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