The absolute value of int_{10}^{19} frac{sin | vedclass

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(C) We use the property $\left| \int_{a}^{b} f(x) dx \right| \le \int_{a}^{b} |f(x)| dx$.Given the integral $I = \int_{10}^{19} \frac{\sin x}{1 + x^8}

Vedclass · Accepted Answer

$10^{-7}$

Vedclass · Answer

(C) We use the property $\left| \int_{a}^{b} f(x) dx \right| \le \int_{a}^{b} |f(x)| dx$.Given the integral $I = \int_{10}^{19} \frac{\sin x}{1 + x^8} dx$,we have:$|I| \le \int_{10}^{19} \frac{|\sin x|}{1 + x^8} dx$.Since $|\sin x| \le 1$ and $1 + x^8 > x^8$,we get:$|I| Evaluating the integral:$\int_{10}^{19} x^{-8} dx = \left[ \frac{x^{-7}}{-7} \right]_{10}^{19} = \frac{1}{7} \left( \frac{1}{10^7} - \frac{1}{19^7} \right)$.Since $\frac{1}{7} $|I| < \frac{1}{10^7} = 10^{-7}$.

The absolute value of $\int_{10}^{19} \frac{\sin x}{1 + x^8} dx$ is less than:

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