Definite Integration Question 8

Question 8 - 25 January - Shift 2

If $\int _{\frac{1}{3}}^{3}|\log _{e} x| dx=\frac{m}{n} \log _{e}(\frac{n^{2}}{e})$, where $m$ and $n$ are coprime natural numbers, then $m^{2}+n^{2}-5$ is equal to

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Answer: 20

Solution:

Formula: Propeties of modulus, Integration By Part, Properties of logarithmic function

$\int _{\frac{1}{3}}^{3}|\ell n nx| dx=\int _{\frac{1}{3}}^{1}(-\ell nx) dx+\int_1^{3}(\ell nx) dx$

$=-[x \ell n x-x] _{1 / 3}^{1}+[x \ell n x-x]_1^{3}$

$=-[-1-(\frac{1}{3} \ln \frac{1}{3}-\frac{1}{3})]+[3 \ln 3-3-(-1)]$

$=[-\frac{2}{3}-\frac{1}{3} \ln \frac{1}{3}]+[3 \ln 3-2]$

$=-\frac{4}{3}+\frac{8}{3} \ln 3$

$=\frac{4}{3}(2 \ln 3-1)$

$=\frac{4}{3}(\ln \frac{9}{e})$

$=\frac{m}{n} \log _{e}(\frac{n^{2}}{e})$

On comparing, we get

$\therefore m=4, n=3$

Now, $m^{2}+n^{2}-5=16+9-5=20$