Recently, I thought of the following question. Take the standard parabola function, y = x^2:
Is it possible to mathematically show that the area above the area within the parabola (i.e. the shaded area in red) is always less than the area not-within the parabola?
Looking at a visual picture, this makes sense. If you were to enumerate the 4 quadrants from the counterclockwise direction, the parabola does not cover any quadrants 3 and 4 all together. Furthermore, there area regions in quadrant 1 and quadrant 2 that are not covered by the parabola. Thus, logically - for this picture that I have drawn, it seems like the area within the parabola must be less than the area that is not-within the parabola.
But in math, this is not a sufficient way to prove a statement whatsoever!
I first tried to quantify the above logic: Suppose the square grid that I drew has an area of 1: this means that the coordinates on the extremities of the square are (-0.5,0.5), (0.5,0.5), (0.5,-0.5), (-0.5, -0.5).
If you were to integrate this parabola from x = -0.5 to x = 0.5, the area would be: 2* ((0.5^3/3) - (-0.5^3/3)) = 0.166. The area of the quadrants 3 and 4 are 0.5. This means that the area within the parabola is 1 - (0.166 + 0.5) = 0 .344, and the area not-within the parabola is 0.666.
Thus, as expected - it would appear that for this isolated case that I drew, the area contained within the parabola is in fact less than the area not-within the parabola.
But could this now be proven for all cases?
Would proof by induction be useful for doing this?
Thanks!
Not known where the limit horizontal lines are initially drawn, area depends on it.
Ignoring quadrants 3 and 4 considering top half of unit square only,small square has area under curve:
$$ A=\int_0^\frac12 x^2 dx=\dfrac{1}{24} $$
Yellow area= $\dfrac{1}{12} $
White area = $\dfrac12 - \dfrac{1}{12}=\dfrac{5}{12}$ not true.
So the bottom border should be drawn even farther below, by an amount upto the line $ y= -(\dfrac{5}{12}-\dfrac{1}{12})=-\dfrac{1}{3}.$