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

= $${{11} \over 6} + 8$$

= $${{59} \over 6}$$