Applications of integration: ASRJC 2024 Promo Q3

(a)

$$\begin{aligned} \text{Area}&=\int_0^1x^2\,\mathrm{d} x-\frac12\left(\frac12\right)(1) \qquad\text{OR}\qquad =\int_0^1x^2\,\mathrm{d} x-\int_{1/2}^1(2x-1)\,\mathrm{d} x\\ &=\left[\frac{x^3}{3}\right]_0^1-\frac14\\ &=\frac13-\frac14\\ &=\frac1{12}\text{ units}^2 \end{aligned}$$

Alternative method $$\begin{aligned} \text{Area}&=\int_0^1\frac12(y+1)-\sqrt y\,\mathrm{d} y\\ &=\left[\frac14y^2+\frac12y-\frac23y^{3/2}\right]_0^1\\ &=\frac14+\frac12-\frac23\\ &=\frac1{12}\text{ units}^2 \end{aligned}$$

(b)

$$\begin{aligned} \text{Volume}&=\pi\int_0^1(x^2)^2\,\mathrm{d} x-\frac13\pi(1)^2\left(\frac12\right) \quad\text{OR}\quad =\pi\int_0^1(x^2)^2\,\mathrm{d} x-\pi\int_{1/2}^1(2x-1)^2\,\mathrm{d} x\\ &=\pi\left[\frac{x^5}{5}\right]_0^1-\frac16\pi\\ &=\frac\pi5-\frac\pi6\\ &=\frac\pi{30}\text{ units}^3 \end{aligned}$$