2008 H2 Mathematics Paper 2 Question 2

Definite Integrals: Areas and Volumes

Answers

(i)

0.999 units2{0.999 \textrm{ units}^2}

(ii)

415π units3{\frac{4}{15} \pi \textrm{ units}^3}

(iii)

x=23{x=\frac{2}{3}}

Full solutions

(i)

Area of R=2∫01x1−x  dx=0.999 units2 (3 sf)  ■\begin{align*} & \textrm{Area of } R \\ & = 2 \int_0^1 \sqrt{x\sqrt{1-x}} \; \mathrm{d}x \\ & = 0.999 \textrm{ units}^2 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(ii)

dudx=−1\frac{\mathrm{d}u}{\mathrm{d}x} = -1
When x=0,{x=0, } u=1{u = 1}
When x=1,{x=1, } u=0{u = 0}
Volume obtained =π∫01x1−x  dx=π∫10(1−u)u(−1)  du=π∫01(1−u)u  du=π∫01(u12−u32)  du=π[23u32−25u52]01=π(23−25)=415π units3  ■\begin{align*} & \textrm{Volume obtained } \\ & = \pi \int_0^1 x \sqrt{1-x} \; \mathrm{d}x \\ & = \pi \int_1^0 (1-u) \sqrt{u} (-1) \; \mathrm{d}u \\ & = \pi \int_0^1 (1-u) \sqrt{u} \; \mathrm{d}u \\ & = \pi \int_0^1 \left( u^{\frac{1}{2}} - u^{\frac{3}{2}} \right) \; \mathrm{d}u \\ & = \pi \left[ \frac{2}{3} u^{\frac{3}{2}} - \frac{2}{5} u^{\frac{5}{2}} \right]_0^1 \\ & = \pi \left( \frac{2}{3} - \frac{2}{5} \right) \\ & = \frac{4}{15} \pi \textrm{ units}^3 \; \blacksquare \end{align*}

(iii)

y2=x1−xy4=x2(1−x)y4=x2−x34y3dydx=2x−3x2\begin{gather*} y^2 = x \sqrt{1-x} \\ y^4 = x^2 (1-x) \\ y^4 = x^2 - x^3 \\ 4y^3 \frac{\mathrm{d}y}{\mathrm{d}x} = 2x - 3x^2 \\ \end{gather*}
At maximum point, dydx=0{\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = 0}
2x−3x2=0x(2−3x)=0x=23  ■ or x=0 (NA)\begin{gather*} 2x - 3x^2 = 0 \\ x(2-3x) = 0 \\ x = \frac{2}{3} \; \blacksquare \textrm{ or } x = 0 \textrm{ (NA)} \end{gather*}