Question

Evaluate the following: $\int\limits_{2}^{3}{\dfrac{1}{x}}dx$.
(a) ${{\log }_{e}}\dfrac{3}{2}$
(b) ${{\log }_{e}}\dfrac{5}{2}$
(c) ${{\log }_{e}}\dfrac{2}{3}$
(d) ${{\log }_{e}}\dfrac{5}{2}$

Answer 1

Hint: In order to solve the above types of integration, we need to solve some simple steps. The first step to solving such types of the equation is to remove the limits and simply integrate it independently. After integrating it independently, we will obtain some values. Now, after obtaining the result, we will have to apply the upper limit and lower limits. On applying the upper limit and lower limit, we will have to evaluate the entire equation in order to obtain the required result for the given integration.

Complete step-by-step solution:
Here, in order to integrate $\dfrac{1}{x}$ independently without the upper and lower limits, we can make use of a formula given by $\int{\dfrac{1}{x}}dx={{\log }_{e}}x+c$ where c is the integration constant. This will give us the expression for the independent integration of $\dfrac{1}{x}$. Now, we can simply apply the lower limit and upper limit to the result obtained above after the independent integration. Further, on applying the lower limit and upper limit, we can use the formula of ‘Logarithm’ to solve the expression. The formula that can be used to solve the expression after applying the limits is given by ${{\log }_{e}}a-{{\log }_{e}}b={{\log }_{e}}\left( \dfrac{a}{b} \right)$.
Here, we have to evaluate the integral $I=\int\limits_{2}^{3}{\dfrac{1}{x}}dx\text{ (say)}..............\text{(i)}$. We can find the integral by removing the limits and simply integrate it independently. So, the integration becomes,
${{\text{I}}_{\text{1}}}=\int{\dfrac{1}{x}}dx............(ii)$
Now, in order to solve the integral ${{\text{I}}_{\text{1}}}$, we can apply the formula $\int{\dfrac{1}{x}}dx={{\log }_{e}}x+c$ where c is the integration constant.
Then, using the above formula in equation (ii), we get,
$\Rightarrow {{\text{I}}_{\text{1}}}=\int{\dfrac{1}{x}}dx$
$\Rightarrow {{\text{I}}_{\text{1}}}={{\log }_{e}}x.............(iii)$
Now, after obtaining the result for independent integration at equation (iii), we can apply the upper limit and lower limit as asked in question to equation (iii), we get,
$\Rightarrow \text{I=}\left[ {{\log }_{e}}x \right]_{2}^{3}$
We can easily evaluate this expression by using the logarithmic formula which is given by ${{\log }_{e}}a-{{\log }_{e}}b={{\log }_{e}}\left( \dfrac{a}{b} \right)$, so, we get,
$\Rightarrow \text{I=}{{\log }_{e}}3-{{\log }_{e}}2$
$\therefore \text{I=}{{\log }_{e}}\dfrac{3}{2}$
Hence, the integral of $\int\limits_{2}^{3}{\dfrac{1}{x}}dx$ is ${{\log }_{e}}\dfrac{3}{2}$.
Thus, the correct option is an option (a).

Note: Students often make mistakes while using the integral formula $\int{\dfrac{1}{x}}dx={{\log }_{e}}x+c$. Students must remember this formula as it reduces time to solve these types of questions. Besides, the other logarithmic property used here ${{\log }_{e}}a-{{\log }_{e}}b={{\log }_{e}}\left( \dfrac{a}{b} \right)$ is very important property. Students often fail to use this property and leave the solution without solving the expression.