Question:
For the rectangle $I=[0,1]\times [0,1] $ in the plane $\Bbb R^2$ ,
Define the function $f:I\to\Bbb R$ by $f(x,y)=x^2y$ for $(x,y)\in I$
Use the Darboux Sum Convergence critation to evaluate $\int_{I}f$
Solution trail:
Let $Q_n$ be partition of $[0,1]$ observed by dividing $[0,1]$ into $n$ equal pieces.
Let $P_n=(Q_n,Q_n)$ and $Q_n=\{\frac{1}{n}\}$
Then, $\vert\vert P\vert\vert=\sqrt{||Q_n||^2+||Q_n||^2}=\sqrt{\frac{1}{n^2}+\frac{1}{n^2}}=\frac{2}{n}\to 0$ as $n\to \infty$
Since $f(x,y)=x^2y$ is continuous on I, $\int _{I} f$ exists.
Then, $\int_{I}f=\lim U(f,P_n)=\lim L(f,P_n)$ as $n\to\infty$
$$U(f,P_n)=\sum_{J in P}M(f,J)volJ$$
$J$ in $P_n$ is of the form $[\frac{k-1}{n}, \frac{k}{n}]\times [\frac{\ell -1}{n},\frac{\ell}{n}]$ for $1\le k,\ell \le n$
There are $n^2$ such intervals,
$M(f,J)=\frac{k^2\ell}{n^3}$
$vol(J)=1/n\cdot1/n$
$U(f,P_n)=\sum{\frac{k^2\ell}{n^3}\frac{1}{n^2}}=\frac{1}{n^5}\sum_{k=1}^{n}{k^2}\sum_{\ell=1}^{n}{\ell}=\frac{1}{n^5}(\frac{n(n+1)}{2})^3$
I think, the question is solved in that way. However I have just solved such a question one time. Thus, I could not see my mistakes and drawbacks. The result of my solution is false i guess. Please can one help me to correct my solution trail Thank you:)
I think $M(f,J)$ should be $$\frac{kl}{n^2} \text{, not } \frac{k^2 l }{n^3}.$$