1. **State the problem:** We are given the curve $y = \csc^2(x)$ and a region $R$ bounded by this curve, the $x$-axis, and the vertical lines $x=1$ and $x=2$. We need to fill in missing values in the table for $y$ at $x=1.2, 1.4, 1.6, 1.8$ and find the area of region $R$. 2. **Calculate missing values:** Recall $y = \csc^2(x) = \frac{1}{\sin^2(x)}$. We calculate $y$ for each $x$ using this formula and round to 3 decimal places. - For $x=1.2$: $y = \csc^2(1.2) = \frac{1}{\sin^2(1.2)}$. Calculate $\sin(1.2) \approx 0.932$, so $y \approx \frac{1}{0.932^2} = \frac{1}{0.868} \approx 1.152$. - For $x=1.4$: $\sin(1.4) \approx 0.985$, so $y \approx \frac{1}{0.985^2} = \frac{1}{0.970} \approx 1.031$. - For $x=1.6$: $\sin(1.6) \approx 0.999$, so $y \approx \frac{1}{0.999^2} = \frac{1}{0.998} \approx 1.002$. - For $x=1.8$: $\sin(1.8) \approx 0.974$, so $y \approx \frac{1}{0.974^2} = \frac{1}{0.949} \approx 1.053$. 3. **Fill in the table:** | x | 1 | 1.2 | 1.4 | 1.6 | 1.8 | 2 | |-----|-------|---------|---------|---------|---------|---------| | y | 1.412 | 1.152 | 1.031 | 1.002 | 1.053 | 1.209 | 4. **Find the area of region $R$:** The area under the curve from $x=1$ to $x=2$ is given by the definite integral $$\text{Area} = \int_1^2 \csc^2(x) \, dx.$$ Recall that $\int \csc^2(x) \, dx = -\cot(x) + C$. So, $$\text{Area} = [-\cot(x)]_1^2 = -\cot(2) + \cot(1).$$ Calculate $\cot(1) = \frac{\cos(1)}{\sin(1)} \approx \frac{0.540}{0.841} = 0.642$. Calculate $\cot(2) = \frac{\cos(2)}{\sin(2)} \approx \frac{-0.416}{0.909} = -0.458$. Therefore, $$\text{Area} = -(-0.458) + 0.642 = 0.458 + 0.642 = 1.100.$$ 5. **Final answer:** - Missing values: $1.152, 1.031, 1.002, 1.053$ - Area of region $R \approx 1.100$ (to 3 decimal places)