If $$\int\limits_{{1 \over 3}}^3 {|{{\log }_e}x|dx = {m \over n}{{\log }_e}\left( {{{{n^2}} \over e}} \right)} $$, where m and n are coprime natural numbers, then ${m^2} + {n^2} - 5$ is equal to _____________.
Answer (integer)
20
Solution
$I=\int\limits_{\frac{1}{3}}^{3}|\ln x| d x=-\int\limits_{\frac{1}{3}}^{1} \ln x d x+\int\limits_{1}^{3} \ln x d x$
<br/><br/>
$$
\begin{aligned}
& \left.=-[x \ln x-x]_{\frac{1}{3}}^{1}+x \ln x-x\right]_{1}^{3} \\\\
& =-\left[(0-1)-\left(\frac{1}{3} \ln 3-\frac{1}{3}\right)\right]+[(3 \ln 3-3)-(0-1)] \\\\
& =\frac{2}{3}-\frac{1}{3} \ln 3+3 \ln 3-2
\end{aligned}
$$<br/><br/>
$$
\begin{aligned}
& =\frac{8}{3} \ln 3-\frac{4}{3} \\\\
& =\frac{4}{3}(2 \ln 3-\ln e) \\\\
& =\frac{4}{3} \ln \left(\frac{3^{2}}{e}\right) \\\\
& m=4, m^{m}=3 \\\\
& m^{2}+n^{2}-5=20
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals
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