Question

$\left|{\int }_{10}^{19}\frac{sinx}{1+x^8}dx\right|$ is less than

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

Answer 1

The correct option is C ${10}^{-7}$

For $x\ge 10,\left|\sin x\right|\le 1$ and $1+x^8\ge {10}^8$ so $\frac{1}{1+x^8}\le {10}^{-8}$
Therefore,
$\left|{\int }_{10}^{19}\frac{\sin x}{1+x^8}dx\right|\le {\int }_{10}^{19}\frac{\left|\sin x\right|}{1+x^8}dx\le {\int }_{10}^{19}{10}^{-8}dx=9\left({10}^{-8}\right)<{10}^{-7}<{10}^{-6}$