Area in between curves

[akoaysigod]

Suppose the graph of a cubic polynomial intersects the parabola y=x^2 when x=0, x=a, and x=b where 0 < a < b. If the two regions between the curves have the same area, how is b related to a?

So far I know the equation of the polynomial will be f(x)=x(x-a)(x-b) because it gives the zeroes of the equation

in the problem. I'm just not sure where to go from here. If I set it up into an integral that would find the area it would be

int[0,a] f(x)-x^2 + int[a,b] x^2-f(x) where f(x)=x(x-a)(x-b)

Which doesn't get me anywhere, and I'm not even sure I can assume which of those functions comes first. Any hints? I'd like to be able to do this on my own but I'm not entirely sure which directions I should I take with this. I could also set the two equal to each other but that also doesn't get me anywhere.

Thanks
Also, is there anyway for me to type integrals and such so they look like they would hand written? Writing it out like this is confusing to look at.

 

[Deleted member 4993]

Suppose the graph of a cubic polynomial intersects the parabola y=x^2 when x=0, x=a, and x=b where 0 < a < b. If the two regions between the curves have the same area, how is b related to a?

So far I know the equation of the polynomial will be f(x)=x(x-a)(x-b) because it gives the zeroes of the equation

No - "zeroes" of a function (not equation) are the points of intersection with "x-axis"

In your case x = 0, a & b are the points of intersection of f(x) with y = x^2.

from that you know

f(0) = 0

f(a) = a² and

f(b) = b²

from the above you could deduce

f(x) = Cx(x-a)(x-b) + x²

and the other given condition

\(\displaystyle \int_0^a [f(x) - x^2] dx \, = \, \int_a^b [x^2 - f(x)] dx\)

\(\displaystyle \int_0^b Cx(x-a)(x-b) dx \, = \, 0\)

\(\displaystyle \int_0^b x(x-a)(x-b) dx \, = \, 0\)

Now continue....



in the problem. I'm just not sure where to go from here. If I set it up into an integral that would find the area it would be

int[0,a] f(x)-x^2 + int[a,b] x^2-f(x) where f(x)=x(x-a)(x-b)

Which doesn't get me anywhere, and I'm not even sure I can assume which of those functions comes first. Any hints? I'd like to be able to do this on my own but I'm not entirely sure which directions I should I take with this. I could also set the two equal to each other but that also doesn't get me anywhere.

Thanks
Also, is there anyway for me to type integrals and such so they look like they would hand written? Writing it out like this is confusing to look at.

 

[soroban]

Hello, akoaysigod!

I agree with Subhotosh, but I got a different (and strange) answer . . .

\(\displaystyle \text{Suppose the graph of a cubic polynomial intersects the parabola }y\:=\:x^2\)
. . \(\displaystyle \text{when: }\,x=0,\;x=a,\;x=b\,\text{ where }0 < a < b.\)

\(\displaystyle \text{If the two regions between the curves have the same area, how is }b\text{ related to }a\,?\)

\(\displaystyle \text{Let the cubic function be }f(x).\)

\(\displaystyle \text{We have: }\;\int^a_0\bigg[f(x) - x^2\bigg]\,dx \;=\;\int^b_a\bigg[x^2 - f(x)\bigg]\,dx\)

. . \(\displaystyle \int^a_0\!\!f(x)\,dx \;- \int^a_0\!\! x^2\,dx \;\;=\;\;\int^b_a\!\!x^2\,dx \;- \int^b_a\!\!f(x)\,dx\)

. . \(\displaystyle \underbrace{\int^a_0\!\!f(x)\,dx \;+ \int^b_a\!\!f(x)\,dx} \;\;=\;\;\underbrace{\int^a_0\!\! x^2\,dx \;+ \int^b_a\!\!x^2\,dx}\)

. . . . . . \(\displaystyle \underbrace{\int^b_0f(x)\,dx}_{\text{area under cubic}} \quad\quad\;=\;\; \quad \underbrace{\int^b_0 x^2\,dx}_{\text{area under parabola}}\)



The graph looks like this:

         |                         *
    o    |                          o
         |    .*.  
         | .*:::::*               o
         o:::::::::*           .o:|
        *|   o::::::*     ..o:::* |
         |          .o.::::::::*  |
       * |           | *:::::*    |
         |           |    *       |
    -----+-----------+------------+-------
      *  |           a            b
         |

Then the area under the cubic:

         |                         *
         |
         |   .*.
         | *:::::*                *
         *:::::::::*             .|
        *|::::::::::*           *:|
         |:::::::::::*         *::|
       * |:::::::::::::*     *::::|
         |::::::::::::::::*:::::::|
    -----+-----------+------------+-------
      *  |           a            b
         |

equals the area under the parabola:

         |
    o    |                          o
         |
      o  |                        o
         o.                    .o:|
         |:::o ..         ..o ::::|
         |:::::::::::o::::::::::::|
         |::::::::::::::::::::::::|
         |::::::::::::::::::::::::|
    -----+-----------+------------+-------
         |           a            b
         |

\(\displaystyle \text{I would conclude that }a\text{ does not contribute to the problem.}\)
. . \(\displaystyle \text{But that can't be right . . .}\)

\(\displaystyle \text{So what's going on?}\)

 

[Deleted member 4993]

I get:

\(\displaystyle \int_0^a [f(x) - x^2] dx \, = \, \int_a^b [x^2 - f(x)] dx\)

\(\displaystyle \int_0^b Cx(x-a)(x-b) dx \, = \, 0\)

\(\displaystyle \int_0^b x(x-a)(x-b) dx \, = \, 0\)

\(\displaystyle \int_0^b [x^3 - x^2(a+b) + abx] dx \, = \, 0\)

\(\displaystyle \frac {b^4}{4} - \frac{b^3}{3}(a+b) + ab\frac{b^2}{2} \, = \, 0\)

\(\displaystyle -\frac {b^4}{12} \, + \, \frac{ab^3}{6} \, = \, 0\)

\(\displaystyle b \, = \, 2a\)

 

[akoaysigod]

Actually I thought about it more, wouldn't it be a -x^2 since you'd set them equal to find out where they intersect? Why is it a +x^2? Although that does give an answer, which I assumed was going to be the answer from the start. I don't think that can be put into an integral, however, I don't fully understand why it's a +x^2.

If this is the case then I get b = 2a-8

 

[akoaysigod]

Never mind, I think I could still use an explanation as to why the x^2 is added to the function it makes sense, otherwise f(a)=0 and that wouldn't fit the original statements of the problem. How did you know to add it though?

 

[mmm4444bot]

The graph looks like this:

         |                         *
    o    |                          o
         |    .*.  
         | .*:::::*               o
         o:::::::::*           .o:|
        *|   o::::::*     ..o:::* |
         |          .o.::::::::*  |
       * |           | *:::::*    |
         |           |    *       |
    -----+-----------+------------+-------
      *  |           a            b
         |

Soroban:

It looks to me like you shifted the graph of y = x^2.

The parabola vertex should be at the origin.

MY EDIT: Deleted reference to vertical shift, since there is a horizontal shift, also.

 

[mmm4444bot]

I get …

\(\displaystyle b \, = \, 2a\)

Subhotosh:

Perhaps, we need to take cases, in this exercise?

I have not felt well, the last few days, so I'm lacking motivation to do anything in-depth.

However, I played around with some concrete values (rounded) on Maple V, and came up with a seeming contradiction to b = 2a.

g(x) = x^2

f(x) = -x(x - 10)(x - 23.562)

a = 10.8541

b = 21.7079

Area between curves on [0, a] = 3469.8393

Area between curves on [a, b] = 3469.5169

But 23.562 is not twice 10.

Of course, I might be screwing up something(s), too.

:| (? We need a smilie that looks like Mr. Yuk feels, heh, heh.)

DOUBLE-CLICK IMAGE, TO EXPAND

[attachment=0:2z9apzuw]EqualRegions.JPG[/attachment:2z9apzuw]

 

[Deleted member 4993]

Re:

I get …

\(\displaystyle b \, = \, 2a\)

Subhotosh:


Area between curves on [0, a] = 3469.8393

Area between curves on [a, b] = 3469.5169

But 23.562 is not twice 10.

And the areas are not exactly equal.

 

[BigGlenntheHeavy]

\(\displaystyle f(x) \ = \ x^{2}. \ If \ g(x) \ = \ \frac{x^{3}}{2}-\frac{x^{2}}{2}+x \ and \ a \ = \ 1, \ b \ =2, \ then\)

\(\displaystyle \int_ {0}^{a}[g(x)-f(x)]dx \ = \ \int_{a}^{b}[f(x)-g(x)]dx \ = \ \frac{1}{8}.\)

See graph below. Note: How is b related to a? b=2a as Subhotosh Khan analyzed the problem. Red curve is graph of f(x)=x^2

[attachment=0:3c94malq]abc.jpg[/attachment:3c94malq]

 

[BigGlenntheHeavy]

Is yaoka the name of your dog, or is that the name for dog in your language.

 

[mmm4444bot]

And the areas are not exactly equal.

I agree.

My brute-force guess of 23.562 for b (when a = 10) results in a difference of roughly 1/3rd of a square unit.

Is this the difference to which you refer?

Given motivation, I could reduce the difference to roughly 1/10000000000th of a square unit (or less), but, in order for that to happen, b would need to be larger than even 23.562.

I agree that "a cubic polynomial" can be found where b = 2a, but it seems to me that this is not a general result. :?

What am I misinterpreting, in the given exercise?

 

[BigGlenntheHeavy]

Here is another one

\(\displaystyle f(x) \ = \ x^{2} \ and \ g(x) \ = \ \frac{x^{3}}{8}+\frac{x^{2}}{4}+x, a \ =2, \ b \ = \ 4.\)

\(\displaystyle \int_{0}^{a}[{g(x)-f(x)]dx \ = \ \int_{a}^{b}[f(x)-g(x)]dx \ = \ \frac{1}{2}\)

[attachment=0:3h2wiusx]zzz.jpg[/attachment:3h2wiusx]

 

[akoaysigod]

Is yaoka the name of your dog, or is that the name for dog in your language.

Its "I am God" in English and tagalog, it started off as just a video game tag from when I was much younger and I've kind of grown attached to it, so I use it for just about any username, whether appropriate or not.

Also, I will post the answer to this as soon as I get it if anyone is still interested although it looks like the 2a=b is right.

 

[mmm4444bot]

… it looks like the 2a=b is right.

It looks like 2a = b in some cases.