Starting from the definition, calculate the following

Starting from the definition, calculate the following integral: $$ \iint_{\substack{0 \leqslant x \leqslant 1 \\ 0 \leqslant y \leqslant 1}} xy \, dx \, dy. $$

Step 1 · Understanding the definition

The definition of a double integral over a rectangle $R=[a,b]\times[c,d]$ is:

$$ \iint_R f(x,y)\,dA = \lim_{n,m\to\infty}\sum_{i=1}^{n}\sum_{j=1}^{m} f(x_i^*,y_j^*)\,\Delta x\,\Delta y, $$

where the region is partitioned into rectangles, and we sum the function values times the area of each small rectangle. For our problem the region is $R=[0,1]\times[0,1]$ and the function is $f(x,y)=xy$.

Step 2 · Set up the Riemann sum

Partition $[0,1]$ for $x$ into $n$ equal subintervals: $\Delta x = \frac1n$. Partition $[0,1]$ for $y$ into $m$ equal subintervals: $\Delta y = \frac1m$. Using right endpoints:

$$ x_i=\frac{i}{n},\ i=1,\dots,n, \qquad y_j=\frac{j}{m},\ j=1,\dots,m. $$

Step 3 · Write the Riemann sum

$$ S_{n,m}=\sum_{i=1}^{n}\sum_{j=1}^{m} f(x_i,y_j)\,\Delta x\,\Delta y = \sum_{i=1}^{n}\sum_{j=1}^{m}\left(\frac{i}{n}\right)\left(\frac{j}{m}\right)\cdot\frac1n\cdot\frac1m = \sum_{i=1}^{n}\sum_{j=1}^{m}\frac{ij}{n^2m^2} = \frac{1}{n^2m^2}\sum_{i=1}^{n}\sum_{j=1}^{m} ij. $$

Step 4 · Separate the sums

$$ S_{n,m}=\frac{1}{n^2m^2}\sum_{i=1}^{n} i\sum_{j=1}^{m} j = \frac{1}{n^2m^2}\left(\sum_{i=1}^{n} i\right)\left(\sum_{j=1}^{m} j\right). $$

Step 5 · Use the formula for the sum of the first $k$ integers

Recall $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$, so:

$$ \sum_{i=1}^{n} i = \frac{n(n+1)}{2}, \qquad \sum_{j=1}^{m} j = \frac{m(m+1)}{2}. $$

Step 6 · Substitute into the Riemann sum

$$ S_{n,m}=\frac{1}{n^2m^2}\cdot\frac{n(n+1)}{2}\cdot\frac{m(m+1)}{2} = \frac{n(n+1)\cdot m(m+1)}{4n^2m^2} = \frac{n+1}{4n}\cdot\frac{m+1}{m} = \frac{(n+1)(m+1)}{4nm}. $$

Step 7 · Take the limit

For simplicity, let $n=m$: $S_{n,n}=\dfrac{(n+1)^2}{4n^2}=\dfrac14+\dfrac{2}{4n}+\dfrac{1}{4n^2}$, so

$$ \lim_{n\to\infty} S_{n,n} = \frac14+0+0=\frac14. $$

More generally, for independent $n,m\to\infty$:

$$ \lim_{n,m\to\infty}\frac{(n+1)(m+1)}{4nm} = \lim_{n\to\infty}\frac{1+1/n}{4}\cdot\lim_{m\to\infty}(1+1/m) = \frac14\cdot 1 = \frac14. $$

Step 8 · Verification using iterated integrals

By Fubini's theorem:

$$ \iint_{\substack{0\leqslant x\leqslant 1\\ 0\leqslant y\leqslant 1}} xy\,dx\,dy = \int_0^1\int_0^1 xy\,dx\,dy. $$

Integrating with respect to $x$ first: $\int_0^1 xy\,dx = y\left[\frac{x^2}{2}\right]_0^1=\frac{y}{2}$. Then $\int_0^1\frac{y}{2}\,dy=\frac12\left[\frac{y^2}{2}\right]_0^1=\frac14$. This confirms the result. ✓

Step 9 · Alternative order of integration

$$ \int_0^1\int_0^1 xy\,dy\,dx: \quad \int_0^1 xy\,dy=x\left[\frac{y^2}{2}\right]_0^1=\frac{x}{2}, \qquad \int_0^1\frac{x}{2}\,dx=\frac14. $$

Again we get $\frac14$. ✓

Summary

Key steps: set up the Riemann sum $S_{n,m}=\sum_{i=1}^{n}\sum_{j=1}^{m}\frac{ij}{n^2m^2}$; separate the sums; apply $\sum_{k=1}^n k=\frac{n(n+1)}2$; simplify to $S_{n,m}=\frac{(n+1)(m+1)}{4nm}$; take the limit to get $\frac14$; and verify with iterated integrals in both orders.

Additional Notes

• The double integral is computed from first principles by partitioning the region, evaluating $f$ at sample points, summing areas times values, and taking the limit as the partition becomes finer.
• This definition reduces to iterated single integrals by Fubini's theorem, which is far more practical for computation.
• The value $\frac14$ is the volume under $z=xy$ over the unit square.