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Applications of integration: NJC 2023 Promo Q3

National Junior College2023 PromoPromo Q3●●○ Standard9 marks

What this question tests

Areas and volumes by integration.

Question

. It is given that f(x)=xe−xk\mathrm{f}(x) = x\mathrm{e}^{-\frac{x}{k}}, where kk is a positive constant.

  1. Find the range of values of xx for which the curve with equation y=f(x)y = \mathrm{f}(x) is concave upwards. [4]

Assume for the remainder of this question that k=2k = 2.

  1. RR is the region bounded by the curve, the xx-axis, the line x=4x = 4 and the line x=nx = n, where n>4n > 4. The solid obtained when rotating RR about the xx-axis through 2π2\pi radians has a volume of VV units3^3. Find the exact value that VV converges to as nn approaches infinity. [5]

[You may use the result that xne−x approaches 0 as x approaches infinity for all real values of n.]\left[\text{You may use the result that } x^n\mathrm{e}^{-x} \text{ approaches } 0 \text{ as } x \text{ approaches infinity for all real values of } n.\right]

Show full worked solution▾

(i)

For the curve to be concave upwards, the gradient must be increasing, i.e. f′′(x)>0\mathrm f''(x)>0.

f(x)=xe−x/k\mathrm f(x)=x\mathrm{e}^{-x/k} f′(x)=x(−1ke−xk)+e−xk=e−xk(1−xk)f′′(x)=e−xk(−1k)+(1−xk)(−1ke−xk)=−e−xkk(1+1−xk)=−e−xkk(2k−xk).\begin{aligned} \mathrm{f}'(x) &= x\left(-\frac{1}{k}\mathrm{e}^{-\frac{x}{k}}\right) + \mathrm{e}^{-\frac{x}{k}} = \mathrm{e}^{-\frac{x}{k}}\left(1 - \frac{x}{k}\right)\\ \mathrm{f}''(x) &= \mathrm{e}^{-\frac{x}{k}}\left(-\frac{1}{k}\right) + \left(1 - \frac{x}{k}\right)\left(-\frac{1}{k}\mathrm{e}^{-\frac{x}{k}}\right)\\ &= -\frac{\mathrm{e}^{-\frac{x}{k}}}{k}\left(1 + 1 - \frac{x}{k}\right)\\ &= -\frac{\mathrm{e}^{-\frac{x}{k}}}{k}\left(\frac{2k - x}{k}\right). \end{aligned} Since e−xkk>0\dfrac{\mathrm{e}^{-\frac{x}{k}}}{k} > 0, −e−xkk(2k−xk)>02k−x<0x>2k.\begin{aligned} -\frac{\mathrm{e}^{-\frac{x}{k}}}{k}\left(\frac{2k - x}{k}\right) &> 0\\ 2k - x &< 0\\ x &> 2k. \end{aligned}

(ii)

Volume of solid obtained =π∫2k∞y2 dx=π∫4∞x2e−2x/2 dx=−π∫4∞x2(−e−x) dx=−π{[x2e−x]4∞−∫4∞2xe−x dx}=−π[0−16e−4]−2π∫4∞x(−e−x) dx=16πe−4−2π{[xe−x]4∞−∫4∞e−x dx}=16πe−4−2π[0−4e−4]−2π∫4∞−e−x dx=16πe−4+8πe−4−2π[e−x]4∞=24πe−4−2π[0−e−4]=24πe−4+2πe−4=26πe−4.\begin{aligned} &=\pi\int_{2k}^{\infty}y^2\,\mathrm{d} x\\ &=\pi\int_4^{\infty}x^2\mathrm{e}^{-2x/2}\,\mathrm{d} x\\ &=-\pi\int_4^{\infty}x^2(-\mathrm{e}^{-x})\,\mathrm{d} x\\ &=-\pi\left\{[x^2\mathrm{e}^{-x}]_4^{\infty}-\int_4^{\infty}2x\mathrm{e}^{-x}\,\mathrm{d} x\right\}\\ &=-\pi[0-16\mathrm{e}^{-4}]-2\pi\int_4^{\infty}x(-\mathrm{e}^{-x})\,\mathrm{d} x\\ &=16\pi\mathrm{e}^{-4}-2\pi\left\{[x\mathrm{e}^{-x}]_4^{\infty}-\int_4^{\infty}\mathrm{e}^{-x}\,\mathrm{d} x\right\}\\ &=16\pi\mathrm{e}^{-4}-2\pi[0-4\mathrm{e}^{-4}]-2\pi\int_4^{\infty}-\mathrm{e}^{-x}\,\mathrm{d} x\\ &=16\pi\mathrm{e}^{-4}+8\pi\mathrm{e}^{-4}-2\pi[\mathrm{e}^{-x}]_4^{\infty}\\ &=24\pi\mathrm{e}^{-4}-2\pi[0-\mathrm{e}^{-4}]\\ &=24\pi\mathrm{e}^{-4}+2\pi\mathrm{e}^{-4}\\ &=26\pi\mathrm{e}^{-4}. \end{aligned}

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