# Area/riemann sums questions

1. Aug 4, 2011

### Rear Naked

1) if you have 2 line segments making a right angle, and connect the endpoints with a line segment with an outward curve relative to the vertex, will the area inside always be irrational?

2) if you have 2 line segments making a right angle, and connect the endpoints with a line segment with an inward curve relative to the vertex, will the area inside always be irrational?

3) if we are able to construct the Riemann sum partition points and C values within at any points we wish, is it possible to exactly replicate the value of the integral of a curve?

4) if that same curve fluctuates between concave and convex, the area could be rational right?

Basically, can orange area ever equal the green area?

I'm only talking about curves here, not linear functions.

Thanks

2. Aug 4, 2011

### gb7nash

No. Look at the area bounded by the curve y = x2, y-axis and y = 1. This area is equal to:

$$1*1 - \int_0^1 \! x^2 dx = 1 - \frac{1}{3} = \frac{2}{3}$$

Also, the area under the curve follows the description of 2) and as you can see, the area is also rational.

I don't see any reason why this wouldn't be possible, but this will almost never happen.

Sure.

Last edited: Aug 4, 2011
3. Aug 4, 2011

### Rear Naked

i think you made a mistake when copying number 3?

Why do you say it would almost never happen?

If we just chose the C value inside the riemann partition endpoints at random, or chose them to be directly in the center, I see why you would say that...but can we choose that C value, (ie height of rectangle) to be any value, and thus the height could be any value, and thus the areas could be any value, and thus there should always be a C value where the those two colored areas would be equal? Perhaps not ALWAYS, but quite possibly.

Thanks for your help by the way!

4. Aug 4, 2011

Fixed

5. Aug 4, 2011

### gb7nash

Well, you didn't say that at first! I assumed you were talking about the standard left,right and midpoint riemann sums. If you're allowed to choose how far along the curve each rectangle is, you can get an exact area.

If you want to really simplify this, we only need one rectangle to calculate the exact area of a continuous function:

http://archives.math.utk.edu/visual.calculus/5/average.1/index.html

This is commonly referred to as the mean value theorem for integrals. Basically, if you're looking at the area of a curve from a to b, there exists a c in [a,b] such that f(c)(b-a) is the area of the curve. In this sense, you only need one rectangle and the height of the rectangle depending on what c is.

6. Aug 5, 2011

### Rear Naked

I did say that at first haha

thanks