The area bounded by the curves $$y^2 = 4 + x$$ and $$x + 2y = 4$$, is
  • A
    $$9$$
  • B
    $$18$$
  • C
    $$72$$
  • D
    $$36$$

Solution


Area when $$x > 0 = \displaystyle 2\int_{-4}^{0}\sqrt{4+x}$$
$$=\left.\begin{matrix}\dfrac{4}{3}(x+4)^{\frac{3}{2}}\end{matrix}\right|_{-4}^{0}$$

$$=\dfrac{4}{3}\times 8-\dfrac{4}{3}\times 0$$

$$=\dfrac{32}{3}$$
Area when $$x > 0$$
$$\displaystyle \int_{0}^{12}(\dfrac{4-x}{2})-\int_{0}^{12}(-\sqrt{4}+x)$$

$$\displaystyle \int_{0}^{12}\frac{4-x}{2}+\int_{0}^{12}\sqrt{4+x}$$

$$=-12+\dfrac{112}{3}$$

$$\therefore $$ Total area $$=-12+\dfrac{144}{3}=36$$