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How do I integrate the function $f(x,y)=15(x^{2}+y^{2})$, in $Q=[0,1]\times[0,1]$ via Riemann sum?

I tried to get the partition $$0=x_{0}<x_{1}<\ldots<x_{n}=1\quad\text{and}\quad 0=y_{0}<y_{1}<\ldots<y_{n}=1$$ and the $$\Delta x=\Delta y=\frac{1}{n}$$ So, \begin{align*} \int\int_{Q}15(x^{2}+y^{2})dxdy & =\lim_{n\to\infty}\sum_{i=0}^{n}\sum_{j=0}^{n}f(i,j)\Delta x \Delta y \\ & = 15\cdot\lim_{n\to\infty}\sum_{i=0}^{n}\sum_{j=0}^{n}(i^{2}+j^{2})\Delta x\Delta y \end{align*} When I compute this double sum, I get $$\frac{1}{3}n(n+1)^{2}(2n+1)$$ But when I substitute it and $\Delta x=\Delta y=1/n$, my limit goes to infinity. How do I fix it?

Thanks!

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The correct Riemann sum is

$$S_n = \frac{15}{n^2}\sum_{i=1}^{n}\sum_{j=1}^{n}[(i/n)^2+(j/n)^2]=\frac{15}{n^4}\sum_{i=1}^{n}\sum_{j=1}^{n}(i^2+j^2)=\frac{15}{n^4}\frac{2n^2(n+1)(2n+1)}{6}.$$

Note that

$$\sum_{i=1}^{n}\sum_{j=1}^{n}(i^2+j^2)=\sum_{i=1}^{n}\left[ni^2+\frac{n(n+1)(2n+1)}{6}\right]=\frac{2n^2(n+1)(2n+1)}{6}.$$

Then

$$\lim_{n \rightarrow \infty}S_n = 5\lim_{n \rightarrow \infty}(1+1/n)(2+1/n)=10$$

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  • $\begingroup$ Why do sum with $i/n$? $\endgroup$ – mvfs314 Jul 11 '14 at 0:07
  • $\begingroup$ You have to partition the interval $[0,1]$ using points between $0$ and $1$ such as $1/n$, $2/n$, ... , $n/n$ $\endgroup$ – RRL Jul 11 '14 at 0:09
  • $\begingroup$ Ok, but the result it shouldn't be $\frac{15}{n^{3}}\frac{2n(n+1)^{2}(2n+1)}{6}$ ? $\endgroup$ – mvfs314 Jul 11 '14 at 0:11
  • $\begingroup$ Use $x_i = i/n$ and $y_j = j/n$. Then you will have $\Delta x = 1/n$ and $\Delta y = 1/n$ and all points $(x_i,y_j) \in [0,1]^2$. $\endgroup$ – RRL Jul 11 '14 at 0:18

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