Question

The area bounded by region {(x, y) : |x + ay + a2| ≤ 1 ∀ a ∈ [0, 1]} is A, then [A] is (where [.] denotes greatest integer function)

Answer 1

4

Answer 2

We need the area of

$$R=\{(x,y) : |\,x+ay+a^2|\le1 \quad \forall\,a\in[0,1]\}.$$

That is, for every $a\in[0,1]$,

$$-1\le x+ay+a^2\le 1.$$

Define

$$f(a)=x+ay+a^2.$$

Since $f(a)$ is quadratic (with positive coefficient of $a^2$) it is convex. Thus, the maximum on $[0,1]$ occurs at the endpoints $a=0$ and $a=1$ and the minimum occurs either at the vertex or at an endpoint.

Upper bound ($f(a)\le1$):

1. At $a=0$:
     $f(0)=x\le1$          (1)

2. At $a=1$:
     $f(1)=x+y+1\le1 \implies x+y\le0$   (2)

Lower bound ($f(a)\ge-1$):

The vertex is at

$$a^*=-\frac{y}{2}.$$

We consider three cases.

Case I: For $y\in[-2,0]$
Here, $a^*\in[0,1]$. Then the minimum occurs at $a^*$:

$$f(a^*)=x - \frac{y^2}{4}\ge -1 \implies x\ge -1+\frac{y^2}{4}. \quad (3)$$

Also, at $a=0$: $x\ge -1$ (weaker than (3) since $-1+\frac{y^2}{4}\ge-1$).

Thus for $y\in[-2,0]$ the conditions are:

$$-1+\frac{y^2}{4}\le x\le \min\{1,\,-y\} \quad \text{(since (2) says $x\le -y$)}.$$

Case II: For $y>0$
Now $a^*<0$ so the minimum is at $a=0$:

$$x\ge -1.$$

And maximum gives (2):

$$x+y\le0\implies x\le -y.$$

Also (1) gives $x\le1$ but $x\le -y$ is stronger for $y>0$.
For a nonempty interval we need $-1\le -y$ i.e. $y\le1$.
So for $y\in[0,1]$:

$$-1\le x\le -y.$$

Case III: For $y<-2$
Now $a^*>1$ so the minimum is at $a=1$:

$$f(1)=x+y+1\ge -1 \implies x\ge -2-y.\quad (4)$$

The upper bound from (1) is $x\le1$. Also (2) gives $x\le -y$; but for $y<-2$, note that $-y>2$ so the $x\le1$ condition is stricter.
Thus, for $y<-2$:

$$-2-y\le x\le 1,$$

but this interval is nonempty only if $-2-y\le 1$ i.e. $y\ge -3$.
Hence for $y\in[-3,-2]$ we have the region.

Area computation:

1. For $y\in[-2,0]$:
    $x$ runs from $L(y)=-1+\frac{y^2}{4}$ to $U(y)=-y$.
   Area $A_1$:

   $$A_1=\int_{y=-2}^{0}\Bigl[(-y)-\Bigl(-1+\frac{y^2}{4}\Bigr)\Bigr]dy =\int_{-2}^{0}\Bigl(1-y-\frac{y^2}{4}\Bigr)dy.$$

   Antiderivative:

   $$\int\Bigl(1-y-\frac{y^2}{4}\Bigr)dy = y-\frac{y^2}{2}-\frac{y^3}{12}.$$

   Evaluate from $-2$ to $0$:

   $$\Bigl[ y-\frac{y^2}{2}-\frac{y^3}{12}\Bigr]_{-2}^{0} = \Bigl[0\Bigr]-\Bigl[ -2-\frac{4}{2}-\frac{(-8)}{12}\Bigr].$$

   At $y=-2$:
   $-2-\;2+\frac{8}{12}=-4+\frac{2}{3}=-\frac{10}{3}$.
   Thus, $A_1=0-(-\frac{10}{3})=\frac{10}{3}$.

2. For $y\in[0,1]$:
    $x$ runs from $-1$ to $-y$.
   Area $A_2$:

   $$A_2=\int_{0}^{1}\Bigl[(-y)-(-1)\Bigr]dy=\int_{0}^{1}(1-y)dy =\left[y-\frac{y^2}{2}\right]_0^1=1-\frac{1}{2}=\frac{1}{2}.$$

3. For $y\in[-3,-2]$:
    $x$ runs from $L(y)=-2-y$ to $U(x)=1$.
   Area $A_3$:

   $$A_3=\int_{y=-3}^{-2}\Bigl[1-(-2-y)\Bigr]dy=\int_{-3}^{-2}(3+y)dy.$$

   Antiderivative:

   $$\int (3+y)dy=3y+\frac{y^2}{2}.$$

   Evaluate from $-3$ to $-2$:

   $$\Bigl[3y+\frac{y^2}{2}\Bigr]_{-3}^{-2} = \Bigl[ -6+\frac{4}{2}\Bigr] - \Bigl[ -9+\frac{9}{2}\Bigr] = \Bigl[-6+2\Bigr] -\Bigl[-9+4.5\Bigr] = (-4) - (-4.5)=0.5.$$

Total Area:

$$A=A_1+A_2+A_3=\frac{10}{3}+\frac{1}{2}+\frac{1}{2}=\frac{10}{3}+1=\frac{13}{3}\approx 4.3333.$$

Taking the greatest integer, we have $[A]=4$.

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Summary of the Answer and Meta-information:

• Explanation (minimal):
  For each $a\in[0,1]$, the inequality $|x+ay+a^2|\le1$ gives two conditions: one from the maximum (at $a=0,1$) and one from the minimum (at the vertex or endpoint). Dividing into cases $y\in[-2,0],\; y\in[0,1]$, and $y\in[-3,-2]$ we find the region in the $xy$-plane. Integrating the length in $x$ over the corresponding $y$-intervals gives total area $A=\frac{13}{3}$ so [A]=4.