Answer:
[tex]16.5 \ un^2[/tex]
Step-by-step explanation:
Plot the region, bounded by given curves
[tex]y^2=x\\ \\y-4=x\\ \\y=-2\\ \\y=1[/tex]
(see attached diagram)
The area of this region is
[tex]A=\int\limits^1_{-2} (y^2-(y-4)) \, dy[/tex]
Evaluate it:
[tex]A=\int\limits^1_{-2} (y^2-y+4) \, dy\\ \\=\left(\dfrac{y^3}{3}-\dfrac{y^2}{2}+4y\right)\big |^1_{-2} \\ \\=\left(\dfrac{1^3}{3}-\dfrac{1^2}{2}+4\cdot 1\right)-\left(\dfrac{(-2)^3}{3}-\dfrac{(-2)^2}{2}+4\cdot (-2)\right)\\ \\=\dfrac{1}{3}-\dfrac{1}{2}+4+\dfrac{8}{3}+\dfrac{4}{2}+8\\ \\=3+12+1.5\\ \\=16.5\ un^2[/tex]
