Question

The absolute value of $\underset{10}{\overset{19}{\int }}\frac{(\sin x)dx}{\left(1+x^8\right)}$ is less than

• A. $10^{-10}$
• B. $10^{-11}$
• C. $10^{-7}$
• D. $10^{-9}$

Answer 1

Answer: (C)

$\left|\underset{10}{\overset{19}{\int }}\frac{\sin x}{1+x^8}\right|\le \underset{10}{\overset{19}{\int }}\frac{|\sin x|}{1+x^8}dx\le \underset{10}{\overset{19}{\int }}\frac{dx}{1+x^8}<\underset{10}{\overset{19}{\int }}\frac{dx}{x^8}={\left[\frac{x^{-7}}{-7}\right]}_{10}^{19}$
$=-\frac{1}{7}\left[19^{-7}-10^{-7}\right]=\frac{1}{7}\left[10^{-7}-19^{-7}\right]<10^{-7}$