Applications of integration: DHS 2024 Promo Q11

(a)

$$\begin{aligned} \sqrt{15x+4}&=x^2-2\\ 15x+4&=(x^2-2)^2\\ 15x+4&=x^4-4x^2+4\\ x(x^3-4x-15)&=0\\ x^3-4x-15&=0\qquad \therefore x\ne0\ \text{(does not satisfy original eqn)}\\ (x-3)(x^2+3x+5)&=0 \end{aligned}$$ Since $x^2+3x+5=\left(x+\dfrac32\right)^2+\dfrac{11}{4}>0$ for all x,

$\therefore x=3,y=7$

$\therefore(3,7)$ is the only point of intersection between the 2 curves.

(b)

Area of the region R $$\begin{aligned} &=\int_0^3\sqrt{15x+4}-(x^2-2)\,\mathrm{d} x\\ &=\left[\frac{(15x+4)^{3/2}}{\frac32(15)}-\frac{x^3}{3}+2x\right]_0^3\\ &=\left[\frac2{45}(45+4)^{3/2}-\frac{27}{3}+6-\frac2{45}(4)^{3/2}\right]\\ &=\left[\frac{686}{45}-3-\frac{16}{45}\right]\\ &=\frac{551-16}{45}\\ &=\frac{535}{45}\\ &=\frac{107}{9}\ \text{unit}^2 \end{aligned}$$

11(c)

Volume of revolution $$\begin{aligned} &=\pi\left[\int_{-2}^{7}y+2\,\mathrm{d} y-\frac1{225}\int_2^7(y^2-4)^2\,\mathrm{d} y\right]\\ &=91.7\ \text{units}^3 \end{aligned}$$

Alternative (markers’ reference only)

Volume of revolution $$\begin{aligned} &=\pi\left[\int_{-2}^{7}y+2\,\mathrm{d} y-\frac1{225}\int_2^7(y^2-4)^2\,\mathrm{d} y\right]\\ &=\pi\left[\left[\frac{y^2}{2}+2y\right]_{-2}^{7} -\frac1{225}\int_2^7 y^4-8y^2+16\,\mathrm{d} y\right]\\ &=\pi\left[\frac{81}{2}-\frac1{225} \left[\frac{y^5}{5}-\frac{8y^3}{3}+16y\right]_2^7\right]\\ &=\pi\left[\frac{81}{2}-\frac1{225}\left(\frac{38381}{15}-\frac{256}{15}\right)\right]\\ &=\pi\left[\frac{81}{2}-\frac{305}{27}\right]\\ &=\frac{1577}{54}\pi\ \text{units}^3 \end{aligned}$$