Question

If \(\int\limits_{\frac{1}{3}}^{3}\left|\log _e x\right| d x=\frac{m}{n} \log e\left(\frac{n^2}{v}\right)\), where \(m\) and \(n\) are coprime natural numbers, then \(m^2+n^2-5\) is equal to _____

Answer 1

Correct Answer: 20

The given integral is:

\( \int_{\frac{1}{3}}^3 \ln(x) dx = \int_{\frac{1}{3}}^1 \ln(x) dx + \int_1^3 \ln(x) dx. \)

For the first integral:

\( \int_{\frac{1}{3}}^1 \ln(x) dx = -[x \ln(x) - x]_{\frac{1}{3}}^1 = - \left[ - \frac{2}{3} + \ln \left( \frac{1}{3} \right) \right]. \)

For the second integral:

\( \int_1^3 \ln(x) dx = [x \ln(x) - x]_1^3 = \frac{8}{3} - \ln \left( \frac{9}{e} \right). \)

Combining both results:

\( \int_{\frac{1}{3}}^3 \ln(x) dx = \frac{4}{3} \ln(3) - \frac{8}{3} + 3 \ln(3) - 2. \)

Thus:

\( m=4, n=3 \implies m^2 + n^2 - 5 = 16 + 9 - 5 = 20. \)

Answer 2

The correct answer is 20.
$\underset{\frac{1}{3}}{\overset{3}{\int }}|\ell nx|dx=\underset{\frac{1}{3}}{\overset{1}{\int }}(-\ell nx)dx+\underset{1}{\overset{3}{\int }}(\ell nx)dx$
$=-[x\ell nx-x{]}_{1/3}^1+[x\ell nx-x{]}_1^3$
$=-\left[-1-\left(\frac{1}{3}\ell n\frac{1}{3}-\frac{1}{3}\right)\right]+[3\ell n3-3-(-1)]$
$=\left[-\frac{2}{3}-\frac{1}{3}\ell n\frac{1}{3}\right]+[3\ell n3-2]$
$=-\frac{4}{3}+\frac{8}{3}\ell n3$
$=\frac{4}{3}(2\ell n3-1)$
$=\frac{4}{3}\left(\ell n\frac{9}{e}\right)$
$\therefore m=4,n=3$
Now, $m^2+n^2-5=16+9-5=20$