(a) Using integration, find the area of the region {(x, y): 4 x^2+9 y^2 ≤ 36., 2 x+3 y ≥ 6}.

\[\begin{aligned} & =\frac{2}{3}\left[\frac{x}{2} \sqrt{9-x^{2}}+\frac{9}{2} \sin ^{-1} \frac{x}{3}+\frac{(3-x)^{2}}{2}\right]_{0}^{3} \\ & =\frac{2}{3}\left[\frac{9 \pi}{4}-\frac{9}{2}\right] \text { or } \frac{3 \pi}{2}-3 \end{aligned} \]

Correct figure

Required area \(\displaystyle =\left|\int_{-2}^{-1}(x+1) d x\right|+\int_{-1}^{3}(x+1) d x\)

\[\begin{aligned} & =-\left.\frac{(x+1)^{2}}{2}\right|_{-2} ^{-1}+\left.\frac{(x+1)^{2}}{2}\right|_{-1} ^{3} \\ & =\frac{1}{2}+8 \\ & =\frac{17}{2} \end{aligned} \]