Question

Choose the correct answer for the following definite integral :
 $
  \int\limits_1^e {\log xdx} = \_\_\_\_\_\_\_\_\_ \\
  A.{\text{ }}e + 1 \\
  B.{\text{ }}e - 1 \\
  C.\,{\text{ }}e + 2 \\
  D.{\text{ }}1 \\
 $

Answer 1

Hint- In order to solve this question, we will use integration by parts i.e. $\int {f\left( x \right) \times g\left( x \right)} dx = f\left( x \right)\int {g\left( x \right)dx} - \int {\left( {f'\left( x \right)\int {g\left( x \right)dx} } \right)} dx$

Complete step-by-step answer:
Integration by parts- Integration by parts formula is used for integrating the product of two functions.
This method is used to find the integrals by reducing them into standard forms. i.e.
$\int {f\left( x \right) \times g\left( x \right)} dx = f\left( x \right)\int {g\left( x \right)dx} - \int {\left( {f'\left( x \right)\int {g\left( x \right)dx} } \right)} dx$
We can choose first and second function by ILATE rule,
The integral of the two functions is taken by, considering the left term as first function and second term as second function. This method is called the ILATE rule. Usually, the preference order of this rule is based on some functions such as INVERSE, ALGEBRAIC, LOGARITHM, TRIGONOMETRIC, EXPONENT.

Now given that,
$\int\limits_1^e {\log xdx} $
Let’s find the value of $\int {\log xdx} $
$\int {\log xdx} = \int {\left( {\log x} \right)1 \times dx} $
Now using integration by parts
${1^{st}}$ function=$f\left( x \right) = \log x$ & ${2^{nd}}$ function=$g\left( x \right) = 1$
Now we know that
$\int {f\left( x \right) \times g\left( x \right)} dx = f\left( x \right)\int {g\left( x \right)dx} - \int {\left( {f'\left( x \right)\int {g\left( x \right)dx} } \right)} dx$
Putting the value of $f\left( x \right)$ and $g\left( x \right)$
$
  \int {\left( {\log x} \right)1 \times dx} = \log x\int {1 \times dx - \int {\left( {\dfrac{{d\left( {\log x} \right)}}{{dx}}\int {1 \times dx} } \right)dx} } \\
   = \left( {\log x} \right)x - \int {\dfrac{1}{x}} \times x \times dx \\
   = x\log x - \int {1 \times dx} \\
   = x\log x - x + c \\
 $
Now put the value of $\int {\log xdx} $ in the given question we get,
$
  \left| {x\log x - x} \right|_1^e \\
   = e\log e - e - \left( {1 \times \log 1 - 1} \right) \\
 $
Now we know that the value of $\log e = 1$ and $\log 1 = 0$
Now putting the value of $\log e = 1$ and $\log 1 = 0$ we get,
or$
  \left| {x\log x - x} \right|_1^e = e \times 1 - e - \left( {1 \times 0 - 1} \right) \\
  \left| {x\log x - x} \right|_1^e = e - e + 1 \\
   = 1 \\
 $
Thus, the correct option is $\left( {D.} \right)$
Note- Whenever we face such types of questions the key concept is that we should do the integration of a given function and then put the values of limits. Here in this question we simply find the value of $\int {\log xdx} $ and then we put the values of limit and thus we get our desired answer.