If int limits_{frac{1}{3}}^3 |log_e x| dx = f | vedclass

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(A) We evaluate the integral $I = \int_{\frac{1}{3}}^3 |\log_e x| dx$. Since $\log_e x < 0$ for $x \in [\frac{1}{3}, 1)$ and $\log_e x \ge 0$ for $x \

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(A) We evaluate the integral $I = \int_{\frac{1}{3}}^3 |\log_e x| dx$. Since $\log_e x $I = \int_{\frac{1}{3}}^1 -\log_e x dx + \int_1^3 \log_e x dx$Using the formula $\int \log_e x dx = x \log_e x - x$,we get:$I = -[x \log_e x - x]_{\frac{1}{3}}^1 + [x \log_e x - x]_1^3$$I = -[(1 \log_e 1 - 1) - (\frac{1}{3} \log_e \frac{1}{3} - \frac{1}{3})] + [(3 \log_e 3 - 3) - (1 \log_e 1 - 1)]$$I = -[-1 - (-\frac{1}{3} \log_e 3 - \frac{1}{3})] + [3 \log_e 3 - 3 + 1]$$I = -[-1 + \frac{1}{3} \log_e 3 + \frac{1}{3}] + [3 \log_e 3 - 2]$$I = -[-\frac{2}{3} + \frac{1}{3} \log_e 3] + 3 \log_e 3 - 2$$I = \frac{2}{3} - \frac{1}{3} \log_e 3 + 3 \log_e 3 - 2 = \frac{8}{3} \log_e 3 - \frac{4}{3} = \frac{4}{3} (2 \log_e 3 - 1) = \frac{4}{3} \log_e (\frac{3^2}{e}) = \frac{4}{3} \log_e (\frac{9}{e})$.Comparing this with $\frac{m}{n} \log_e (\frac{n^2}{e})$,we get $m = 4$ and $n = 3$.These are coprime natural numbers.Thus,$m^2 + n^2 - 5 = 4^2 + 3^2 - 5 = 16 + 9 - 5 = 20$.

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

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