Discussion Forum



1)

The area ( in sq units) of the region {(x,y):x ≥ 0, x +y ≤ 3, x2 ≤ 4y and y ≤ 1+√x } is 


A) $ \frac{59}{12}$

B) $ \frac{3}{2}$

C) $ \frac{7}{3}$

D) $ \frac{5}{2}$

Answer:

Option D

Explanation:

Required area

        $=\int_{0}^{1} (1+\sqrt{x})dx +\int_{1}^{2}(3-x)dx-\int_{0}^{2}\frac{x^{2}}{4}dx  $

   19112019643_area.PNG

   = $=[ x+\frac{x^{\frac{3}{2}}}{3/2}]_{0}^{1}+[3x-\frac{x^{2}}{2}]_{1}^{2}- [\frac{x^{3}}{12}]_{0}^{2}$

   $=(1+\frac{2}{3})+(6-2-3+\frac{1}{2})-(\frac{8}{12})$

   $=\frac{5}{3}+\frac{3}{2}-\frac{2}{3}=1+\frac{3}{2}=\frac{5}{2} sq. units$