Question

The absolute value of ${\int }_{10}^{19}\frac{\sin x}{1+x^8}dx$ is

• A. less than ${10}^{-7}$
• B. more than ${10}^{-7}$
• C. less than ${10}^{-6}$
• D. more than ${10}^{-6}$

Answer 1

The correct option is A

less than ${10}^{-7}$

$\left|{\int }_{10}^{19}\frac{\sin x}{1+x^8}dx\right|\le {\int }_{10}^{19}\frac{\left|\sin x\right|}{1+x^8}dx$
Now, $\sin x\le 1$ and $1+x^8>x^8$
$\therefore \frac{1}{1+x^8}<\frac{1}{x^8}\Rightarrow \frac{\left|\sin x\right|}{1+x^8}<\frac{1}{x^8}\text{So,}\left|{\int }_{10}^{19}\frac{\sin x}{1+x^8}dx\right|\le {\int }_{10}^{19}\frac{\left|\sin x\right|}{1+x^8}dx<{\int }_{10}^{19}\frac{1}{x^8}dx<{\int }_{10}^{19}\frac{1}{{10}^8}dx=9\times {10}^{-8}<10\times {10}^{-8}={10}^{-7}$
Hence, the absolute value of ${\int }_{10}^{19}\frac{\sin x}{1+x^8}dx$ is less than ${10}^{-7}$.