Area under the curve problem, but there's a twist

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In summary, the conversation discusses a problem involving finding the area under a curve and the difficulty of using the traditional method for finding the area. Several suggestions are given, including using the inverse function, mirroring the function, and flipping the function upside down. The conversation also mentions using Adobe Illustrator to create a chart.
  • #1
relativitydude
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I have a unique area under a curve problem. Say we have a function y = x^2. Ok, so, the area under that curve is just the integral of x^2 from 0 to some value of x. That's easy. Let's call this area A.

Now normally if we wanted to find the area under a height, we would simply mulitply height by the value of x and subtract the integrated form to find the are above the curve to that height. Let's call this area B.

I can't use this method. I can't simply say h*x minus the integral of whatever to find area B. Is there another way to find area B while knowing the function but not involving area A in any way?
 
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  • #2
try making this confusing question a little more precise. an example could help.
 
  • #3
I included a diagram. I need to find area A and area B separately. I can't find area B by subtracting A out of it and conversely I can't find area A by subtracting B out of it.
 

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  • #4
I can't find area B by subtracting A out of it

Why can't you?
 
  • #5
Hurkyl said:
Why can't you?

I just cant.
 
  • #6
relativitydude said:
I just cant.
Why not? Try it for some value of x and h. What's the problem? On the other hand, maybe you could try expressing the f(x) in terms of y and integrating wrt y instead to find B directly.

PS. What did you use to create the picture? It looks rather neat.
 
  • #7
I can't do it the normal way because my problem that this assists boils down to 1=1, a true yet uninspiring finding. I will try it with your suggestions, Ethereal.

That chart? Eh, that's nothing. I made it in Adobe Illustrator. You can draw ANYTHING in that program.
 
  • #8
If you found one of them via the integral, couldn't you find the other by subtracting?
 
  • #9
Hurkyl said:
If you found one of them via the integral, couldn't you find the other by subtracting?

I said I can't use the subtraction method.
 
  • #10
relativitydude said:
I said I can't use the subtraction method.

Do you mean you don't know how? Or aren't you allowed to use it for solving the problem?

If you're looking for another way to find area B, here's a hint:
Think of the line that gives the height as your x-axis and stand on your head.
 
  • #11
I am not allowed.

Thanks Gilleo, will do.
 
  • #12
wouldn't the inverse function of x^2 have the same area if you integrated along the x-axis and used the height as the upper bound limit?
 
  • #13
Here's how

Find [itex] f^{-1}(x) [/itex]. In this case, you are using [itex] f(x) = x^2 [/tex]. The inverse of this function is [itex] f(x) = \sqrt{x} [/itex]. You can now integrate, but make sure to adjust your limits of integration. If you wanted from 0 to 3 of the original function, that would translate to 0 to 9 of the inverse. Just plug in the original limits to the inverse (y) to obtain your new limits.

Hope this helps!
 
  • #14
I don't know if inversion will help. Inversion mirrors the function in (through?) the line y=x.
My idea was to mirror the function in the x-axis, then add the height h to the function and then integrate. (or alternatively subtract the height from the function and then mirror it).

But the calculation would be exactly the same if you were calculating the area of the rectangle and subtract area A. So it seems silly not to use this method and do use the above method.

Subtract height: f(x)-h
Mirror: h-f(x)
Integrate: [itex]h\Delta x - \int f(x)dx[/itex]

It's the same :)
 
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  • #15
Well basically it is the same exact thing as what you planed on doing. The area between the function and the y-axis is the same as its inverse function and the x-axis.
 
  • #16
ok I'm not sure if anyone mentioned htis but...

lets say the equation was y = x^2... you wanted to find area B

could you not find the integral from 0 -> x by changing the equation ... by flipping it upside down (reflect upon x-axis) and shift it up h units?

thus you get
y = -x^2 + h

and find integral from 0-->x now?

correct me if I'm wrong
 

1. What is the "area under the curve" problem?

The "area under the curve" problem is a mathematical concept that involves finding the area of a region bounded by a curve on a graph and the x-axis. This problem is commonly encountered in calculus and is used to calculate the total change or accumulated value of a function over a specific interval.

2. What is the twist in the "area under the curve" problem?

The twist in the "area under the curve" problem refers to a modification or variation of the original problem. This could include changing the limits of integration, adding a second curve, or using a different function to calculate the area.

3. How is the "area under the curve" problem used in science?

The "area under the curve" problem is commonly used in science to analyze and interpret data from experiments or observations. It can be used to calculate the rate of change of a variable over time, the total amount of a substance present in a sample, or the effectiveness of a treatment.

4. What are some real-life applications of the "area under the curve" problem?

The "area under the curve" problem has numerous real-life applications, including in economics, physics, chemistry, and biology. It is used to analyze stock market trends, calculate the work done by a varying force, determine the concentration of a chemical in a solution, and measure the growth rate of a population.

5. What are some common techniques used to solve the "area under the curve" problem?

There are several techniques for solving the "area under the curve" problem, including the Riemann sum method, the trapezoidal rule, and the Simpson's rule. These methods involve dividing the area into smaller, simpler shapes and using mathematical formulas to calculate their individual areas, which are then summed to find the total area under the curve.

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