Official solution
When y = 4 then,
x2 + 3x = 4
\[ \Rightarrow \] x2 + 3x - 4 = 0
\[ \Rightarrow \] x2 + 4x - x - 4 = 0
\[ \Rightarrow \] x(x + 4) - (x + 4) = 0
\[ \Rightarrow \] (x + 4)(x - 1) = 0
\[ \Rightarrow \] x = 1, - 4
As 0 \[ \le \] x \[ \le \] 3,
so possible value of x = 1.
\[ \therefore \] y = x2 + 3x parabola cut the line y = 4 at x = 1.
Required area
= \[\int\limits_0^1 {\left( {{x^2} + 3x} \right)} dx\] + 2 \[ \times \] 4
= \[\left[ {{{{x^3}} \over 3} + 3\left( {{{{x^2}} \over 2}} \right)} \right]_0^1\] + 8
= \[{\left( {{1 \over 3} + {3 \over 2}} \right)}\] + 8
= \[{{11} \over 6} + 8\]
= \[{{59} \over 6}\]