TCX2101 | CT2 Helpsheet (A4 Double-Sided)

A4 double-sided helpsheet for CT2 (16 Mar). Focus: Improper Integrals + Volume by Rotation.

Improper Integrals (4.8)

TypeDefinitionKey Rule
I (infinite limit)$\int_a^\infty f,dx = \lim_{b\to\infty}\int_a^b f,dx$Replace $\infty$ with $b$, take limit
II (discontinuity)$\int_a^b f,dx = \lim_{c\to a^+}\int_c^b f,dx$Replace bad bound with $c$, take limit
Interior at $d$MUST split → $\int_a^d + \int_d^b$One diverges → whole thing DIVERGES

$p$-Test & Convergence

FunctionAt $\infty$ ($\int_1^\infty$)At $0$ ($\int_0^1$)
$1/x^2$ $(p{=}2)$ConvergesDiverges
$1/x$ $(p{=}1)$DivergesDiverges
$1/\sqrt{x}$ $(p{=}\frac{1}{2})$DivergesConverges
$\ln(x)$Always converges
$\sin(x)/x$Always converges ($\to 1$)

Rule: At $\infty$: $p > 1$ converges. At $0$: $p < 1$ converges. Directions flip!

Traps:

  • $\int_{-1}^1 \frac{1}{x^2}dx$ — interior blow-up at $0$ → divergent
  • $\sin(x)/x \to 1$ → bounded → convergent (no computation needed)
  • Write “DIVERGES” not “$= \infty$”
  • Odd/Even: $f$ odd on $[-a,a]$: $\int = 0$ · $f$ even: $\int = 2\int_0^a$. Must verify both halves converge first.

Volume (4.9–4.10)

MethodFormulaWhen
Disk$\pi\int_a^b [f(x)]^2;dx$Solid, no hole, $\perp$ to axis
Washer$\pi\int_a^b [R(x)^2 - r(x)^2];dx$Two curves, hole in middle
Shell$2\pi\int_a^b (\text{radius})(\text{height});dx$Parallel to axis
Cross-section$\int_a^b A(x);dx$Non-revolution
Cross-section ($s$ = side)Area
Square$s^2$
Equil. triangle$\frac{\sqrt{3}}{4}s^2$
Isosc. right triangle (leg $s$)$\frac{1}{2}s^2$
Semicircle (dia $s$)$\frac{\pi}{8}s^2$
Circle (dia $s$)$\frac{\pi}{4}s^2$
AxisMethodShiftedMethodShift radius
$x$-axisDisk/Washer$y=k$Disk/Washer$f(x)-k$ or $k-f(x)$
$y$-axisShell$x=k$Shell$\lvert x-k\rvert$

Standard Antiderivatives/Integral (+ C)

$f(x)$$\int f\,dx$$f(x)$$\int f\,dx$$f(x)$$\int f\,dx$
$x^n$$\frac{x^{n+1}}{n+1}$$\frac{1}{x}$$\ln|x|$$e^x$$e^x$
$e^{-x}$$-e^{-x}$$\sin x$$-\cos x$$\cos x$$+\sin x$
$\sec^2 x$$\tan x$$\csc^2 x$$-\cot x$$\sec x\tan x$$\sec x$
$\csc x\cot x$$-\csc x$$\tan x$$\ln|\sec x|$$\cot x$$\ln|\sin x|$
$\frac{1}{1+x^2}$$\tan^{-1} x$$\frac{1}{\sqrt{1-x^2}}$$\sin^{-1} x$$\ln(t)$$t\ln t - t$
$\frac{1}{a^2+x^2}$$\frac{1}{a}\tan^{-1}\frac{x}{a}$$\frac{1}{\sqrt{a^2-x^2}}$$\sin^{-1}\frac{x}{a}$Trap: $\div$ NEW exp $(n{+}1)$, not old!

Differentiation ↔ Integration

RuleDifferentiationReverse (Integration)
Product ↔ IBP$(uv)’ = u’v + uv'$$\int u,dv = uv - \int v,du$
Chain ↔ u-sub$[f(g(x))]’ = f’(g(x)) \cdot g’(x)$$\int f(g(x)) \cdot g’(x),dx = \int f(u),du$
Quotient$(\frac{u}{v})’ = \frac{u’v - uv’}{v^2}$rewrite as product: $u \cdot v^{-1}$

IBP: Choose u (LIATE) → find du, v → apply $uv - \int v,du$

U-sub: Spot inner $u=g(x)$ → compute $du=g’(x),dx$ → replace all $x$ → integrate → sub back

u-sub bounds: set $u=g(x)$, then $x=a \to u=g(a)$, $x=b \to u=g(b)$. Convert IMMEDIATELY.

Trig Identities

$\sin^2 x + \cos^2 x = 1$$1 + \tan^2 x = \sec^2 x$$1 + \cot^2 x = \csc^2 x$
$\sin^2 u = \frac{1-\cos 2u}{2}$$\cos^2 u = \frac{1+\cos 2u}{2}$$\sin 2u = 2\sin u\cos u$

Limits

$\tan^{-1}(\infty) = \frac{\pi}{2}$$\tan^{-1}(0) = 0$$\tan^{-1}(-\infty) = -\frac{\pi}{2}$$e^{-\infty} = 0$$e^{\infty} = \infty$
$\ln(0^+) = -\infty$$\frac{1}{\infty} = 0$$\frac{1}{0^+} = \infty$$\frac{1}{0^-} = -\infty$$e^0 = 1$
$\ln(1) = 0$$\lim_{x\to 0^+} x\ln x = 0$$\lim_{x\to\infty} \frac{\ln x}{x} = 0$$\frac{\sin x}{x} \to 1\ (x{\to}0)$$\lim_{x\to\infty} x e^{-x} = 0$

Integration Techniques

TechniqueFormulaNotes
IBP$\int u,dv = uv - \int v,du$LIATE: Log > InvTrig > Algebra > Trig > Exp
PF: $(x-a)$$\frac{A}{x-a} \to A\ln\lvert x-a\rvert$Single linear factor
PF: $(x-a)^n$$\frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + \cdots + \frac{A_n}{(x-a)^n}$Each power gets its own term
PF: $(x^2+bx+c)$$\frac{Ax+B}{x^2+bx+c}$Irreducible quadratic → numerator is linear
Trig sub ($a^2+x^2$)$x=a\tan\theta$$\sqrt{a^2-x^2} \to x=a\sin\theta$ · $\sqrt{x^2-a^2} \to x=a\sec\theta$
L’Hôpital$\frac{0}{0}$ or $\frac{\infty}{\infty}$ → $\lim \frac{f’}{g’}$Diff top & bottom separately (NOT quotient rule)
FTC$\frac{d}{dx}\int_a^{g(x)} f(t),dt = f(g(x)) \cdot g’(x)$Chain rule on upper bound
Differentiability$f’(a) = \lim_{h\to 0} \frac{f(a+h) - f(a)}{h}$Piecewise: left & right limits must match
Area$\int_a^b [\text{top} - \text{bottom}],dx$Sketch first to identify which is top

FDT: $f’$ changes $+\to-$ → local max · $-\to+$ → local min · same sign → no extremum. SDT inconclusive ($f’’=0$) → fall back to FDT.


Common Errors

ErrorFix
Evaluate across interior discontinuityScan denominator for zeros → split
$\sqrt{\text{negative}}$ = NOT REAL$\lvert x-a\rvert$ left of $a$ → use $(a-x)$, chain rule $(-1)$
Optimization: max or min?$f’’ < 0$ → max (concave down) · $f’’ > 0$ → min (concave up)
Forgot shifted axis radiusRadius $= f(x) - k$, not $f(x)$

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