1. ## Changing integration limits

Hi
Am a bit rusty on integration.

Was wondering,

Original integration:

integrate 1/x between limits 1 and 2

Problem:

need limits to be 0 and 1. what do i change the 1/x when i change the limits?

cheers

2. Have you considered u = x-1? du = dx

Changing the limits:
1-1 = 0
2-1 = 1

$\displaystyle \int_{1}^{2}\frac{1}{x}dx = \int_{0}^{1}\frac{1}{u+1}du$

3. Thanks for quick replies guys.
TKHUNNY i think that might be right.

Galactacus, yeah its a unusual format for a question. The aim is to apply monte carlo simulation to evaluate the integral, but this particular monte carlo simulation technique is based on evaluating the integral between 0 and 1.
So a standard question gives you an integral with the "wrong" limits and you have to change these limits to 0 and 1 to enable monte carlo

4. Hello, wjn1982!

Original integration: .$\displaystyle \int^2_1\frac{1}{x}\,dx$

Problem: need limits to be 0 and 1.
How do i change the 1/x when i change the limits?

If I understand the problem, we have a horizontal translation . . .

The original integral represents the area under $\displaystyle y \,=\,\frac{1}{x}\,\text{ from }x = 1\text{ to } x = 2$
Code:
            |*
|
| *
|  *
|     *.
|     |:::::*
|     |:::::|
- - - + - - + - - + - -
|     1     2

To maintain the same area, using limit $\displaystyle x = 0\text{ to }x = 1$
. . the graph looks like this:
Code:
       *    |
|
*   |
*  |
*.
|:::::*
|:::::|
- - - + - - + - -
0     1

The graph is moved one unit to the left.

Therefore, the function would be: .$\displaystyle y \:=\:\frac{1}{x+1}$

5. Great, thanks!!