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

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


\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.

  • $\begingroup$ $|1 + x^8| \ge |x|^8 - 1 \ge 10^8 - 1$ $\endgroup$ – user296602 Apr 15 '18 at 4:12
  • $\begingroup$ What's the reason for -1,I included my question,formatted properly and didn't even ask for the whole solution.I only asked about my specific problem.I think I have followed all the rules and regulations provided my mathematics stackexchange. $\endgroup$ – Calculus Programmer Apr 15 '18 at 18:04

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

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

  • $\begingroup$ I can't understand your last line. $\endgroup$ – Calculus Programmer Apr 15 '18 at 4:57
  • $\begingroup$ @CalculusProgrammer : That's just the fact that if $f(x) \le M$ for all values of $x$ and $a<b$ then $\displaystyle \int_a^b f(x) \, dx \le M(b-a). \qquad$ $\endgroup$ – Michael Hardy Apr 15 '18 at 16:43


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