Calculus II — Integration

Question 1

(a) Disk method about $x$-axis: $V = \pi\int_1^\infty \left(\frac1x\right)^2 dx = \pi\int_1^\infty x^{-2}\,dx.$ This is an improper (Type I) integral. (1) $= \pi \lim_{t\to\infty}\left[-\frac1x\right]_1^t = \pi\lim_{t\to\infty}\left(1 - \frac1t\right) = \pi(1-0).$ (2 for limit evaluation) $V = \pi.$ Finite. (1)

(b) Area $= \displaystyle\int_1^\infty \frac1x\,dx = \lim_{t\to\infty}[\ln x]_1^t = \lim_{t\to\infty}\ln t = \infty.$ (2) The area is infinite (the $p$-integral with $p=1$ diverges). This is the classic Gabriel's Horn: finite volume, infinite surface/cross-sectional area. (1)

(c) For $x\ge 1$, $-1\le\sin x\le 1$ so $0 \le 2+\sin x \le 3$, giving $0 \le \frac{2+\sin x}{x^2} \le \frac{3}{x^2}.$ (1) Since $\displaystyle\int_1^\infty \frac{3}{x^2}\,dx = 3\cdot 1 = 3$ converges ($p=2>1$), (1) by the Comparison Test the given integral converges. (1)

Question 2

(a) Split odd power of sine: $\sin^3 x = \sin x(1-\cos^2 x)$. Let $u=\cos x$, $du=-\sin x\,dx$. (1) Limits: $x=0\Rightarrow u=1$; $x=\pi/2\Rightarrow u=0$. $\int_0^{\pi/2}(1-\cos^2x)\cos^2x\sin x\,dx = -\int_1^0 (1-u^2)u^2\,du = \int_0^1(u^2-u^4)\,du.$ (2) $= \left[\frac{u^3}{3}-\frac{u^5}{5}\right]_0^1 = \frac13-\frac15 = \frac{2}{15}.$ (1)

(b) Write numerator to match derivative of denominator: $\frac{d}{dx}(x^2+2x+5)=2x+2$. $x+4 = \tfrac12(2x+2) + 3.$ (1) $\int\frac{\frac12(2x+2)}{x^2+2x+5}dx + \int\frac{3}{(x+1)^2+4}dx.$ First integral $= \frac12\ln(x^2+2x+5)$. (1) Second: $\int\frac{3}{(x+1)^2+2^2}dx = \frac{3}{2}\arctan\frac{x+1}{2}$. (1) $= \frac12\ln(x^2+2x+5) + \frac32\arctan\frac{x+1}{2} + C.$ (1)

(c) Integration by parts twice (LIATE: algebraic $x^2$ = $u$). $u=x^2,\ dv=e^x dx \Rightarrow du=2x\,dx,\ v=e^x$: $\int_0^1 x^2 e^x dx = [x^2 e^x]_0^1 - 2\int_0^1 x e^x dx.$ (1) For $\int x e^x dx$: $u=x,\ dv=e^x dx$: $=[xe^x]_0^1 - \int_0^1 e^x dx = (e-0) - (e-1) = 1.$ (2) So $\int_0^1 x^2 e^x dx = (e - 0) - 2(1) = e - 2.$ (1)

Question 3

(a) Intersection: $x^2 = 2-x^2 \Rightarrow 2x^2=2 \Rightarrow x=\pm1$. (1) On $[-1,1]$, top curve $y=2-x^2$, bottom $y=x^2$. $A = \int_{-1}^{1}\big[(2-x^2)-x^2\big]dx = \int_{-1}^1 (2-2x^2)\,dx.$ (1) By symmetry $= 2\int_0^1(2-2x^2)dx = 2\left[2x - \frac{2x^3}{3}\right]_0^1 = 2\left(2-\frac23\right) = 2\cdot\frac43 = \frac{8}{3}.$ (2)

(b) Revolving about $y=-1$. Outer radius = (top curve) $-(-1)$, inner radius = (bottom) $-(-1)$: $R_{out} = (2-x^2)+1 = 3-x^2, \qquad R_{in} = x^2+1.$ (2) $V = \pi\int_{-1}^1\big[(3-x^2)^2 - (x^2+1)^2\big]dx.$ (1) Expand: $(3-x^2)^2 = 9 - 6x^2 + x^4$; $(x^2+1)^2 = x^4 + 2x^2 + 1$. Difference $= 9-6x^2 - 2x^2 - 1 = 8 - 8x^2$. (1) $V = \pi\int_{-1}^1(8-8x^2)dx = \pi\cdot 2\int_0^1(8-8x^2)dx = 2\pi\left[8x - \frac{8x^3}{3}\right]_0^1.$ $= 2\pi\left(8-\frac83\right) = 2\pi\cdot\frac{16}{3} = \frac{32\pi}{3}.$ (2)

Question 4

(a) Let $x = 3\tan\theta$, $dx = 3\sec^2\theta\,d\theta$; $9+x^2 = 9\sec^2\theta$. (1) $(9+x^2)^{3/2} = 27\sec^3\theta$. (1) $\int\frac{3\sec^2\theta}{27\sec^3\theta}d\theta = \frac{1}{9}\int\cos\theta\,d\theta = \frac19\sin\theta + C.$ (2) With $\tan\theta = x/3$, $\sin\theta = \dfrac{x}{\sqrt{9+x^2}}$: $= \frac{x}{9\sqrt{9+x^2}} + C.$ (1)

(b) Set $\dfrac{3x^2+2x+1}{(x-1)(x^2+1)} = \dfrac{A}{x-1} + \dfrac{Bx+C}{x^2+1}.$ (1) $3x^2+2x+1 = A(x^2+1) + (Bx+C)(x-1)$. $x=1$: $6 = 2A \Rightarrow A=3$. (1) Compare $x^2$: $3 = A+B \Rightarrow B=0$. Compare constant: $1 = A - C \Rightarrow C = 2$. (1) $\int\frac{3}{x-1}dx + \int\frac{2}{x^2+1}dx = 3\ln|x-1| + 2\arctan x + C.$ (2)

Question 5

(a) $y' = \dfrac{x^2}{2} - \dfrac{1}{2x^2}$. (1) $(y')^2 = \frac{x^4}{4} - \frac12 + \frac{1}{4x^4}.$ $1+(y')^2 = \frac{x^4}{4} + \frac12 + \frac{1}{4x^4} = \left(\frac{x^2}{2} + \frac{1}{2x^2}\right)^2.$ (2) So $\sqrt{1+(y')^2} = \dfrac{x^2}{2}+\dfrac{1}{2x^2}$ (positive on $[1,2]$). (1) $L = \int_1^2\left(\frac{x^2}{2}+\frac{1}{2x^2}\right)dx = \left[\frac{x^3}{6} - \frac{1}{2x}\right]_1^2 = \left(\frac86-\frac14\right) - \left(\frac16-\frac12\right).$ $= \left(\frac{4}{3}-\frac14\right) - \left(\frac16-\frac12\right) = \frac{13}{12} - \left(-\frac13\right) = \frac{13}{12}+\frac{4}{12} = \frac{17}{12}.$ (1)

(b) Average value $= \dfrac{1}{2-1}\displaystyle\int_1^2\left(\frac{x^3}{6}+\frac{1}{2x}\right)dx.$ (1) $\int_1^2\left(\frac{x^3}{6}+\frac{1}{2x}\right)dx = \left[\frac{x^4}{24} + \frac12\ln x\right]_1^2 = \left(\frac{16}{24}+\frac12\ln2\right)-\left(\frac{1}{24}+0\right).$ (1) $= \frac{15}{24} + \frac12\ln 2 = \frac58 + \frac{\ln 2}{2}.$ Average value $= \dfrac58 + \dfrac{\ln 2}{2}$. (1)

[
  {"claim":"Q1a volume = pi","code":"x=symbols('x'); V=pi*integrate(1/x**2,(x,1,oo)); result = simplify(V-pi)==0"},
  {"claim":"Q2a sin^3 cos^2 integral = 2/15","code":"x=symbols('x'); I=integrate(sin(x)**3*cos(x)**2,(x,0,pi/2)); result = simplify(I-Rational(2,15))==0"},
  {"claim":"Q2c x^2 e^x from 0 to 1 = e-2","code":"x=symbols('x'); I=integrate(x**2*exp(x),(x,0,1)); result = simplify(I-(E-2))==0"},
  {"claim":"Q3b volume = 32pi/3","code":"x=symbols('x'); V=pi*integrate((3-x**2)**2-(x**2+1)**2,(x,-1,1)); result = simplify(V-Rational(32,3)*pi)==0"},
  {"claim":"Q5a arc length = 17/12","code":"x=symbols('x'); yp=x**2/2-1/(2*x**2); L=integrate(sqrt(1+yp**2),(x,1,2)); result = simplify(L-Rational(17,12))==0"},
  {"claim":"Q5b average value = 5/8 + ln2/2","code":"x=symbols('x'); A=integrate(x**3/6+1/(2*x),(x,1,2)); result = simplify(A-(Rational(5,8)+log(2)/2))==0"}
]