Question

The area of the region given by A={(x,y); $x^2$≤y≤min{x+2,4−3x}} is

• A. $\frac{31}{8}$
• B. $\frac{17}{6}$
• C. $\frac{19}{6}$
• D. $\frac{27}{8}$

Answer 1

Answer: (B)

$\frac{17}{6}$

Answer 2

A = {(x , y) : x 2 ≤ _y _≤ min {x + 2, 4 – 3 x}
So, the area of the required region
A=$\int_{-1}^{\frac{1}{2}}$(x+2−x2)dx+$\int_{1}^{\frac{1}{2}}$(4−3x−x2)dx
=[$\frac{x^2}{2}+2x-\frac{x^3}{3}$]$^{\frac{1}{2}}$-1+[4x−$\frac{3x^2}{2}$−$\frac{x^3}{3}$]$^{\frac{1}{2}}$1
=($\frac{1}{8}$+1−$\frac{1}{24}$)−($\frac{1}{2}$−2+$\frac{1}{3}$)+(4−$\frac{3}{2}$−$\frac{1}{3}$)−(2−$\frac{3}{8}$−$\frac{1}{24}$)=$\frac{17}{6}$