Question

If the area of the region {(x,y)∈R2:y2≥4x,|y|≤x2+2} is Δ, then find the value of [Δ]. [Note: [S] denotes greatest integer less than or equal to S.]

Answer 1

The area of the region is infinite (diverges) so that [Δ] is not finite.

Answer 2

We will show that when the two inequalities

$y^2\ge 4x$ and $|y|\le x^2+2$

are “combined” the resulting set is not a “closed‐bounded” (i.e. compact) set. In other words the “region” has an infinite area.

Let us explain briefly.

Step 1. Rewrite the first inequality as

$y^2\ge4x \Longrightarrow x\le\frac{y^2}4.$

Thus every point in the region must have its $x$–coordinate less than or equal to $y^2/4$.

Step 2. The second inequality is

$|y|\le x^2+2,$

which for every fixed $x$ tells us that

$- (x^2+2)\le y\le x^2+2.$

In particular when $x$ is very “negative” (say $x\to -\infty$) the number $x^2+2$ becomes arbitrarily large so that for a large range of $y$ the inequality is satisfied.

Step 3. In fact for any given real $y$ the condition $x\le y^2/4$ places an upper‐bound on $x$ while the inequality $|y|\le x^2+2$ (when $|y|<2$) is automatically true for all $x$ (since then $x^2+2>0$). In particular if you choose any $y$ with $|y|<2$ then the $x$–coordinate is allowed to be any number on the half–infinite interval $(-\infty,\,y^2/4]$. (Even for $|y|\ge2$ one finds at least one “branch” of the $x$–values which extends to $-\infty.$)

Step 4. Hence the region “stretches infinitely far” in the negative $x$–direction. One may “slice” the region by horizontal lines (i.e. fix a $y$–value) and readily check that for many values of $y$ the corresponding “slice” is an interval of infinite length. (For example, when $|y|<2$ the “slice” is

$(-\infty,\;y^2/4],$

which is unbounded.) Therefore when one “integrates” to compute the area the $x$–integration always extends over an interval of infinite length.

Conclusion. The “area” of the region is infinite (i.e. $\Delta=+\infty$) and consequently the greatest integer function $[\Delta]$ is not a finite number.

Explanation (minimal):

1. The inequality $y^2\ge4x$ is equivalent to $x\le \tfrac{y^2}{4}$.
2. For any fixed $y$ with $|y|<2$ the second inequality $|y|\le x^2+2$ holds automatically for all $x$ because $x^2+2>0$.
3. Thus for such $y$ the $x$–coordinate is free to run all the way from $-\infty$ up to $y^2/4$.
4. This shows that the region extends infinitely in the negative $x$–direction and so its area is unbounded (infinite).