Integration based on modulus

Question

Let $\displaystyle I= \int^{19}_{10} \frac{\sin x}{1+x^8} \, dx$

I have to prove that $|I|<10^{-7}$

MyApproach:

\begin{align} |I| & = \left|\int^{19}_{10} \frac{\sin x}{1+x^8} \, dx\right| \\[10pt] & \le \int^{19}_{10} \left| \frac{\sin x}{1+x^8}\right| \, dx \end{align}

because $|\sin x|\le 1$

$$\le \int^{19}_{10} \frac{1}{|1+x^8|} \, dx$$

Now I can't proceed from here because i don't know the denominator's modulus. I would be interested to know how to continue solving this problem from my last step.Any help would be appreciated.

Answer 1

If $10\le x\le 19$ then $1+10^8 \le 1+x^8 \le 1+19^8.$

Therefore $\dfrac 1 {1+10^8} \ge \dfrac 1 {1+x^8} \ge \dfrac 1 {1+19^8},$ and so

$$ (19-10) \cdot \frac 1 {1+10^8} \ge \int_{10}^{19} \frac{dx}{1+x^8}. $$

Answer 2

$I<red<red+blue=\dfrac{19-10}{10^8+1}<\dfrac{10}{10^8}=\dfrac{1}{10^7}$