FAD1014 — 1-Hour Mechanical Speed Drill

Fast. Repetitive. Builds instant recall. No deep theory — just write the answer and move.

Targets: Part A basics (standard integrals, substitution, binomial, conic ID) + Maclaurin Series (leaked).

Rules: 5 min per mini-section. If stuck >30s, skip and check solution. Do each section twice if time.

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BLOCK 1: STANDARD INTEGRALS — MUST BE INSTANT (5 min)

Q1

$$\int x^5,dx$$

Q2

$$\int \frac{1}{x},dx$$

Q3

$$\int e^{3x},dx$$

Q4

$$\int \sin 2x,dx$$

Q5

$$\int \cos 4x,dx$$

Q6

$$\int \sec^2 x,dx$$

Q7

$$\int \frac{1}{\sqrt{1-x^2}},dx$$

Q8

$$\int \frac{1}{1+x^2},dx$$

Q9

$$\int \frac{1}{\sqrt{4-x^2}},dx$$

Q10

$$\int (2x+3)^5,dx$$

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BLOCK 2: U-SUBSTITUTION (5 min)

Q11

$$\int x,e^{x^2},dx$$

Q12

$$\int \frac{x}{x^2+1},dx$$

Q13

$$\int x^2 \cos(x^3),dx$$

Q14

$$\int \frac{2x}{\sqrt{x^2+4}},dx$$

Q15

$$\int x\sqrt{x^2+1},dx$$

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BLOCK 3: DEFINITE INTEGRALS (5 min)

Q16

$$\int_0^{\pi/2} \sin x,dx$$

Q17

$$\int_0^1 e^{2x},dx$$

Q18

$$\int_1^e \frac{1}{x},dx$$

Q19

$$\int_0^{\pi/4} \sec^2 x,dx$$

Q20

$$\int_0^2 (3x^2 + 2x),dx$$

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BLOCK 4: AREA UNDER CURVE (5 min)

Q21

Area under $y = x^2$ from $x=0$ to $x=2$.

Q22

Area under $y = e^x$ from $x=0$ to $x=\ln 3$.

Q23

Area under $y = \sin x$ from $x=0$ to $x=\pi$.

Q24

Area under $y = \frac{1}{x}$ from $x=1$ to $x=e$.

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BLOCK 5: BINOMIAL I (5 min)

Q25

State $\binom{7}{2}$, $\binom{8}{3}$, $\binom{10}{4}$.

Q26

Expand $(2x+1)^4$ completely.

Q27

Expand $(x-2)^5$ completely.

Q28

Find coefficient of $x^3$ in $(1+2x)^6$.

Q29

Find coefficient of $x^2$ in $(x-3)^5$.

Q30

Find coefficient of $x^4$ in $(x^2-2)^6$.

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BLOCK 6: SERIES CONVERGENCE / METHOD OF DIFFERENCES (5 min)

Q31

Does $a_n = \frac{2n^2+3}{5n^2-2n+1}$ converge? Find the limit if so.

Q32

Does $\sum_{n=1}^\infty \frac{n}{2n+1}$ converge or diverge?

Q33

Evaluate $\sum_{r=1}^{20} (3r-1)$.

Q34

Evaluate $\sum_{r=1}^{n} \frac{1}{r(r+1)}$.

Q35

Evaluate $\sum_{r=1}^{n} \frac{2}{r(r+2)}$.

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BLOCK 7: CONIC ID — QUICK FIRE (5 min)

Q36

$(x-2)^2 = 8(y+3)$. State vertex, focus, directrix, direction.

Q37

$y^2 = -12(x-1)$. State vertex, focus, directrix, direction.

Q38

$\frac{x^2}{25} + \frac{y^2}{9} = 1$. State centre, vertices, foci. Horizontal or vertical?

Q39

$\frac{(x+1)^2}{4} + \frac{(y-2)^2}{16} = 1$. State centre, vertices, foci.

Q40

$\frac{x^2}{16} - \frac{y^2}{9} = 1$. State centre, vertices, foci, asymptotes.

Q41

$\frac{(y-1)^2}{4} - \frac{(x+2)^2}{9} = 1$. State centre, vertices, asymptotes.

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BLOCK 8: MACLAURIN — STANDARD FORMS (5 min)

Q42

$$e^x = ?$$

Q43

$$\sin x = ?$$

Q44

$$\cos x = ?$$

Q45

$$\ln(1+x) = ?$$

Q46

$$\frac{1}{1-x} = ?$$

Q47

$$\sinh x = ?$$

Q48

$$\cosh x = ?$$

Q49

$$\arctan x = ?$$

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BLOCK 9: MACLAURIN — SUBSTITUTION (8 min)

Q50

$$e^{-2x}$$

Q51

$$e^{x^2}$$

Q52

$$\sin(3x^2)$$

Q53

$$\cos(2x^3)$$

Q54

$$\ln(1+4x)$$

Q55

$$x e^{-x}$$

Q56

$$x^2 \cos x$$

Q57

$$\frac{1}{1+3x}$$

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BLOCK 10: MACLAURIN — MULTIPLICATION & INTEGRATION (7 min)

Q58

Find up to $x^3$: $e^x \sin x$

Q59

Find up to $x^4$: $e^{-x} \cos x$

Q60

Use series to approximate: $\displaystyle\int_0^{0.5} e^{-x^2},dx$ (3 d.p.)

Q61

Find series for $\displaystyle\int \frac{\sin x}{x},dx$ up to $x^5$ term.

Q62

Derive $\tan x$ up to $x^3$ using $f(0), f'(0), f''(0), f'''(0)$.

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SOLUTIONS

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BLOCK 1

Q1

$$\frac{x^6}{6} + C$$

Q2

$$\ln|x| + C$$

Q3

$$\frac13 e^{3x} + C$$

Q4

$$-\frac12 \cos 2x + C$$

Q5

$$\frac14 \sin 4x + C$$

Q6

$$\tan x + C$$

Q7

$$\sin^{-1} x + C$$

Q8

$$\tan^{-1} x + C$$

Q9

$$\sin^{-1}\left(\frac{x}{2}\right) + C$$

Q10

$\frac{1}{12}(2x+3)^6 + C$ (let $u=2x+3$, $du=2dx$)

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BLOCK 2

Q11

$u=x^2$, $du=2x,dx$ → $\frac12\int e^u,du = \frac12 e^{x^2} + C$

Q12

$u=x^2+1$, $du=2x,dx$ → $\frac12\int\frac{du}{u} = \frac12\ln(x^2+1) + C$

Q13

$u=x^3$, $du=3x^2,dx$ → $\frac13\int\cos u,du = \frac13\sin(x^3) + C$

Q14

$u=x^2+4$, $du=2x,dx$ → $\int u^{-1/2},du = 2\sqrt{x^2+4} + C$

Q15

$u=x^2+1$, $du=2x,dx$ → $\frac12\int u^{1/2},du = \frac13(x^2+1)^{3/2} + C$

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BLOCK 3

Q16

$$[-\cos x]_0^{\pi/2} = 0 - (-1) = 1$$

Q17

$$\left[\frac12 e^{2x}\right]_0^1 = \frac12 e^2 - \frac12 = \frac12(e^2-1)$$

Q18

$$[\ln x]_1^e = 1 - 0 = 1$$

Q19

$$[\tan x]_0^{\pi/4} = 1 - 0 = 1$$

Q20

$$[x^3 + x^2]_0^2 = (8+4) - 0 = 12$$

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BLOCK 4

Q21

$$\int_0^2 x^2,dx = \left[\frac{x^3}{3}\right]_0^2 = \frac83$$

Q22

$$\int_0^{\ln 3} e^x,dx = [e^x]_0^{\ln 3} = 3 - 1 = 2$$

Q23

$$\int_0^\pi \sin x,dx = [-\cos x]_0^\pi = 1 - (-1) = 2$$

Q24

$$\int_1^e \frac1x,dx = [\ln x]_1^e = 1 - 0 = 1$$

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BLOCK 5

Q25

$\binom{7}{2}=21$, $\binom{8}{3}=56$, $\binom{10}{4}=210$

Q26

$$(2x)^4 + 4(2x)^3(1) + 6(2x)^2(1)^2 + 4(2x)(1)^3 + 1^4$$ $$= 16x^4 + 32x^3 + 24x^2 + 8x + 1$$

Q27

$$x^5 + 5x^4(-2) + 10x^3(-2)^2 + 10x^2(-2)^3 + 5x(-2)^4 + (-2)^5$$ $$= x^5 - 10x^4 + 40x^3 - 80x^2 + 80x - 32$$

Q28

$$\binom{6}{3}(1)^3(2x)^3 = 20 \cdot 8x^3 = 160$$

Q29

$\binom{5}{2}x^2(-3)^3$... Wait: $(x-3)^5$, term with $x^2$: need $r=3$ (since $b=-3$, $a=x$, $n=5$, term $r$ has $a^{5-r}b^r$) $$\binom{5}{3}x^2(-3)^3 = 10 \cdot (-27)x^2 = -270$$

Q30

$(x^2-2)^6$, need $(x^2)^2 = x^4$ → $r=4$? No: $a=x^2$, $b=-2$, need $a^{6-r} = (x^2)^{6-r} = x^{12-2r} = x^4$ → $12-2r=4$ → $r=4$ $$\binom{6}{4}(x^2)^2(-2)^4 = 15 \cdot 16 = 240$$

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BLOCK 6

Q31

$\lim_{n\to\infty} \frac{2n^2+3}{5n^2-2n+1} = \frac{2}{5}$ → Converges

Q32

$\lim_{n\to\infty} \frac{n}{2n+1} = \frac12 \neq 0$ → Diverges (nth term test)

Q33

$$3\sum r - \sum 1 = 3\cdot\frac{20\cdot21}{2} - 20 = 630 - 20 = 610$$

Q34

$\frac{1}{r(r+1)} = \frac1r - \frac1{r+1}$ → $1 - \frac1{n+1} = \frac{n}{n+1}$

Q35

$$\frac{2}{r(r+2)} = \frac1r - \frac1{r+2}$$ Telescopes: $1 + \frac12 - \frac1{n+1} - \frac1{n+2} = \frac32 - \frac1{n+1} - \frac1{n+2}$

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BLOCK 7

Q36

Vertex $(2,-3)$. $4a=8$ → $a=2$. Opens up. Focus $(2,-1)$. Directrix $y=-5$.

Q37

Vertex $(1,0)$. $4a=12$ → $a=3$. Opens left. Focus $(-2,0)$. Directrix $x=4$.

Q38

Centre $(0,0)$. $a=5$, $b=3$ (horizontal). $c=\sqrt{25-9}=4$. Vertices $(\pm5,0)$. Foci $(\pm4,0)$.

Q39

Centre $(-1,2)$. $a=2$, $b=4$ (vertical). $c=\sqrt{16-4}=2\sqrt3$. Vertices $(-1,6)$ and $(-1,-2)$. Foci $(-1,2\pm2\sqrt3)$.

Q40

Centre $(0,0)$, horizontal. $a=4$, $b=3$. $c=\sqrt{16+9}=5$. Vertices $(\pm4,0)$. Foci $(\pm5,0)$. Asymptotes $y = \pm\frac34 x$.

Q41

Centre $(-2,1)$, vertical. $a=2$, $b=3$. $c=\sqrt{4+9}=\sqrt{13}$. Vertices $(-2,3)$ and $(-2,-1)$. Asymptotes $y-1 = \pm\frac23(x+2)$.

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BLOCK 8

Q42

$$1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$$

Q43

$$x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$$

Q44

$$1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$$

Q45

$$x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \quad (-1 < x \leq 1)$$

Q46

$$1 + x + x^2 + x^3 + \cdots \quad (|x| < 1)$$

Q47

$$x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots$$

Q48

$$1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots$$

Q49

$$x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots$$

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BLOCK 9

Q50

$e^{-2x} = 1 - 2x + \frac{4x^2}{2!} - \frac{8x^3}{3!} + \frac{16x^4}{4!} + \cdots$ $= 1 - 2x + 2x^2 - \frac{4x^3}{3} + \frac{2x^4}{3} + \cdots$

Q51

$$e^{x^2} = 1 + x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} + \cdots$$

Q52

$\sin(3x^2) = 3x^2 - \frac{(3x^2)^3}{3!} + \cdots$ → up to $x^4$: just $3x^2$

Q53

$\cos(2x^3) = 1 - \frac{(2x^3)^2}{2!} + \cdots$ → up to $x^4$: just $1$

Q54

$\ln(1+4x) = 4x - \frac{16x^2}{2} + \frac{64x^3}{3} - \frac{256x^4}{4} + \cdots$ $= 4x - 8x^2 + \frac{64x^3}{3} - 64x^4 + \cdots$

Q55

$x e^{-x} = x\left(1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \cdots\right)$ $= x - x^2 + \frac{x^3}{2} - \frac{x^4}{6} + \cdots$

Q56

$x^2 \cos x = x^2\left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots\right)$ $= x^2 - \frac{x^4}{2} + \cdots$ (up to $x^4$)

Q57

$\frac{1}{1+3x} = \frac{1}{1-(-3x)} = 1 + (-3x) + (-3x)^2 + (-3x)^3 + \cdots$ $= 1 - 3x + 9x^2 - 27x^3 + \cdots$

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BLOCK 10

Q58

$$e^x\sin x = (1 + x + \frac{x^2}{2} + \frac{x^3}{6})(x - \frac{x^3}{6})$$

$x$: $x$ $x^2$: $x^2$ $x^3$: $-\frac{x^3}{6} + \frac{x^3}{2} = \frac{x^3}{3}$

$$\boxed{x + x^2 + \frac{x^3}{3} + \cdots}$$

Q59

$$e^{-x}\cos x = (1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \frac{x^4}{24})(1 - \frac{x^2}{2} + \frac{x^4}{24})$$

Up to $x^4$: $1$: $1$ $x$: $-x$ $x^2$: $-\frac12 + \frac12 = 0$ $x^3$: $\frac{x^3}{6} + \frac{x^3}{2} = -\frac16 + \frac12 = \frac13 x^3$... wait let me be careful.

$$1\cdot1 = 1$$ $$1(-\frac{x^2}{2}) + (-x)(1) = -\frac{x^2}{2} - x$$ $$1(\frac{x^4}{24}) + (-x)(-\frac{x^2}{2}) + (\frac{x^2}{2})(1) = \frac{x^4}{24} + \frac{x^3}{2} + \frac{x^2}{2}$$

Actually let me be systematic:

• Constant: $1\cdot1 = 1$
• $x$: $(-x)(1) = -x$
• $x^2$: $1(-\frac{x^2}{2}) + (\frac{x^2}{2})(1) = -\frac12 + \frac12 = 0$
• $x^3$: $(-x)(-\frac{x^2}{2}) + (-\frac{x^3}{6})(1) = \frac{x^3}{2} - \frac{x^3}{6} = \frac{x^3}{3}$
• $x^4$: $1(\frac{x^4}{24}) + (\frac{x^2}{2})(-\frac{x^2}{2}) + (\frac{x^4}{24})(1) = \frac1{24} - \frac14 + \frac1{24} = \frac1{24} - \frac6{24} + \frac1{24} = -\frac4{24} = -\frac16$

$$\boxed{1 - x + \frac{x^3}{3} - \frac{x^4}{6} + \cdots}$$

Q60

$$e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \cdots$$

$$\int_0^{0.5} e^{-x^2},dx \approx \int_0^{0.5} \left(1 - x^2 + \frac{x^4}{2}\right)dx$$

$$= \left[x - \frac{x^3}{3} + \frac{x^5}{10}\right]_0^{0.5}$$

$$= 0.5 - \frac{0.125}{3} + \frac{0.03125}{10}$$

$= 0.5 - 0.041667 + 0.003125 = \boxed{0.461}$ (3 d.p.)

Q61

$$\frac{\sin x}{x} = \frac{x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots}{x} = 1 - \frac{x^2}{6} + \frac{x^4}{120} - \cdots$$

$$\int \frac{\sin x}{x},dx = \int \left(1 - \frac{x^2}{6} + \frac{x^4}{120} - \cdots\right)dx$$

$$= x - \frac{x^3}{18} + \frac{x^5}{600} - \cdots + C$$

Q62

$$f(x) = \tan x$$

$$f(0) = 0$$ $f'(x) = \sec^2 x$, $f'(0) = 1$ $f''(x) = 2\sec^2 x \tan x$, $f''(0) = 0$ $f'''(x) = 4\sec^2 x \tan^2 x + 2\sec^4 x$, $f'''(0) = 2$

$$\tan x = x + \frac{x^3}{3} + \cdots$$

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BLOCK 11: PART B WILDCARDS (15 min)

Integration by Parts (6 min)

Q63

$$\int x e^x,dx$$

Q64

$$\int x \ln x,dx$$

Q65

$$\int x^2 \ln x,dx$$

Q66

$$\int_0^1 x e^x,dx$$

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Homogeneous DE (5 min)

Q67

$$\frac{dy}{dx} = \frac{x^2 + y^2}{xy}$$

Q68

$$\frac{dy}{dx} = \frac{x^2 + 2xy}{y^2}, \quad y(1) = 1$$

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Trig Substitution or Parametric — Pick One (4 min)

Pick the one you're more comfortable with.

Q69a (Trig Sub)

$$\int \frac{dx}{\sqrt{4 - x^2}}$$

Q70a (Trig Sub)

$$\int \frac{dx}{\sqrt{x^2 + 9}}$$

OR

Q69b (Parametric)

$x = t^2$, $y = t^3$. Find $\frac{dy}{dx}$ and tangent at $t = 1$.

Q70b (Parametric)

$x = \cos t$, $y = \sin t$. Find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$.

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SOLUTIONS — BLOCK 11

Integration by Parts

Q63

$u = x$, $dv = e^x,dx$ → $du = dx$, $v = e^x$ $$\int x e^x,dx = x e^x - \int e^x,dx = \boxed{x e^x - e^x + C}$$

Q64

$u = \ln x$, $dv = x,dx$ → $du = \frac1x,dx$, $v = \frac{x^2}{2}$ $$\int x\ln x,dx = \frac{x^2}{2}\ln x - \int \frac{x}{2},dx = \boxed{\frac{x^2}{2}\ln x - \frac{x^2}{4} + C}$$

Q65

$u = \ln x$, $dv = x^2,dx$ → $du = \frac1x,dx$, $v = \frac{x^3}{3}$ $$\int x^2\ln x,dx = \frac{x^3}{3}\ln x - \int \frac{x^2}{3},dx = \boxed{\frac{x^3}{3}\ln x - \frac{x^3}{9} + C}$$

Q66

$$\int_0^1 x e^x,dx = \left[x e^x - e^x\right]_0^1 = (e - e) - (0 - 1) = \boxed{1}$$

Homogeneous DE

Q67

$y = vx$, $\frac{dy}{dx} = v + x\frac{dv}{dx}$

$$v + x\frac{dv}{dx} = \frac{x^2 + v^2 x^2}{x \cdot vx} = \frac{1 + v^2}{v}$$

$$x\frac{dv}{dx} = \frac{1 + v^2}{v} - v = \frac{1}{v}$$

$v,dv = \frac{dx}{x}$ → $\frac{v^2}{2} = \ln|x| + C$

$$\boxed{y^2 = 2x^2(\ln|x| + C)}$$

Q68

$y = vx$, $v + x\frac{dv}{dx} = \frac{x^2 + 2vx^2}{v^2 x^2} = \frac{1 + 2v}{v^2}$

$$x\frac{dv}{dx} = \frac{1 + 2v}{v^2} - v = \frac{1 + 2v - v^3}{v^2}$$

This gets messy. Let's try: actually $\frac{x^2 + 2xy}{y^2} = \frac{x^2 + 2x(vx)}{(vx)^2} = \frac{x^2(1 + 2v)}{v^2 x^2} = \frac{1 + 2v}{v^2}$

$$x\frac{dv}{dx} = \frac{1 + 2v}{v^2} - v = \frac{1 + 2v - v^3}{v^2}$$

This is separable but the integration is hard. Let me use a simpler example instead.

Actually, replace Q68 with this:

Q68 (revised)

$$\frac{dy}{dx} = \frac{y}{x} + \frac{x}{y}, \quad y(1) = 1$$

$y = vx$, $v + xv' = v + \frac{1}{v}$

$xv' = \frac{1}{v}$ → $v,dv = \frac{dx}{x}$ → $\frac{v^2}{2} = \ln|x| + C$

$$\frac{y^2}{2x^2} = \ln|x| + C$$

$y(1) = 1$ → $\frac12 = C$

$$\boxed{y^2 = 2x^2\left(\ln|x| + \frac12\right)}$$

Trig Substitution

Q69a

$x = 2\sin\theta$, $dx = 2\cos\theta,dθ$ $$\int \frac{2\cos\theta,dθ}{2\cos\theta} = \int dθ = \theta + C = \boxed{\sin^{-1}\left(\frac{x}{2}\right) + C}$$

Q70a

$x = 3\tan\theta$, $dx = 3\sec^2\theta,dθ$ $$\int \frac{3\sec^2\theta,dθ}{3\sec\theta} = \int \sec\theta,dθ = \ln|\sec\theta + \tan\theta| + C$$ $$= \boxed{\ln\left|\frac{x}{3} + \frac{\sqrt{x^2+9}}{3}\right| + C = \ln\left|x + \sqrt{x^2+9}\right| + C'}$$

Parametric

Q69b

$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3t^2}{2t} = \frac{3t}{2}$$

At $t=1$: $x=1$, $y=1$, slope $= \frac32$

Tangent: $y - 1 = \frac32(x - 1)$ → $\boxed{3x - 2y = 1}$

Q70b

$$\frac{dy}{dx} = \frac{\cos t}{-\sin t} = \boxed{-\cot t}$$

$$\frac{d^2y}{dx^2} = \frac{d}{dt}(-\cot t) \big/ \frac{dx}{dt} = \frac{\csc^2 t}{-\sin t} = \boxed{-\csc^3 t}$$

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APPENDIX: MUST-KNOW FORMULAS (CUT OUT + KEEP)

Part A Essentials

Formula	Answer
$\int x^n,dx$	$\frac{x^{n+1}}{n+1} + C$
$\int \frac1x,dx$	$\ln
$\int e^{ax},dx$	$\frac1a e^{ax} + C$
$\int \sin(ax),dx$	$-\frac1a \cos(ax) + C$
$\int \cos(ax),dx$	$\frac1a \sin(ax) + C$
$\int \sec^2 x,dx$	$\tan x + C$
$\int \frac{dx}{\sqrt{a^2-x^2}}$	$\sin^{-1}(\frac{x}{a}) + C$
$\int \frac{dx}{a^2+x^2}$	$\frac1a\tan^{-1}(\frac{x}{a}) + C$

Binomial

$$(a+b)^n = \sum_{r=0}^n \binom{n}{r} a^{n-r}b^r$$

Maclaurin

$$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n$$

Conic Quick Reference

Shape	$c^2$ formula
Ellipse	$c^2 = a^2 - b^2$
Hyperbola	$c^2 = a^2 + b^2$

Parabola form	Opens	Focus	Directrix
$(x-h)^2 = 4a(y-k)$	Up	$(h,k+a)$	$y=k-a$
$(x-h)^2 = -4a(y-k)$	Down	$(h,k-a)$	$y=k+a$
$(y-k)^2 = 4a(x-h)$	Right	$(h+a,k)$	$x=h-a$
$(y-k)^2 = -4a(x-h)$	Left	$(h-a,k)$	$x=h+a$