# Calculate Integral of $f(x,y)=(x+y)^2$ on the square $[0,1]\times [0,1]$ with Riemann sums

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?

• I would like to see how you go from summatories to k-formulas. Jun 28 at 18:01
• 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
– LEY
Jun 28 at 18:03

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$$\frac23+\frac12=\frac76$$

• At which position exactly. I can't see it :/ Jun 28 at 18:12

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.

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