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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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:

Similar Questions

$\int_{0}^{\pi} e^{\sin^2 x} \cos^3 x \, dx$ is equal to

$\int_{-\pi}^\pi \frac{2 x(1+\sin x)}{1+\cos ^2 x} d x=$

$\int_0^{2 \pi} (\sin x + |\sin x|) \, dx =$

Let $[t]$ denote the greatest integer $\leq t$. Then $\frac{2}{\pi} \int_{\pi/6}^{5\pi/6} (8[\operatorname{cosec} x] - 5[\cot x]) \, dx$ is equal to

$\int_0^\pi \frac{x \, dx}{4 \cos^2 x + 9 \sin^2 x} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module