Question

How to evaluate the definite integral of $\ln x$ from $0$ to $1$?

Answer 1

Hint: To solve this question we need to know the concept of definite integral. We are given with the limits of the integral for the given function. For integrating the given function,$\ln x$ the question will be solved using the method of integration by parts. So the formula used here will be $\int udv=uv-\int vdu$, where $u$ and $v$ are the two functions.

Complete step by step answer:
To solve the integration of the function we should know the formula of the integration. The question is $\int \ln xdx$ with limit as $[0,1]$, which could be written as
$\underset{0}{\overset{1}{\int }}\ln xdx$
The formula used to integrate the question $\ln x$ is $\int udv=uv-\int vdu$ is , if the function is $\ln x$ is $u$ and the function $dv=dx$. Mathematically it will be written as:
$\Rightarrow \underset{0}{\overset{1}{\int }}\ln xdx$
On applying the formula we get $intudv=uv-\int vdu$ , where $u=\ln x$ on differentiating both side of the function $\ln x$ we get:
$\Rightarrow du={\displaystyle \frac{1}{x}}dx$
Also the term $dv$ is considered as $dx$, which means:
$\Rightarrow dv=dx$
The above equation on integrating both side becomes:
$\Rightarrow v=x$
Considering the formula the equation becomes:
$\Rightarrow uv-\int vdu$
On substituting the values on the above formula we get:
$\Rightarrow {[x\ln x]}_0^1-\underset{0}{\overset{1}{\int }}dx$
Integration of $dx$ is $x$.
$\Rightarrow {[x\ln x-x]}_0^1$
On substituting the values on the function with the limits we get:
$\Rightarrow [1\times \ln 1-1]-[0\times \ln 0-0]$
The value of $\ln 1=0$ and $\ln 0$ is not defined. Putting the values in the above expression, we get:
$\Rightarrow [0-1]-[0-0]$
On further calculation it becomes:
$\Rightarrow -1$

$\therefore$ The integration of the function $\ln x$ from $[0,1]$ is $-1$.

Note: To solve this question we need to remember the formulas for integration, so that we do not make mistakes while writing the formulas. The above question has been solved by the method of integration by parts. The concept of definite integral is very useful and has a large number of applications. It is used to find the area of the enclosed figure etc.