. IrishBoy123
    • one year ago
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    \(C = 0 \implies I = 0\)

  79. anonymous
    • one year ago
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    I said -0.5 not 0.5

  80. anonymous
    • one year ago
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    You do not know what bounded means right?

  81. anonymous
    • one year ago
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    you did not say bounded x=0

  82. anonymous
    • one year ago
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    |dw:1439518158177:dw|

  83. anonymous
    • one year ago
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  84. anonymous
    • one year ago
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    @IrishBoy123 It is g(x)-f(x) not f(x)-g(x) -0.5 is on top of sin(x) when bounded

  85. anonymous
    • one year ago
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    cannot open it.

  86. anonymous
    • one year ago
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  87. phi
    • one year ago
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    For what it is worth

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