We have, $y=-x^2$ and $x+y+2=0$

$$\begin{array}{lrl} \Rightarrow & -x-2 & =-x^2 \Rightarrow x^2-x-2=0 \\ \Rightarrow & x^2+x-2 x-2 & =0 \Rightarrow x(x+1)-2(x+1)=0 \\ \Rightarrow & (x-2)(x+1) & =0 \Rightarrow x=2,-1 \end{array}$$

$$ \begin{aligned} \therefore \quad \text { Area of shaded region, }A & =\left|\int_{-1}^2\left(-x-2+x^2\right) d x\right|=\left|\int_{-1}^2\left(x^2-x-2\right) d x\right| \\ & =\left|\left[\frac{x^3}{3}-\frac{x^2}{2}-2 x\right]_{-1}^2\right|=\left|\left[\frac{8}{3}-\frac{4}{2}-4+\frac{1}{3}+\frac{1}{2}-2\right]\right| \\ & =\left|\frac{16-12-24+2+3-12}{6}\right|=\left|-\frac{27}{6}\right|=\frac{9}{2} \text { sq units } \end{aligned}$$

Given equation of the curves are $y=\sqrt{x}$ and $x=2 y+3$ in the first quadrant.

$$\begin{aligned} &\text { On solving both the equations for } y \text {, we get }\\ &\begin{array}{r} & y =\sqrt{2 y+3} \\ \Rightarrow & y^2 =2 y+3 \\ \Rightarrow & y^2-2 y-3 =0 \\ \Rightarrow & y^2-3 y+y-3 =0 \\ \Rightarrow & y(y-3)+1(y-3) =0 \\ \Rightarrow & (y+1)(y-3) =0 \\ \Rightarrow & y =-1,3 \end{array} \end{aligned}$$

$$\begin{aligned} &\therefore \text { Required area of shaded region, }\\ &\begin{aligned} A & =\int_0^3\left(2 y+3-y^2\right) d y=\left[\frac{2 y^2}{2}+3 y-\frac{y^3}{3}\right]_0^3 \\ & =\left[\frac{18}{2}+9-9-0\right]=9 \text { sq units } \end{aligned} \end{aligned}$$

We have, $y^2=2 x$ and $x^2+y^2=4 x$

$$\begin{array}{lr} \Rightarrow & x^2+2 x=4 x \\ \Rightarrow & x^2-2 x=0 \\ \Rightarrow & x(x-2)=0 \\ \Rightarrow & x=0,2 \\ \text { Also, } & x^2+y^2=4 x \\ \Rightarrow & x^2-4 x=-y^2 \\ \Rightarrow & x^2-4 x+4=-y^2+4 \\ \Rightarrow & (x-2)^2-2^2=-y^2 \end{array}$$

$$\begin{aligned} \therefore \text { Required area } & =2 \cdot \int_0^2\left[\sqrt{2^2-(x-2)^2}-\sqrt{2 x}\right] d x \\ & =2\left[\left[\frac{x-2}{2} \cdot \sqrt{2^2-(x-2)^2}+\frac{2^2}{2} \sin ^{-1}\left(\frac{x-2}{2}\right)\right]_0^2-\left[\sqrt{2} \cdot \frac{x^{3 / 2}}{3 / 2}\right]_0^2\right] \\ & =2\left[\left(0+0-1 \cdot 0+2 \cdot \frac{\pi}{2}\right)-\frac{2 \sqrt{2}}{3}\left(2^{3 / 2}-0\right)\right] \\ & =\frac{4 \pi}{2}-\frac{8 \cdot 2}{3}=2 \pi-\frac{16}{3}=2\left(\pi-\frac{8}{3}\right) \text { sq units } \end{aligned}$$

$$\begin{aligned} \text { Required area }=\int_0^{2 \pi} \sin x d x & =\int_0^\pi \sin x d x+\left|\int_\pi^{2 \pi} \sin x d x\right| \\ & =-[\cos x]_0^\pi+\left|[-\cos x]_\pi^{2 \pi}\right| \\ & =-[\cos \pi-\cos 0]+|-[\cos 2 \pi-\cos \pi]| \end{aligned}$$

$$\begin{aligned} & =-[-1-1]+|-(1+1)| \\ & =2+2=4 \text { sq units } \end{aligned}$$

Let we have the vertices of a $\triangle A B C$ as $A(-1,1), B(0,5)$ and $C(3,2)$.

$\therefore \quad$ Equation of $A B$ is $y-1=\left(\frac{5-1}{0+1}\right)(x+1)$

$$ \begin{array}{lr} \Rightarrow & y-1=4 x+4 \\ \Rightarrow & y=4 x+5\quad\text{.... (i)} \end{array}$$

and $\quad$ equation of $B C$ is $y-5=\left(\frac{2-5}{3-0}\right)(x-0)$

$$\begin{array}{lr} \Rightarrow & y-5=\frac{-3}{3}(x) \\ \Rightarrow & y=5-x\quad\text{.... (ii)} \end{array}$$

Similarly, $\quad$ equation of $A C$ is $y-1=\left(\frac{2-1}{3+1}\right)(x+1)$

$$\begin{array}{lc} \Rightarrow & y-1=\frac{1}{4}(x+1) \\ \Rightarrow & 4 y=x+5\quad\text{.... (iii)} \end{array}$$

$\therefore \quad$ Area of shaded region $=\int_{-1}^0\left(y_1-y_2\right) d x+\int_0^3\left(y_1-y_2\right) d x$

$$\begin{aligned} & =\int_{-1}^0\left[4 x+5-\frac{x+5}{4}\right] d x+\int_0^3\left[5-x-\frac{x+5}{4}\right] d x \\ & =\left[\frac{4 x^2}{2}+5 x-\frac{x^2}{8}-\frac{5 x}{4}\right]_{-1}^0+\left[5 x-\frac{x^2}{2}-\frac{x^2}{8}-\frac{5 x}{4}\right]_0^3 \\ & =\left[0-\left(4 \cdot \frac{1}{2}+5(-1)-\frac{1}{8}+\frac{5}{4}\right)\right]+\left[\left(15-\frac{9}{2}-\frac{9}{8}-\frac{15}{4}\right)-0\right] \\ & =\left[-2+5+\frac{1}{8}-\frac{5}{4}+15-\frac{9}{2}-\frac{9}{8}-\frac{15}{4}\right] \\ & =18+\left(\frac{1-10-36-9-30}{8}\right) \\ & =18+\left(-\frac{84}{8}\right)=18-\frac{21}{2}=\frac{15}{2} \text { sq units } \end{aligned}$$