Calculus II — Integration

Question 1 (12)

(a) By definition (2): $g'(x)=\lim_{h\to0}\frac{g(x+h)-g(x)}{h}.$

(b) (5) Since $g(x)=\int_a^x f$, $g(x+h)-g(x)=\int_a^{x+h}f-\int_a^{x}f=\int_x^{x+h}f(t)\,dt.$ $f$ continuous on $[x,x+h]$ ⇒ by the Extreme Value Theorem attains min $m$ and max $M$ there. So (for $h>0$) $m\,h \le \int_x^{x+h}f(t)\,dt \le M\,h \quad\Rightarrow\quad m \le \frac{g(x+h)-g(x)}{h}\le M.$ (1 mark single integral, 2 EVT bounds, 2 dividing by $h$; symmetric argument for $h<0$.)

(c) (3) As $h\to0$, the interval shrinks to $\{x\}$; by continuity both $m\to f(x)$ and $M\to f(x)$. By the Squeeze Theorem, $g'(x)=\lim_{h\to0}\frac{g(x+h)-g(x)}{h}=f(x).$

(d) (2) FTC Part 2: if $F$ is any antiderivative of continuous $f$, then $\int_a^b f=F(b)-F(a)$. Proof: $g(x)=\int_a^x f$ is an antiderivative by Part 1, so $F=g+C$. Then $F(b)-F(a)=g(b)-g(a)=\int_a^b f-0=\int_a^b f$.

Question 2 (10)

(a) (3) Product rule: $(uv)'=u'v+uv'$. Integrate both sides over the relevant interval: $uv=\int u'v\,dx+\int uv'\,dx \;\Rightarrow\; \int u\,v'\,dx=uv-\int u'v\,dx,$ i.e. with $dv=v'dx$, $du=u'dx$: $\int u\,dv=uv-\int v\,du$.

(b) (2) LIATE = Log, Inverse trig, Algebraic, Trig, Exponential; pick $u$ as whichever comes first. Rationale: earlier types simplify (differentiate) toward a constant while later types integrate cleanly, so $\int v\,du$ becomes simpler.

(c) (5) $u=x,\ dv=e^{2x}dx \Rightarrow du=dx,\ v=\tfrac12 e^{2x}$. $\int_0^1 xe^{2x}dx=\Big[\tfrac{x}{2}e^{2x}\Big]_0^1-\int_0^1\tfrac12 e^{2x}dx=\tfrac12 e^{2}-\Big[\tfrac14 e^{2x}\Big]_0^1.$ $=\tfrac12 e^2-\tfrac14 e^2+\tfrac14=\tfrac14 e^2+\tfrac14=\frac{e^2+1}{4}.$ (2 setup, 2 boundary term, 1 final.)

Question 3 (10)

(a) (5) Partition $[a,b]$; on subinterval the chord length is $\sqrt{\Delta x_i^2+\Delta y_i^2}=\sqrt{1+(\Delta y_i/\Delta x_i)^2}\,\Delta x_i$. By the Mean Value Theorem there is $x_i^*$ with $\Delta y_i/\Delta x_i=f'(x_i^*)$. Sum: $L\approx\sum\sqrt{1+f'(x_i^*)^2}\,\Delta x_i \xrightarrow{\ \|P\|\to0\ }\int_a^b\sqrt{1+(f'(x))^2}\,dx.$

(b) (5) $y=\tfrac23(x^2+1)^{3/2}\Rightarrow y'=\tfrac23\cdot\tfrac32(x^2+1)^{1/2}\cdot2x=2x\sqrt{x^2+1}.$ $1+(y')^2=1+4x^2(x^2+1)=4x^4+4x^2+1=(2x^2+1)^2.$ $L=\int_0^1(2x^2+1)\,dx=\Big[\tfrac23x^3+x\Big]_0^1=\tfrac23+1=\frac{5}{3}.$

Question 4 (9)

(a) (2) Use $x=2\tan\theta$ (form $a\tan\theta$ with $a=2$): it clears $\sqrt{x^2+4}$ since $x^2+4=4\sec^2\theta$.

(b) (7) $dx=2\sec^2\theta\,d\theta$, $\sqrt{x^2+4}=2\sec\theta$, $x^2=4\tan^2\theta$. $\int\frac{2\sec^2\theta\,d\theta}{4\tan^2\theta\cdot2\sec\theta}=\int\frac{\sec\theta}{4\tan^2\theta}d\theta=\frac14\int\frac{\cos\theta}{\sin^2\theta}d\theta.$ Let $w=\sin\theta$: $=\frac14\int w^{-2}dw=-\frac{1}{4\sin\theta}+C.$ Triangle: opposite $=x$, adjacent $=2$, hyp $=\sqrt{x^2+4}$, so $\sin\theta=\dfrac{x}{\sqrt{x^2+4}}$. $\boxed{\,-\frac{\sqrt{x^2+4}}{4x}+C\,}$ (2 setup, 2 simplify, 2 integrate, 1 back-substitute.)

Question 5 (10)

(a) (4) $\int_1^t x^{-2}dx=\big[-x^{-1}\big]_1^t=1-\tfrac1t$. Then $\int_1^\infty x^{-2}dx=\lim_{t\to\infty}\Big(1-\tfrac1t\Big)=1<\infty \Rightarrow \text{converges to }1.$

(b) (3) For $x\ge1$, $x^2+e^x>x^2$, so $0<\dfrac{1}{x^2+e^x}<\dfrac{1}{x^2}$. Since $\int_1^\infty x^{-2}dx$ converges (part a), by the comparison test the given integral converges. (Also acceptable: compare with $e^{-x}$.)

(c) (3) The comparison test bounds the tail integral: $\int f\le\int g$ only forces convergence if $\int g$ is finite. A pointwise bound $f\le g$ with divergent $\int g$ gives no information — a smaller function than a divergent one may still diverge. Convergence must be inherited from a convergent dominating integral (and the inequality must hold on the whole tail, i.e. eventually, with $f\ge0$).

Question 6 (9)

(a) (5) Shell method about $y$-axis: radius $x$, height $x^2$, $x\in[0,2]$. $V=\int_0^2 2\pi x\cdot x^2\,dx=2\pi\int_0^2 x^3dx=2\pi\cdot\frac{2^4}{4}=2\pi\cdot4=8\pi.$

(b) (4) Washer in $y$: for $y\in[0,4]$, outer radius $R=2$, inner radius $x=\sqrt{y}$. $V=\int_0^4 \pi\big(2^2-(\sqrt{y})^2\big)\,dy=\pi\int_0^4(4-y)\,dy.$ (Check: $=\pi[4y-\tfrac{y^2}{2}]_0^4=\pi(16-8)=8\pi$, consistent.)

[
  {"claim":"Q2c: integral of x e^{2x} from 0 to 1 = (e^2+1)/4","code":"x=symbols('x'); v=integrate(x*exp(2*x),(x,0,1)); result = simplify(v-(exp(2)+1)/4)==0"},
  {"claim":"Q3b: arc length = 5/3","code":"x=symbols('x'); y=Rational(2,3)*(x**2+1)**Rational(3,2); L=integrate(sqrt(1+diff(y,x)**2),(x,0,1)); result = simplify(L-Rational(5,3))==0"},
  {"claim":"Q4: antiderivative check d/dx(-sqrt(x^2+4)/(4x)) = 1/(x^2 sqrt(x^2+4))","code":"x=symbols('x',positive=True); F=-sqrt(x**2+4)/(4*x); result = simplify(diff(F,x)-1/(x**2*sqrt(x**2+4)))==0"},
  {"claim":"Q5a: improper integral of 1/x^2 on [1,inf) = 1","code":"x=symbols('x'); result = integrate(1/x**2,(x,1,oo))==1"},
  {"claim":"Q6a: shell volume = 8 pi","code":"x=symbols('x'); V=integrate(2*pi*x*x**2,(x,0,2)); result = simplify(V-8*pi)==0"},
  {"claim":"Q6b: washer volume = 8 pi","code":"y=symbols('y'); V=integrate(pi*(4-y),(y,0,4)); result = simplify(V-8*pi)==0"}
]