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.

1. anonymous

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

2. anonymous

@Hero do you know how to do these ?

3. anonymous

oh okay thanks for stopping by either way :)

4. anonymous

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

5. anonymous

The way you got the y by itself is correct.

6. anonymous

7. anonymous

which function is on top of the bounded region?

8. anonymous

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

9. anonymous

Does the problem also say bounded by x=0?

10. anonymous

no copyed and pasted exactly what was given to me .

11. anonymous

i'd assume it is a given though right?

12. anonymous

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

13. 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 $\int\limits_{-3}^{0}\sqrt{2x+6}$

14. anonymous

Know how I got that?

15. anonymous

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

16. 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?

17. anonymous

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

18. anonymous

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

19. anonymous

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

20. anonymous

lol

21. anonymous

So you got to separate it.

22. 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

23. anonymous

im sorry your picture is jiberish XD

24. anonymous

Go to desmos graphing calculator online.

25. anonymous
26. anonymous

i have a graphing calculator, iv graphed it

27. 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)

28. anonymous

yes i understand what your trying to say

29. anonymous

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

30. 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}$

31. anonymous

Actually I was wrong

32. anonymous

-_-

33. anonymous

It should be $\int\limits_{1}^{5}\sqrt{2x+6}-(x-1)dx+\int\limits_{0}^{1}\sqrt{2x+6}-(x-1)dx+\int\limits_{-3}^{0}\sqrt{x^2+6}dx$

34. anonymous

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

35. anonymous

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

36. anonymous

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

37. 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.

38. anonymous

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

39. anonymous

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

40. anonymous

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

41. anonymous

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

42. anonymous

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

43. 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) $\int\limits_{0}^{4}t^2~dt=\frac{ t^3 }{ 3 }|0\rightarrow 4$ $=\frac{ 1 }{ 3 }\left( 4^3-0^3 \right)=\frac{ 64 }{ 3 }$

44. anonymous

similarly solve other integrals

45. anonymous

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

46. anonymous

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

47. anonymous

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

48. 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 $\int_{-2}^4 y+1-\left( \frac{1}{2}y^2 -3\right) \ dy$

49. anonymous

I got area 18 for yours.

50. anonymous

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

51. anonymous

Not a multiple choice right?

52. anonymous

nope, open ended

53. IrishBoy123
54. phi

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

55. anonymous

no need this method will do , thanks again @phi

56. anonymous

What is the graph for x?

57. 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.

58. phi

like this

59. phi

each small rectangle will have area width* height height is dy width is (for a positive number) is right side minus left side.

60. 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)

61. 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?

62. anonymous

Even if you flip it, it still seems less.

63. 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.

64. 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.

65. anonymous

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

66. anonymous

67. anonymous

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

68. anonymous

it increased.

69. anonymous

@phi think you made a mistake.

70. anonymous

the parabola increased increased at y=0

71. anonymous

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

72. anonymous

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

73. anonymous

oops other way around in the upper and lower limits.

74. 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)]

75. anonymous

okay i went to eat what'd i miss?

76. 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)

77. 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

78. IrishBoy123

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

79. IrishBoy123

$$C = 0 \implies I = 0$$

80. anonymous

I said -0.5 not 0.5

81. anonymous

You do not know what bounded means right?

82. anonymous

you did not say bounded x=0

83. anonymous

|dw:1439518158177:dw|

84. anonymous

85. anonymous

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

86. anonymous

cannot open it.

87. anonymous

88. phi

For what it is worth