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

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

Answers (1)

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}

 

 

Posted by

infoexpert26

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