Area between Curves (CIE)

$\def\D{\displaystyle}$(CIE 0606/2012/s/11/Q11 and Solution)
The diagram shows part of the curve $\D y = 9x^2 - x^3,$ which meets the x-axis at the origin $\D O$ and at the point $\D A$. The line $\D y - 2x + 18 = 0$ passes through $\D A$ and meets the y-axis at the point $\D B.$

(i) Show that, for $\D x \ge  0, 9x^2 - x^3 \le  108.$ [4]
(ii) Find the area of the shaded region bounded by the curve, the line $\D AB$ and the y-axis. [6]


*********math solution*************

$y=9 x^{2}-x^{3}$

$\dfrac{d y}{d x}=18 x-3 x^{2}$

For stationary points, $\dfrac{d x}{d x}=0$. Thus $18 x-3 x^{2}=0$

$x(18-3 x)=0$

$x=0$ or $x=6$

$\dfrac{d^{2} y}{d x^{2}}=18-6 x$

When $x=0, \dfrac{d^{2} y}{d x^{2}}=18>0$

and $\quad y=0$

$\therefore(0,0)$ is minimum point

At $x =6, \dfrac{d^{2} y}{d x^{2}}=18-6 \times 6=-18<0 .$ 

and $y =9(6)^{2}-(6)^{3}=108 .$

$\therefore(6,108)$ is maximum point.

Thus. $9 x^{2}-x^{3} \leqslant 108.$ 

ii) When $y=0, \quad 0-2 x+18=0$

$\therefore x=9$

$\therefore A=(9,0) .$

Area of shaded region $=\displaystyle\int_{0}^{9}\left(9 x^{2}-x^{3}\right)-(2 x-18) d$

$=\left[\dfrac{9 x^{3}}{3}-\dfrac{x^{4}}{4}-\dfrac{2 x^{2}}{2}+18 x\right]_{0}^{9}$

$=3(9)^{3}-\dfrac{1}{4}(9)^{4}-9^{2}+18 \times 9$



**********end math solution********************

Post a Comment