Evaluate the double integral of the absolute value of (xy - 1/4) over the unit square 0 <= x <= 1, 0 <= y <= 1.

Step-by-step solution

1. 1

   Locate the curve where the absolute value switches sign. The integrand changes sign exactly on $xy=\tfrac14$, i.e. $y=\dfrac{1}{4x}$. Inside the unit square this branch is only visible when $\dfrac{1}{4x}\le1$, that is when $x\ge\tfrac14$. For $x<\tfrac14$ the whole vertical slice satisfies $xy\le x<\tfrac14$, so no split is needed there.

2. 2

   Handle the easy strip $0\le x\le\tfrac14$. Here $\left|xy-\tfrac14\right|=\tfrac14-xy$, so

   $\int_0^1\left(\frac14-xy\right)dy=\frac14-\frac{x}{2}$

   $I_1=\int_0^{1/4}\left(\frac14-\frac{x}{2}\right)dx=\frac{1}{16}-\frac{1}{64}=\frac{3}{64}$

3. 3

   Split each slice for $\tfrac14\le x\le1$ at $a=\dfrac{1}{4x}$. Below $a$ the integrand is $\tfrac14-xy$, above it is $xy-\tfrac14$. Using $\dfrac{a}{4}=\dfrac{1}{16x}$ and $\dfrac{xa^2}{2}=\dfrac{1}{32x}$, the two pieces are $\dfrac{1}{32x}$ and $\dfrac{x}{2}-\dfrac14+\dfrac{1}{32x}$, so each slice contributes

   $\frac{x}{2}-\frac14+\frac{1}{16x}$

4. 4

   Integrate the slice contribution over $x$.

   $I_2=\int_{1/4}^{1}\left(\frac{x}{2}-\frac14+\frac{1}{16x}\right)dx=\frac{15}{64}-\frac{3}{16}+\frac{\ln4}{16}=\frac{3}{64}+\frac{\ln4}{16}$

   The $\dfrac{1}{16x}$ term is the sole source of the logarithm — it is the signature of integrating across a hyperbolic boundary.

5. 5

   Add the pieces and verify numerically.

   $I=\frac{3}{64}+\frac{3}{64}+\frac{\ln4}{16}=\frac{3}{32}+\frac{\ln2}{8}$

   Numerically this is $0.09375+0.0866434=0.1803933976$. A $3000\times3000$ midpoint grid over the unit square gives $0.1803933889$ — agreement to seven decimal places, the expected accuracy for a grid method on a kinked integrand ✓.