(a)OR(b)

Using integration, find the area of the region $\displaystyle \left\{(x, y): 4 x^{2}+9 y^{2} \leq 36\right.$, $\displaystyle 2 x+3 y \geq 6\}$.

Correct figure Clearly points of intersection are \(\displaystyle (3,0)\) and \(\displaystyle (0,2)\) Required area \(\displaystyle =\frac{2}{3} \int_{0}^{3} \sqrt{9-x^{2}} d x-\frac{2}{3} \int_{0}^{3}(3-x) d x\) \[\begin{aligned} & =\frac{2}{3}\left[\frac{x}{2} \sqrt{9-x^{2}}+\frac{9}{2} \sin ^{-1} \frac{x}{3}+\frac{(3-x)^{2}}{2}\right]_{0}^{3} \\ & =\frac{2}{3}\left[\frac{9 \pi}{4}-\frac{9}{2}\right] \text { or } \frac{3 \pi}{2}-3 \end{aligned} \] Correct figure Required area \(\displaystyle =\left|\int_{-2}^{-1}(x+1) d x\right|+\int_{-1}^{3}(x+1) d x\) \[\begin{aligned} & =-\left.\frac{(x+1)^{2}}{2}\right|_{-2} ^{-1}+\left.\frac{(x+1)^{2}}{2}\right|_{-1} ^{3} \\ & =\frac{1}{2}+8 \\ & =\frac{17}{2} \end{aligned} \]