## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 5 - The Integral - 5.2 The Definite Integral - Exercises - Page 246: 77

#### Answer

$$0.0198 \leq \int_{0.2}^{0.3} \sin x d x \leq 0.0296$$

#### Work Step by Step

Since $\frac{d}{dx}\sin x=\cos x>0$ on $[0.2,0.3]$, then $\sin x$ is increasing and $$f(0.2)\leq \sin x\leq f(0.3)$$ Hence, by the comparison theorem \begin{aligned} m(b-a)& \leq \int_{a}^{b} f(x) d x \leq M(b-a)\\ 0.198(0.3-0.2) & \leq \int_{0.2}^{0.3} \sin x d x \leq 0.296(0.3-0.2) \\ 0.198(0.1) & \leq \int_{0.2}^{0.3} \sin x d x \leq 0.296(0.1) \\ 0.0198 & \leq \int_{0.2}^{0.3} \sin x d x \leq 0.0296 \end{aligned}

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