Verify the relation: -4 < double integral over D

Verify the relation: $$ -4 < \iint_D\left(x+xy-x^2-y^2\right)\,dx\,dy < \frac23, $$ where $D=\{(x,y):0<x<1,\ 0<y<2\}$.

Step 1 · Identify the region

$D=[0,1]\times[0,2]$, area $A=1\times2=2$. The function is $f(x,y)=x+xy-x^2-y^2$.

Step 2 · Find critical points

$$ \frac{\partial f}{\partial x}=1+y-2x=0, \qquad \frac{\partial f}{\partial y}=x-2y=0. $$

From the second equation $x=2y$; substituting into the first: $1+y-4y=0\Rightarrow y=\frac13$, $x=\frac23$. The critical point $\left(\frac23,\frac13\right)$ lies in $D$.

Step 3 · Evaluate $f$ at the critical point

$$ f\!\left(\frac23,\frac13\right) = \frac23+\frac23\cdot\frac13-\left(\frac23\right)^2-\left(\frac13\right)^2 = \frac23+\frac29-\frac49-\frac19 = \frac23-\frac13=\frac13. $$

Step 4 · Second derivative test

$$ f_{xx}=-2,\quad f_{yy}=-2,\quad f_{xy}=1, \qquad H=f_{xx}f_{yy}-f_{xy}^2=4-1=3>0. $$

Since $H>0$ and $f_{xx}<0$, $\left(\frac23,\frac13\right)$ is a local maximum with $f=\frac13$.

Step 5 · Check the boundary

Edge $x=0$: $f(0,y)=-y^2$, ranging from $0$ (at $y=0$) to $-4$ (at $y=2$).
Edge $x=1$: $f(1,y)=y-y^2$, extremum at $y=\frac12$ giving $\frac14$; values $0,\frac14,-2$ at $y=0,\frac12,2$.
Edge $y=0$: $f(x,0)=x-x^2$, extremum at $x=\frac12$ giving $\frac14$; values $0,\frac14,0$ at $x=0,\frac12,1$.
Edge $y=2$: $f(x,2)=3x-x^2-4$, increasing on $[0,1]$ (its critical point $x=\frac32$ is outside); values $-4$ at $x=0$ and $-2$ at $x=1$.

Step 6 · Determine global extrema

Comparing the interior value $\frac13$ with all boundary values $\{-4,0,\frac14,-2\}$:

$$ \max_{(x,y)\in D} f(x,y)=\frac13, \qquad \min_{(x,y)\in D} f(x,y)=-4. $$

Step 7 · Apply the estimation theorem

$$ (-4)\cdot2 \leqslant \iint_D f\,dA \leqslant \frac13\cdot2 \quad\Longrightarrow\quad -8\leqslant I\leqslant\frac23. $$

This is weaker than the stated $-4<I<\frac23$, so we compute the exact value.

Step 8 · Calculate the exact integral

Integrating with respect to $x$ first:

$$ \int_0^1\left(x+xy-x^2-y^2\right)dx = \left[\frac{x^2}{2}+\frac{x^2y}{2}-\frac{x^3}{3}-xy^2\right]_0^1 = \frac16+\frac{y}{2}-y^2. $$

Then integrating with respect to $y$:

$$ \int_0^2\left(\frac16+\frac{y}{2}-y^2\right)dy = \left[\frac{y}{6}+\frac{y^2}{4}-\frac{y^3}{3}\right]_0^2 = \frac13+1-\frac83 = -\frac43. $$

Step 9 · Verify the inequality

$$ -4 < -\frac43 < \frac23 \quad \checkmark $$

since $-4<-1.333$ and $-1.333<0.667$. The inequality is verified.

Summary

Key steps: the interior critical point $\left(\frac23,\frac13\right)$ is a local maximum with $f=\frac13$; the global minimum on the boundary is $f(0,2)=-4$; so $-4\leqslant f\leqslant\frac13$ on $D$; the exact integral is $-\frac43$, which sits strictly between $-4$ and $\frac23$.

Additional Notes

• A full extremum search on a closed rectangle requires checking both interior critical points (via the Hessian) and all four boundary edges.
• The exact value $-\frac43$ falls comfortably inside the estimation-theorem bounds scaled by the area $2$.