Need solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Very short answer type question 36

Answer: $\frac{18}{\log _{e} 3}$

Hint: you must know the rule of integration

Given: $I=\int_{2}^{3} 3^{x} d x$

Solution:

Let $y=3^{x}$            ..............(i)

Taking logarithm on both sides

\begin{aligned} &\log y=\log 3^{x} \\\\ &\log y=x \log 3 \\\\ &y=e^{x \log 3} \end{aligned}        ........(ii)

From eqn (i) and (ii) we can write

$y=3^{x}=e^{x \log 3}$

This means instead of integrating   $3^{x}$  we will integrate  $e^{x \log 3}$

\begin{aligned} &I=\int_{2}^{3} 3^{x} d x \\\\ &=\int_{2}^{3} e^{x \log 3} d x \end{aligned}

\begin{aligned} &=\left[\frac{e^{x \log 3}}{\log 3}\right]_{2}^{3} \\\\ &=\frac{1}{\log 3}\left[e^{3 \log 3}-e^{2 \log 3}\right] \end{aligned}

\begin{aligned} &=\frac{1}{\log 3}\left[e^{\log 3^{3}}-e^{\log 3^{2}}\right] \\\\ &=\frac{1}{\log 3}\left[3^{3}-3^{2}\right] \end{aligned}

\begin{aligned} &=\frac{(27-9)}{\log 3} \\\\ &=\frac{18}{\log 3} \end{aligned}