If _1/3^3 |_ex|,dx = mn_e(n^2e), where m and n are coprime natural numbers, then m^2+n^2-5…

Question

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

Answer: 20

Accepted Answer

$$\int_{1/3}^3|\ln x|\,dx = \int_{1/3}^1(-\ln x)\,dx+\int_1^3\ln x\,dx = \big[x-x\ln x\big]_{1/3}^1 + \big[x\ln x-x\big]_1^3$$

$$= \left(1-\left(\frac13+\frac13\ln3\right)\right) + \left(3\ln3-3+1\right) = \frac43(2\ln3-1) = \frac43\ln\frac9e$$

So $m=4,\,n=3$, and $m^2+n^2-5 = 16+9-5=20$.

Solutions · 1