Question

Estimating the absolute value of the integral ${\int }_{10}^{19}\frac{\sin x}{1+x^8}dx$ we get ${10}^{-m}$.Find $m$?

Answer 1

To find, $I=\left|{\int }_{10}^{19}\frac{\sin x}{1+x^8}dx\right|\le {\int }_{10}^{19}\left|\frac{\sin x}{1+x^8}\right|dx$ (i)
Since, $|\sin x|\le 1$ for $x\ge 10$
The inequality $\left|\frac{\sin x}{1+x^8}\right|\le \frac{1}{|1+x^8|}$ (ii)
Also, $10\le x\le 19⟹1+x^8>{10}^8$
$⟹\frac{1}{1+x^8}<\frac{1}{{10}^8}$ or $\frac{1}{|1+x^8|}<{10}^{-8}$ (iii)
From Equations (ii) and (iii), $|\frac{\sin x}{1+x^8}<{10}^{-8}|$ is fulfilled.
$\left|{\int }_{10}^{19}\frac{\sin x}{1+x^8}dx\right|<{\int }_{10}^{19}{10}^{-8}dx$
$\therefore \left|{\int }_{10}^{19}\frac{\sin x}{1+x^8}dx\right|<(19-10){.10}^{-8}<{10}^{-7}$