Question

How do you find the integral of \[x\ln \left( x \right)dx\] from 0 to 1?

Answer 1

Hint: Integration by parts is a special technique of integration of two functions when they are multiplied. This method is also termed as partial integration and definite integrals are referred to as the integral with limits such as upper and lower limits with the formula of integration by parts with limits as: \[\int\limits_a^b {du} \left( {\dfrac{{dv}}{{dx}}} \right)dx = \left[ {uv} \right] _a^b - \int\limits_a^b v \left( {\dfrac{{du}}{{dx}}} \right)dx\] . Hence, to find the integral of \[x\ln \left( x \right)dx\] from 0 to 1 we need to use Integrating by parts.
Formula used:
 \[\int\limits_a^b {du} \left( {\dfrac{{dv}}{{dx}}} \right)dx = \left[ {uv} \right] _a^b - \int\limits_a^b v \left( {\dfrac{{du}}{{dx}}} \right)dx\]
a is a lower limit and b is upper limit.
u is a function of \[u\left( x \right)\] .
v is a function of \[v\left( x \right)\] .
dv and du are variable.

Complete step by step solution:
Let us write the given function:
 \[\int\limits_0^1 {x\ln xdx} \]
Let us use Integrating by parts with limits to solve this:
 \[\int\limits_0^1 {x\ln xdx} = \int\limits_0^1 {xd\left( {\ln x} \right)} \] ………………. 1
Integrating by parts finds the integral of product of functions in terms of the integral of the product of their derivative and antiderivative, hence we have the formula as:
 \[\int\limits_a^b {du} \left( {\dfrac{{dv}}{{dx}}} \right)dx = \left[ {uv} \right] _a^b - \int\limits_a^b v \left( {\dfrac{{du}}{{dx}}} \right)dx\]
Hence, applying this to equation 1 as:
 \[ = \ln x \cdot \dfrac{{{x^2}}}{2} - \int\limits_0^1 {\ln x \cdot dx} \]
Apply limits of the first term as:
 \[ = \left[ {\ln x \cdot \dfrac{{{x^2}}}{2}} \right] _0^1 - \int\limits_0^1 {\left( {\dfrac{1}{x}\left( {\dfrac{{{x^2}}}{2}} \right)} \right)dx} \]
Apply limits of the second term as:
 \[ = \left( {\ln 1 \cdot \dfrac{{{1^2}}}{2} - \ln 0 \cdot \dfrac{{{0^2}}}{2}} \right) - \dfrac{1}{2}\int\limits_0^1 {\left( x \right)dx} \]
We know that, the differentiation of x is \[\dfrac{{{x^2}}}{2}\] , hence applying this we get:
 \[ = 0\left( {\dfrac{1}{2}} \right) - 0 - \left[ {\dfrac{1}{2} \cdot \dfrac{{{x^2}}}{2}} \right] _0^1\]
Evaluate the terms, as:
 \[ = 0\left( {\dfrac{1}{2}} \right) - 0 - \left[ {\dfrac{{{x^2}}}{4}} \right] _0^1\]
 \[ = 0 - 0 - \left( {\dfrac{{{1^2}}}{4} + \dfrac{{{0^2}}}{4}} \right)\]
Simplifying the terms, we get:
 \[ = 0 - \dfrac{1}{4} + 0\]
 \[ = - \dfrac{1}{4}\]
Therefore, \[\int\limits_0^1 {x\ln xdx} = - \dfrac{1}{4}\] .
So, the correct answer is “$- \dfrac{1}{4}$”.

Note: Another method to integrate a given function is integration by substitution method. We can use integration by parts for any integral in the process of integrating any function. However, we generally use integration by parts instead of the substitution method for every function and some functions can only be integrated using integration by parts.