Integral as a limit of a sum How do I get 
$$
\int_a^b \frac{1}{x}dx = \ln\left(\frac{b}{a}\right)$$
 as a limit of sum. The constant width partition of the interval $(a,b)$ doesn't seem to work.
 A: Take $x_i = a \left( \frac{b}{a} \right)^{i/n}$ then:
$$\lim\limits_{n \to \infty} \sum_{i = 0}^{n-1} \frac{x_{i+1} - x_i}{x_i} = \lim\limits_{n \to \infty} \frac{ \left(\frac{b}{a}\right)^{\frac{1}{n}} - 1 }{\frac{1}{n}} = \ln \frac{b}{a}$$
A: Lets do it for $\int\limits_1^b\frac{1}{x}dx$:
Consider the following partition of our interval  $\,[1,b]$:
$$P_n=\{1, \sqrt[n]{b},\sqrt[n]{b^2},\dots,\sqrt[n]{b^n}=b\}$$
Now, we need to calculate the upper and lower Darboux sums in respect to the $P_n$'s. Notice that $\frac{1}{x}$ is monotonic decreasing on the interval so the $\sup$ and the $\inf$ are obtained on the edges:
$$\mathcal{U}\left(\frac{1}{x},P_n\right)=\sum_{k=1}^n b^{-\frac{k}{n}}(b^{\frac{k}{n}}-b^{\frac{k-1}{n}})=\sum_{k=1}^n 1-b^{-\frac{1}{n}}=n(\sqrt[n]{b}-1)$$
define $g(x)=b^x$:
$$n(\sqrt[n]{b}-1)=\frac{g(\frac{1}{n})-g(0)}{\frac{1}{n}-0}\to_{n\to\infty}g'(0)=b^x\ln b|_{x=0}=\ln b$$
I'll leave it to you to figure out the lower sums as it's practically the same procedure.
A: The constant width partition works fine. No need for fancy partitions. To wit:
Let $x_k = a+\frac{k}{N}(b-a)$, with $k = 0,...,N$. We have, of course, $x_0=a$, $x_N = b$.
Using the Taylor series expansion of $x \mapsto \ln x$, and the mean value theorem, we obtain the bounds, for $x \geq y$:
$$ \ln y + \frac{x-y}{y} - \frac{(x-y)^2}{2 y^2} \leq \ln x \leq \ln y + \frac{x-y}{y}.$$
Letting $y=x_k$, $x=x_{k+1}$, we get:
$$\ln x_k + \frac{1}{N x_k} - \frac{1}{2 N^2 x_k^2} \leq \ln x_{k+1} \leq \ln x_k + \frac{1}{N x_k}.$$
Rearranging gives:
$$- \frac{1}{2 N^2 x_k^2} \leq \ln x_{k+1} - \ln x_k - \frac{1}{N x_k} \leq 0.$$
Now sum the inequality over $k = 0,...,N-1$, to get:
$$- \frac{1}{2 N^2} \sum_{k=0}^{N-1}\frac{1}{x_k^2} \leq \ln \frac{b}{a} - \frac{1}{N}\sum_{k=0}^{N-1} \frac{1}{x_k} \leq 0.$$
Since $x_k\geq a$, we obtain:
$$- \frac{1}{2 N a^2} \leq \ln \frac{b}{a} - \frac{1}{N}\sum_{k=0}^{N-1} \frac{1}{x_k} \leq 0.$$
Now let $N \to \infty$, to obtain the desired result,
$$ \lim_{N \to \infty} \frac{1}{N}\sum_{k=0}^{N-1} \frac{1}{x_k} = \int_a^b \frac{1}{t} dt = \ln \frac{b}{a}.$$
A: $$\int _a^b {f(t) dt}=\lim_{n\to\infty} \frac{(b-a)}{n}\sum \limits_{k=1}^n f(\frac{k(b-a)}{n}+a)$$
$$\int _a^b \frac{1}{t} dt=\lim_{n\to\infty} \frac{(b-a)}{n}\sum \limits_{k=1}^n \frac{n}{k(b-a)+na}=\lim_{n\to\infty} \frac{(b-a)}{1}\sum \limits_{k=1}^n \frac{1}{k(b-a)+na}$$
$$\int _a^b \frac{1}{t} dt=\lim_{n\to\infty} \frac{(b-a)}{n}\sum \limits_{k=1}^n \frac{n}{k(b-a)+na}=\lim_{n\to\infty} \frac{(b-a)}{(b-a)}\sum \limits_{k=1}^n \frac{1}{k+\frac{na}{b-a}}$$
$$\int _a^b \frac{1}{t} dt=\lim_{n\to\infty} \sum \limits_{k=1}^n \frac{1}{k+\frac{n}{(b/a)-1}}$$
$b/a=x$
$$f(x-1)=\int _a^b \frac{1}{t} dt=\lim_{n\to\infty} \sum \limits_{k=1}^n \frac{1}{k+\frac{n}{x-1}}$$
$$f(x)=\lim_{n\to\infty} \sum \limits_{k=1}^n \frac{1}{k+\frac{n}{x}}$$
$$f(x)=\lim_{n\to\infty}\frac{x}{n} \sum \limits_{k=1}^n \frac{1}{1+\frac{kx}{n}}=\lim_{n\to\infty} \frac{x}{n} \sum \limits_{k=1}^n (1-\frac{kx}{n}+\frac{k^2x^2}{n^2}-\frac{k^3x^3}{n^3}+....)=\lim_{n\to\infty} \frac{x}{n} \sum \limits_{k=1}^n 1 -\lim_{n\to\infty} \frac{x^2}{n^2} \sum \limits_{k=1}^n k+\lim_{n\to\infty} \frac{x^3}{n^3} \sum \limits_{k=1}^n k^2-\lim_{n\to\infty} \frac{x^4}{n^4} \sum \limits_{k=1}^n k^3+.....$$
$$f(x)=\lim_{n\to\infty} \frac{x}{n} \sum \limits_{k=1}^n 1 -\lim_{n\to\infty} \frac{x^2}{n^2} \sum \limits_{k=1}^n k+\lim_{n\to\infty} \frac{x^3}{n^3} \sum \limits_{k=1}^n k^2-\lim_{n\to\infty} \frac{x^4}{n^4} \sum \limits_{k=1}^n k^3+.....=\lim_{n\to\infty} \frac{x}{n} n -\lim_{n\to\infty} \frac{x^2}{n^2} (\frac{n^2}{2}+\frac{n}{2})+\lim_{n\to\infty} \frac{x^3}{n^3}   (\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6})-\lim_{n\to\infty} \frac{x^4}{n^4}(\frac{n^4}{4}+\frac{n^3}{2}+\frac{n^2}{4})+....$$

$$\sum \limits_{k=1}^{n}   k^m=\frac{n^{m+1}}{m+1}+a_mn^m+....+a_1n=\frac{n^{m+1}}{m+1}+\sum \limits_{j=1}^m a_jn^j$$  where $a_j$ are constants.where aj are constants. More information about summation http://en.wikipedia.org/wiki/Summation

After solving limits. We get:
$$f(x)=\frac{x}{1} -\frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}+ ....=\sum \limits_{k=1}^{\infty} (-1)^{k+1} \frac{x^k}{k}=\ln(x+1)$$
$$f(x-1)=\ln(x)$$
$b/a=x$
$$\int _a^b \frac{1}{t} dt=f(x-1)=\ln(x)=\ln(b/a)$$
