Have you considered u = x-1? du = dx
Changing the limits:
1-1 = 0
2-1 = 1
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
Hello, wjn1982!
Original integration: .
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 underCode:|* | | * | * | *. | |:::::* | |:::::| - - - + - - + - - + - - | 1 2
To maintain the same area, using limit
. . the graph looks like this:Code:* | | * | * | *. |:::::* |:::::| - - - + - - + - - 0 1
The graph is moved one unit to the left.
Therefore, the function would be: .