Double Integrals — Question 10

Question 10 – Solution

Strategy. Control the variation within each cell by its maximum distance from the midpoint, then add the cellwise error bounds.

Step 1: One-cell error The Lipschitz estimate makes $f$ continuous, hence integrable on the closed square. A grid square has side $1/n$, area $1/n^2$, and maximum distance from its midpoint to a corner $$d_n=\frac{\sqrt 2}{2n}.$$ If $M_{ij}$ is its midpoint, the Lipschitz condition gives $$|f(P)-f(M_{ij})|\le 2d_n=\frac{\sqrt 2}{n}$$ throughout that cell. Its integral differs from midpoint height times area by at most $\sqrt 2/n^3$.

Step 2: Global error There are $n^2$ cells, so the triangle inequality yields $$\boxed{\left|\iint_R f\,dA-S_n\right|\le n^2\frac{\sqrt 2}{n^3}=\frac{\sqrt 2}{n}}.$$ To guarantee strict error below $0.01$, require $n>100\sqrt 2\approx 141.421$. The least integer is therefore .

Step 3: Numerical enclosure For $n=142$, $$\frac{\sqrt 2}{142}\approx 0.009959.$$ Thus $$\boxed{2.731-\frac{\sqrt 2}{142}\le\iint_Rf\,dA\le 2.731+\frac{\sqrt 2}{142}},$$ or approximately $$\boxed{2.72104\le\iint_Rf\,dA\le 2.74096}.$$ The half-diagonal, rather than the full cell diagonal, is essential to the stated constant.