1
$\begingroup$

I need to calculate the integral of $f(x,y)=(x+y)^2$ on the square $[0,1]\times [0,1]$ by using the definition of Riemann Sums. I know with Fubini that it has to be $\frac{7}{6} $ but in my solution I get $\frac{1}{2} $. Where is my mistake?

Let $z_k$ be a partition of $[0,1]\times[0,1]$ with $z_k=z_1 \times z_2$ and $z_1=\{0,1/k,2/k,...,1\}$ and $z_2=\{0,1/k,2/k,...,1\}$

$x_{ij} \in [(i-1)/k,i/k]\times[(j-1)/k,j/k]$ from wehre we choose $x_{ij}=(i/k,j/k)$

Then: \begin{align} S_{z_k}(f)&=\sum_{i,j=1}^k(i/k+j/k)^2\cdot1/k^2\\ &=1/k^4 (\sum_{i=1}^k i^2+2\sum_{i=1}^k \sum_{j=1}^kij+\sum_{j=1}^kj^2\\ &=1/k^4((1/3)(k(k+1)(2k+1)+(1/2)k^2(k+1)^2)\\ &=1/2+5/(3k)+3/(2k^2)+1/(3k^3) \rightarrow 1/2 \end{align} for $k$ to infinity.

Can someone help me?

$\endgroup$
2
  • $\begingroup$ I would like to see how you go from summatories to k-formulas. $\endgroup$
    – Ripi2
    Jun 28 at 18:01
  • $\begingroup$ the last equation is wrong, you miss a $\sum_j$ for $\sum_i i^2$,so is $sum_i$ missing for $\sum_j j^2$,add them and you get2/3+1/2=7/6 $\endgroup$
    – LEY
    Jun 28 at 18:03
1
$\begingroup$

the last equation($\sum_{i,j=1}^k(i/k+j/k)^2\cdot1/k^2\\ =1/k^4 (\sum_{i=1}^k i^2+2\sum_{i=1}^k \sum_{j=1}^kij+\sum_{j=1}^kj^2$) is wrong, RHS you miss a $\sum_j $for $\sum_i i^2$,so is $\sum_i $missing for $\sum_j j^2$,add them and you get$\frac23+\frac12=\frac76 $

$\endgroup$
1
  • $\begingroup$ At which position exactly. I can't see it :/ $\endgroup$
    – Blue2001
    Jun 28 at 18:12
1
$\begingroup$

Starting from your sum $S_{z_{k}}(f)$:

\begin{align} S_{z_{k}}(f) &= \sum_{i,j=1}^{k}(i/k+j/k)^{2}\times 1/k^{2} \tag{1}\\ &= \frac{1}{k^{4}}\left(\sum_{i,j=1}^{k}i^{2} + 2\sum_{i,j=1}^{k} ij + \sum_{i,j=1}^{k} j^{2}\right) \tag{2}\\ &= \frac{1}{k^{4}}\left( k \frac{k(k+1)(2k+1)}{6} +\frac{k^{2}(k+1)^{2}}{2}+k\frac{k(k+1)(2k+1)}{6}\right) \tag{3} \end{align}

After simplifying and letting $k\to \infty$: we obtain $$\frac{1}{3}+\frac{1}{2}+\frac{1}{3}=\frac{7}{6}.$$


To point out where your mistake is: You go from $(1)$ above to $$\frac{1}{k^{4}}\left(\sum_{i=1}^{k}i^{2} + 2\sum_{i,j=1}^{k}ij + \sum_{j=1}^{k} j^{2}\right)$$ which is incorrect because you are missing a summation over $j$ in the first term $"\sum_{i=1}^{k} i^{2}"$ and similarly you miss a summation over $i$ in the last term $"\sum_{j=1}^{k} j^{2}$" (compare with $(2)$). Leaving out these summations have the effect that when $k\to \infty$ neither of the aforementioned terms contribute to the end result.

$\endgroup$
4
  • $\begingroup$ In your third equation with k(k+1)(2k+1)/6 why is there k*k*(k+1)(2k+1)/6. Why are there two k? $\endgroup$
    – Blue2001
    Jun 28 at 18:11
  • 1
    $\begingroup$ We know that $\sum_{i=1}^{k}i^{2} = \frac{k(k+1)(2k+1)}{6}$ so $$\sum_{i,j=1}^{k} i^{2} = \sum_{j=1}^{k} \frac{k(k+1)(2k+1)}{6} = k \frac{k(k+1)(2k+1)}{6}$$ $\endgroup$ Jun 28 at 18:14
  • $\begingroup$ Oh thank you very much !! :)) $\endgroup$
    – Blue2001
    Jun 28 at 18:15
  • $\begingroup$ @Blue2001 Happy to help, sorry for missing your question where your mistake is! I have added it now in case you are still wondering. $\endgroup$ Jun 28 at 18:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.