I just started learning this topic and I'm quite confused about the methods that are being used to create the riemann sums. The method I tried using involves calculating the integral as the sum:

$$\lim_{n\to \infty}\sum_{i=1}^n f(x_i^*)\Delta x$$

where $x_i^*=a+i\Delta x$ and $\Delta x=\frac{b-a}{n}$. Using this I tried evaluating the integral

$$\int_1^2 {\frac{1}{x}}dx$$

$$\Delta x=\frac{2-1}{n}=\frac{1}{n},\hspace{0.3cm} x_i^*=a+i\Delta x=1+\frac{i}{n}=\frac{i+n}{n},\hspace{0.3cm} f(x_i^*)=\frac{n}{n+i}$$

All I need to do is calculate $$\lim_{n\to\infty}\sum_{i=1}^n \frac{1}{n+i}$$

Instead in couple of other books using the same exact example they provide completely different solution.

„Subdivide the interval [1,2] into n parts so that the points of division $x_i$ $(i=0, 1, 2, \ldots, n)$ form the geometric progression:

$$x_0=1,\hspace{0.2cm} x_1=q,\hspace{0.2cm} x_2=q^2,\hspace{0.2cm} x_3=q^3 \hspace{0.2cm}\ldots\hspace{0.2cm} x_n=q^n=2, $$ whence $q=\sqrt[n]2$ “

There is another way to divide this interval. It involves using $x_i=2^{\frac{i}{n}}$.

I just don't understand what's the motivation behind these solutions. Frankly I'm not sure how you calculate $x_i^*$ and $\Delta x$ after doing these interval divisions.

Also using the geometric progression divsion seems to be quite popular in some similar problems.

  • 2
    $\begingroup$ The choice of $x_i, x_i^{*} $ is completely arbitrary as long the maximum of $\Delta x_i=x_i-x_{i-1}$ tends to $0$ as $n\to\infty$. The choice of $x_i$ in arithmetic progression is more popular, but the choice of $x_i$ in geometric progression works very well for integrating powers like $x^k$ (including your case $k=-1$). $\endgroup$ – Paramanand Singh Apr 9 '18 at 2:28

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