Let I = int_{10}^{19} frac{sin x}{1+x^{6}} dx | vedclass

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(C) For $x \in [10, 19]$, we have $|\sin x| \leq 1$ and $1+x^{6} > 10^{6}$.Since $x \geq 10$, $1+x^{6} > 10^{6}$, which implies $\frac{1}{1+x^{6}} < 1

Vedclass · Accepted Answer

$|I| < 10^{-5}$

Vedclass · Answer

(C) For $x \in [10, 19]$, we have $|\sin x| \leq 1$ and $1+x^{6} > 10^{6}$.Since $x \geq 10$, $1+x^{6} > 10^{6}$, which implies $\frac{1}{1+x^{6}} Therefore, $|I| = \left| \int_{10}^{19} \frac{\sin x}{1+x^{6}} dx ight| \leq \int_{10}^{19} \frac{|\sin x|}{1+x^{6}} dx$.Since $|\sin x| \leq 1$ and $\frac{1}{1+x^{6}} $|I| However, checking the options provided and the magnitude, $9 imes 10^{-6} < 10^{-5}$. Thus, $|I| < 10^{-5}$ is the correct bound.

Let $I = \int_{10}^{19} \frac{\sin x}{1+x^{6}} dx$. Then,

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