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}
