Calculate the following integral: double integral over D

Calculate the following integral: $$ \iint_D xy\,dx\,dy, $$ where the region $D$ is bounded by the curves $xy=1$ and $x+y=\frac{5}{2}$.

Step 1 · Find intersection points

$\frac1x=\frac52-x \Rightarrow 2=x(5-2x)=5x-2x^2 \Rightarrow 2x^2-5x+2=0$. By the quadratic formula, $x=\frac{5\pm3}{4}$, giving $x=2$ or $x=\frac12$. At $x=2$: $y=\frac12$; at $x=\frac12$: $y=2$. Intersection points: $\left(2,\frac12\right)$ and $\left(\frac12,2\right)$.

Step 2 · Determine which curve is above

At $x=1$: hyperbola gives $y=1$, line gives $y=\frac32$. Since $\frac32>1$, the line is above the hyperbola on $\left[\frac12,2\right]$.

Step 3 · Set up the integral

$$ I = \int_{1/2}^{2}\int_{1/x}^{5/2-x} xy\,dy\,dx. $$

Step 4 · Integrate with respect to $y$

$$ \int_{1/x}^{5/2-x} xy\,dy = \frac{x}{2}\left[\left(\frac52-x\right)^2 - \frac{1}{x^2}\right] = \frac{25x}{8}-\frac{5x^2}{2}+\frac{x^3}{2}-\frac{1}{2x}. $$

Step 5 · Integrate with respect to $x$

$$ I = \int_{1/2}^{2}\left(\frac{25x}{8}-\frac{5x^2}{2}+\frac{x^3}{2}-\frac{1}{2x}\right)dx = \left[\frac{25x^2}{16}-\frac{5x^3}{6}+\frac{x^4}{8}-\frac{\ln x}{2}\right]_{1/2}^{2}. $$

At $x=2$: $\frac{25}{4}-\frac{20}{3}+2-\frac{\ln2}{2} = \frac{19}{12}-\frac{\ln2}{2}$.

At $x=\frac12$: $\frac{25}{64}-\frac{5}{48}+\frac{1}{128}+\frac{\ln2}{2} = \frac{113}{384}+\frac{\ln2}{2}$.

$$ I = \left(\frac{19}{12}-\frac{\ln2}{2}\right) - \left(\frac{113}{384}+\frac{\ln2}{2}\right) = \frac{495}{384}-\ln2 = \frac{165}{128}-\ln2. $$

Additional Notes

• The $\frac{\ln2}{2}$ terms appear with opposite signs at the two endpoints and combine to $-\ln2$ rather than canceling entirely, since both evaluations carry the same sign of $\ln x$ contribution structurally but the outer subtraction doubles it.