The area (in sq. units) of the region (x, y) / x 0, x+y 3, x^2 4 y . and .y 1+ x is

Step-by-step solution

Given inequalities are aligned & x 0 & x+y 3 & x^2 4 y, & y 1+ x aligned The equalities are aligned & x+y=3 ...(i) & x^2=4 y ...(ii) & y=1+ x ...(iii) aligned from (i) and (iii), we get array ll & 3-x=1+ x & x+ x -2=0 array aligned & ( x +2)( x -1)=0 & x =1 [ x cannot be negative ] & x=1 and y=2 aligned From (i) and (ii), we get array ll & x+ x^2 4 =3 & x^2+4 x-12=0 & (x+6)(x-2)=0 & x=2 & y=1 array [ x 0] Required area aligned & = ₀^1 (1+ x - x^2 4 ) d x+ ₁^2 (3-x- x^2 4 ) d x & = ₀^1(1+ x ) d x+ ₁^2(3-x)- 1 4 ₀^2