2011 H2 Mathematics Paper 2 Question 4

Integration Techniques

Answers

(ai)
14−14(2n2+2n+1)e−2n{\frac{1}{4} - \frac{1}{4}(2n^2+2n+1)\mathrm{e}^{-2n}}
(aii)
∫0∞x2e−2x  dx=14{\displaystyle \int_0^\infty x^{2}\mathrm{e}^{- 2 x} \; \mathrm{d}x = \frac{1}{4}}

(b)

2π2−4π{2 \pi^2 - 4 \pi}

Full solutions

(ai)
∫0nx2e−2x  dx=[−12x2e−2x]0n−∫0n2x(−12e−2x)  dx=−12n2e−2n+0+∫0nxe−2x  dx=−12n2e−2n+[−12xe−2x]0n−∫0n−12e−2x  dx=−12n2e−2n−12ne−2n+∫0n12e−2x  dx=−12n2e−2n−12ne−2n−[14e−2x]0n=−12n2e−2n−12ne−2n−14e−2n+14=14−14(2n2+2n+1)e−2n  ■\begin{align*} & \int_0^n x^{2}\mathrm{e}^{- 2 x} \; \mathrm{d}x \\ & = \left[ -\frac{1}{2} x^{2} \mathrm{e}^{- 2 x} \right]_0^n - \int_0^n 2 x \left(- \frac{1}{2} \mathrm{e}^{- 2 x}\right) \; \mathrm{d}x \\ & = -\frac{1}{2}n^2 \mathrm{e}^{-2n} + 0 + \int_0^n x \mathrm{e}^{- 2 x} \; \mathrm{d}x \\ & = -\frac{1}{2}n^2 \mathrm{e}^{-2n} + \left[ - \frac{1}{2} x \mathrm{e}^{- 2 x} \right]_0^n - \int_0^n - \frac{1}{2} \mathrm{e}^{- 2 x} \; \mathrm{d}x \\ & = -\frac{1}{2}n^2 \mathrm{e}^{-2n} - \frac{1}{2}n \mathrm{e}^{-2n} + \int_0^n \frac{1}{2} \mathrm{e}^{- 2 x} \; \mathrm{d}x \\ & = -\frac{1}{2}n^2 \mathrm{e}^{-2n} - \frac{1}{2}n \mathrm{e}^{-2n} - \left[ \frac{1}{4} \mathrm{e}^{- 2 x} \right]_0^n \\ & = -\frac{1}{2}n^2 \mathrm{e}^{-2n} - \frac{1}{2}n \mathrm{e}^{-2n} - \frac{1}{4} \mathrm{e}^{-2n} + \frac{1}{4} \\ & = \frac{1}{4} - \frac{1}{4}(2n^2+2n+1)\mathrm{e}^{-2n}\;\blacksquare \\ \end{align*}
(aii)
As n→∞,  {n \to \infty, \;} n2e−2n,  {n^2 \mathrm{e}^{-2n}, \;} ne−2n{n \mathrm{e}^{-2n} } and e−2n→0{\mathrm{e}^{-2n} \to 0}
Hence ∫0nx2e−2x→14{\displaystyle \int_0^n x^{2}\mathrm{e}^{- 2 x} \to \frac{1}{4}}
∫0∞x2e−2x  dx=14  ■\int_0^\infty x^{2}\mathrm{e}^{- 2 x} \; \mathrm{d}x = \frac{1}{4} \; \blacksquare

(b)

dxdθ=sec⁡2θ\frac{\mathrm{d}x}{\mathrm{d}\theta} = \sec^2 \theta
When x=0,  θ=0.{x=0, \; \theta = 0. \quad}When x=1,  θ=π4{x=1, \; \theta = \frac{\pi}{4}}
Volume of solid obtained=π∫01(4xx2+1)2  dx=π∫0π4(4tan⁡θtan⁡2θ+1)2sec⁡2θ  dθ=π∫0π416tan⁡2θ(sec⁡2θ)2sec⁡2θ  dθ=16π∫0π4tan⁡2θsec⁡2θ  dθ=16π∫0π4sin⁡2θ  dθ  ■\begin{align*} & \textrm{Volume of solid obtained} \\ & = \pi \int_0^1 \left( \frac{4x}{x^2+1} \right)^2 \; \mathrm{d}x \\ & = \pi \int_0^{\frac{\pi}{4}} \left( \frac{4\tan \theta}{\tan^2 \theta +1} \right)^2 \sec^2 \theta \; \mathrm{d}\theta \\ & = \pi \int_0^{\frac{\pi}{4}} \frac{16 \tan^2 \theta}{(\sec^2 \theta)^2} \sec^2 \theta \; \mathrm{d}\theta \\ & = 16 \pi \int_0^{\frac{\pi}{4}} \frac{\tan^2 \theta}{\sec^2 \theta} \; \mathrm{d}\theta \\ & = 16 \pi \int_0^{\frac{\pi}{4}} \sin^2 \theta \; \mathrm{d}\theta \; \blacksquare \\ \end{align*}
16π∫0π4sin⁡2θ  dθ=16π∫0π41−cos⁡2θ2  dθ=8π[θ−sin⁡2θ2]0π4=8π(π4−sin⁡π22)=2π2−4π  ■\begin{align*} & 16 \pi \int_0^{\frac{\pi}{4}} \sin^2 \theta \; \mathrm{d}\theta \\ & = 16 \pi \int_0^{\frac{\pi}{4}} \frac{1-\cos 2 \theta}{2} \; \mathrm{d}\theta \\ & = 8 \pi \left[ \theta - \frac{\sin 2\theta}{2} \right]_0^{\frac{\pi}{4}} \\ & = 8 \pi \left( \frac{\pi}{4} - \frac{\sin \frac{\pi}{2}}{2} \right) \\ & = 2 \pi^2 - 4 \pi \; \blacksquare \end{align*}