Double Integrals — Question 3

Question 3 – Solution

Strategy. Because $xy$ increases in each coordinate on the unit square, lower-left and upper-right corners produce the lower and upper sums.

Step 1: Lower sum Every cell has area $1/n^2$. Using indices $i,j=0,\ldots,n-1$ at lower-left corners, $$\begin{align*} L_n&=\frac 1{n^2}\sum_{i=0}^{n-1}\sum_{j=0}^{n-1}\frac{i}{n}\frac{j}{n} =\frac 1{n^4}\left(\sum_{i=0}^{n-1}i\right)^2\\ &=\frac 1{n^4}\left(\frac{n(n-1)}2\right)^2 =\frac{(n-1)^2}{4n^2}. \end{align*}$$

Step 2: Upper sum and squeeze Upper-right corners give $$U_n=\frac 1{n^4}\left(\sum_{i=1}^{n}i\right)^2 =\frac{(n+1)^2}{4n^2}.$$ Both sums approach $1/4$. Since $L_n\le\iint_Rxy\,dA\le U_n$, $$\boxed{\iint_Rxy\,dA=\frac 14}.$$

Step 3: Resolution The uncertainty is $$U_n-L_n=\frac{(n+1)^2-(n-1)^2}{4n^2}=\frac 1n.$$ Thus $1/n<0.01$ requires $n>100$, and the least integer is . The strict inequality rules out $n=100$.