# How to calculate the integral as a limit of sum?

1. Jun 22, 2012

### PrakashPhy

1. The problem statement, all variables and given/known data

The problem is to find the integral of function $\frac{1}{x}$ using the definition i.e the area under the curve as a limit of a sum?

2. Relevant equations
$\int_a^b \frac{1}{x} dx = ln(\frac{b}{a})$

3. The attempt at a solution
I tried with dividing the interval [a,b] into n parts such that $h=\frac{b-a}{n}$ but what I got was a too vague expression of the sum.

$\lim_{n \rightarrow ∞} \frac{b-a}{n} = h$
now
$\int_a^b \frac{1}{x} dx = \lim_{h \rightarrow 0}\bigg[ \frac{1}{a}×h+\frac{1}{a+h}×h+\frac{1}{a+2h}×h^2...\bigg]$

how do I proceed now

2. Jun 22, 2012

### algebrat

So is the idea that you should end up with the power series for ln(x)?

3. Jun 23, 2012

### PrakashPhy

Something like that; probably some limit of $ln(x)$ because $\lim_{h\rightarrow 0}$ is there intact throughout.

Last edited: Jun 23, 2012
4. Jun 23, 2012

### SammyS

Staff Emeritus
Maybe that's a typo, but that h should not be squared in your last line.

$\int_a^b \frac{1}{x} dx = \lim_{h \rightarrow 0}\bigg[ \frac{1}{a}×h+\frac{1}{a+h}×h+\frac{1}{a+2h}×h^2...\bigg]$

should be

$\int_a^b \frac{1}{x} dx = \lim_{h \rightarrow 0}\bigg[ \frac{1}{a}×h+\frac{1}{a+h}×h+\frac{1}{a+2h}×h ...\bigg]$