suvadip said:How to evaluate \(\displaystyle \int^1_0 log(1-x) dx \)
I am confused as log is not defined at 0. Please help
suvadip said:How to evaluate \(\displaystyle \int^1_0 log(1-x) dx \)
I am confused as log is not defined at 0. Please help
The purpose of integrating log(1-x) from 0 to 1 is to find the definite integral of the natural logarithm function between the limits of 0 and 1. This can be useful in various mathematical and scientific applications, such as calculating areas under curves and solving differential equations.
The general formula for integrating log(1-x) from 0 to 1 is ∫(log(1-x))dx = (-x + xlog(1-x)) + C, where C is a constant of integration. This formula can be derived using integration by parts.
The integral of log(1-x) from 0 to 1 is directly related to the natural logarithm function, as it is essentially the inverse process of differentiation. The general formula for integrating log(1-x) can be derived from the derivative of the natural logarithm function, which is 1/(1-x).
The integral of log(1-x) from 0 to 1 has several properties, including linearity, meaning that the integral of a sum of functions is equal to the sum of their integrals. It also has the property of integration by parts, which allows for the integration of more complex functions.
The integration of log(1-x) from 0 to 1 has various real-life applications in fields such as physics, engineering, and economics. For example, it can be used to calculate the work done by a variable force, find the center of mass of a continuous distribution, or determine the present value of a future cash flow.