Calculus II — Integration

Question 1

(a) FTC Part 1 proof (7 marks)

• Setup (1): Define $g(x)=\int_a^x f(t)\,dt$. Form the difference quotient: $\frac{g(x+h)-g(x)}{h}=\frac{1}{h}\int_x^{x+h} f(t)\,dt.$ (2 marks)
• MVT for integrals (2): Since $f$ is continuous on $[x,x+h]$, there exists $c_h\in[x,x+h]$ with $\int_x^{x+h}f(t)\,dt=f(c_h)\,h$. Hence the quotient $=f(c_h)$. (3 marks — statement + application)
• Limit (2): As $h\to0$, $c_h\to x$ (squeeze since $x\le c_h\le x+h$), and by continuity $f(c_h)\to f(x)$. Therefore $g'(x)=\lim_{h\to0}f(c_h)=f(x)$. (2 marks)

(b) Midpoint sum, n=2 (5 marks)

• $\Delta x = 1/2$; midpoints $x_1=0.25,\ x_2=0.75$. (2)
• $I\approx \tfrac12\big(e^{-0.0625}+e^{-0.5625}\big)$. (2)
• $e^{-0.0625}=0.939413$, $e^{-0.5625}=0.569783$; sum $=1.509196$; $\times 0.5=0.754598$. $I\approx 0.7546.$ (1) (True value $\approx 0.746824$.)

(c) Pseudocode + error bound (8 marks)

def midpoint(f, a, b, n):
    h = (b - a) / n
    total = 0.0
    for i in range(n):
        xm = a + (i + 0.5) * h
        total += f(xm)
    return h * total

(4 marks: correct h, midpoint x_m = a+(i+0.5)h, loop, return h*sum)

• Error bound: $|E_M|\le \dfrac{K(b-a)^3}{24 n^2}=\dfrac{2\cdot 1}{24 n^2}=\dfrac{1}{12 n^2}$. (2)
• Require $\dfrac{1}{12n^2}<10^{-4}\Rightarrow n^2>\dfrac{10^4}{12}=833.3\Rightarrow n>28.87$. (1)
• So $n\ge 29$ subintervals guarantee the bound. (1)

Question 2

(a) Disk volume (4 marks)

$V=\pi\int_0^4 (\sqrt{x})^2\,dx=\pi\int_0^4 x\,dx=\pi\Big[\tfrac{x^2}{2}\Big]_0^4=\pi\cdot 8=8\pi\ \text{m}^3.$ (setup 2, evaluate 2)

(b) Average value (4 marks)

$\bar r=\frac{1}{4-0}\int_0^4 \sqrt{x}\,dx=\frac14\Big[\tfrac{2}{3}x^{3/2}\Big]_0^4=\frac14\cdot\frac{2}{3}\cdot 8=\frac{16}{12}=\frac{4}{3}\ \text{m}.$ (setup 2, evaluate 1)

• Interpretation (1): $\bar r=4/3\approx1.33$ m is the mean radius; a uniform cylinder of this radius over $[0,4]$ would have the same "average" profile (though not the same volume).

(c) Work to pump out (8 marks)

• Thin vertical disk at $x$, thickness $dx$, radius $r(x)=\sqrt{x}$, so cross-sectional area $\pi x$; mass $dm=\rho\,\pi x\,dx$. (2)
• Each disk of water sits centered on the axis; it must be lifted to the top at height $r_{\max}=2$. Taking the axis as height $0$, the lift distance is $2$ m (measured from axis to top). (2)
• Work element $dW=\rho g (\text{lift})\,dV=1000\cdot 9.8\cdot 2\cdot \pi x\,dx$. (2) $W=1000\cdot9.8\cdot2\cdot\pi\int_0^4 x\,dx=19600\pi\cdot 8=156800\pi\ \text{J}.$ (2) $W=156800\pi\approx 4.926\times10^5\ \text{J}.$

(Accept the model where lift distance = $r_{\max}=2$ used as stated; full marks for consistent setup and evaluation.)

(d) Shell method about y-axis (6 marks)

• Shells: radius $x$, height $\sqrt{x}$, thickness $dx$: $V=2\pi\int_0^4 x\sqrt{x}\,dx=2\pi\int_0^4 x^{3/2}\,dx=2\pi\Big[\tfrac{2}{5}x^{5/2}\Big]_0^4=2\pi\cdot\frac{2}{5}\cdot 32=\frac{128\pi}{5}.$ (setup 2, evaluate 2)
• Disk/washer check in $y$: the region rotated about $y$-axis; for $0\le y\le 2$, outer radius $x=4$, inner radius $x=y^2$: $V=\pi\int_0^2\big(4^2-(y^2)^2\big)\,dy=\pi\int_0^2(16-y^4)\,dy.$ (2 marks for correct washer setup) $=\pi\big[16y-\tfrac{y^5}{5}\big]_0^2=\pi(32-\tfrac{32}{5})=\frac{128\pi}{5}.\ \checkmark$

Question 3

(a) Comparison test (6 marks)

• Since $-1\le\sin x\le 1$, we have $1\le 2+\sin x\le 3$, so $0<\dfrac{2+\sin x}{x^2}\le\dfrac{3}{x^2}$ for $x\ge1$. (2)
• $\int_1^\infty \dfrac{3}{x^2}\,dx=3\big[-\tfrac1x\big]_1^\infty=3(0+1)=3$ converges (p=2>1). (2)
• By the comparison test (nonnegative integrand dominated by convergent integral), the given integral converges. (2)

(b) Trig substitution (7 marks)

• Let $x=2\tan\theta$, $dx=2\sec^2\theta\,d\theta$, $x^2+4=4\sec^2\theta$, $(x^2+4)^{3/2}=8\sec^3\theta$. (2) $\int\frac{2\sec^2\theta}{8\sec^3\theta}\,d\theta=\frac14\int\cos\theta\,d\theta=\frac14\sin\theta+C.$ (2)
• $\sin\theta=\dfrac{x}{\sqrt{x^2+4}}$, so antiderivative $=\dfrac{x}{4\sqrt{x^2+4}}$. (1)
• Evaluate improper: $\displaystyle\lim_{b\to\infty}\frac{x}{4\sqrt{x^2+4}}\Big|_0^b=\frac14-0=\frac14.$ (2) $\int_0^\infty\frac{dx}{(x^2+4)^{3/2}}=\frac14.$

(c) Partial fractions (5 marks)

• $x^2+x-2=(x+2)(x-1)$. Write $\dfrac{3x+1}{(x+2)(x-1)}=\dfrac{A}{x+2}+\dfrac{B}{x-1}$. (2)
• $3x+1=A(x-1)+B(x+2)$. At $x=1$: $4=3B\Rightarrow B=\tfrac43$. At $x=-2$: $-5=-3A\Rightarrow A=\tfrac53$. (2) $\int=\frac53\ln|x+2|+\frac43\ln|x-1|+C.$ (1)

[
  {"claim":"Q1b midpoint sum n=2 for e^{-x^2} on [0,1] equals ~0.7546",
   "code":"val=(Rational(1,2))*(exp(-Rational(1,16))+exp(-Rational(9,16))); result=abs(float(val)-0.7546)<0.001"},
  {"claim":"Q2a disk volume of y=sqrt(x) about x-axis on [0,4] is 8*pi",
   "code":"x=symbols('x'); V=pi*integrate(x,(x,0,4)); result=simplify(V-8*pi)==0"},
  {"claim":"Q2d shell volume about y-axis equals 128*pi/5 and matches washer",
   "code":"x,y=symbols('x y'); Vs=2*pi*integrate(x*sqrt(x),(x,0,4)); Vw=pi*integrate(16-y**4,(y,0,2)); result=simplify(Vs-Rational(128,5)*pi)==0 and simplify(Vs-Vw)==0"},
  {"claim":"Q3b improper integral of 1/(x^2+4)^{3/2} from 0 to oo is 1/4",
   "code":"x=symbols('x',positive=True); I=integrate(1/(x**2+4)**Rational(3,2),(x,0,oo)); result=simplify(I-Rational(1,4))==0"},
  {"claim":"Q3c partial fraction coefficients A=5/3,B=4/3",
   "code":"x=symbols('x'); expr=apart((3*x+1)/(x**2+x-2),x); target=Rational(5,3)/(x+2)+Rational(4,3)/(x-1); result=simplify(expr-target)==0"}
]