MAT1051 · Week 14 · Inverse Substitution and Review

Final review: limits, derivatives, applications and integrals

A one-page summary of the whole semester: the main formula of each topic, decision charts and 12 mixed questions like the final.

limitcontinuityderivativechain ruleimplicit differentiationrelated ratesabsolute extremaoptimizationlinearizationFundamental Theorem of Calculussubstitutionintegration by partspartial fractions

This page is for the last read before the final. I have gathered the most frequently asked part of each week in one place; for details go back to the page of the relevant week. The final exam usually consists of four blocks: limits and continuity, computing derivatives, applications of derivatives, and integrals. At least one question comes from each block, and the application questions (related rates, optimization, curve sketching) carry the most points.

Study suggestion: read the tables below, then before looking at the questions, write your own solution on paper. When you open the answer, look not only at the result but at the order of the steps. In the exam, partial credit comes from the steps.

Block 1: limits and continuity (weeks 3-4)

CaseWhat do you do?
A number after substitutingThe answer is that number
00\tfrac00, polynomialFactor, simplify
00\tfrac00, difference of rootsMultiply by the conjugate
00\tfrac00, trigonometricThe pattern sin⁡uu→1\dfrac{\sin u}{u}\to1
c0\tfrac{c}{0}, c≠0c\ne0Infinite limit; right and left signs
Oscillation times zeroSqueeze Theorem
x→±∞x\to\pm\infty, rationalDegree rule
x→−∞x\to-\infty and a rootx2=−x\sqrt{x^2}=-x
Continuity constantAt the junction left = right = f(a)f(a)
Is there a root?IVT: continuity + opposite signs

In a continuity question write the three conditions explicitly: f(a)f(a) defined, the limit exists, they are equal. In a “removable” discontinuity the limit exists; in a “jump” the left and right differ; in an “infinite” one the limit is infinite.

Block 2: computing derivatives (weeks 5-7)

Derivative from the definition: f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}. In a fraction find a common denominator, in a root multiply by the conjugate, cancel hh.

FunctionDerivativeFunctionDerivative
xnx^nnxn−1nx^{n-1}eue^ueuu′e^uu'
sin⁡u\sin uu′cos⁡uu'\cos uaua^uauln⁡a⋅u′a^u\ln a\cdot u'
cos⁡u\cos u−u′sin⁡u-u'\sin uln⁡u\ln uu′u\dfrac{u'}{u}
tan⁡u\tan uu′sec⁡2uu'\sec^2uarcsin⁡u\arcsin uu′1−u2\dfrac{u'}{\sqrt{1-u^2}}
sec⁡u\sec uu′sec⁡utan⁡uu'\sec u\tan uarctan⁡u\arctan uu′1+u2\dfrac{u'}{1+u^2}

For derivatives that look hard, the order is: first simplify (expand with log rules, divide the fraction), then look at the outer structure (product, quotient or composition), chain rule last. If the base and exponent are both variable, use logarithmic differentiation. If yy cannot be written explicitly, use implicit differentiation: every term with yy is multiplied by y′y'.

Derivative of an inverse function: (f−1)′(b)=1f′(a)(f^{-1})'(b)=\dfrac{1}{f'(a)}, f(a)=bf(a)=b. Find aa first.

Block 3: applications of derivatives (weeks 8-10)

Related rates. Figure, variables, equation, derivative with respect to tt, then the numbers at that moment. Don’t fix a changing quantity before differentiating.

Absolute extrema (closed interval): critical points (f′=0f'=0 or undefined) and endpoints; table of values; the largest and the smallest.

MVT: f′(c)=f(b)−f(a)b−af'(c)=\dfrac{f(b)-f(a)}{b-a}. For an “exactly one root” question: at least one by the IVT, at most one by Rolle.

Curve sketching: domain, intercepts, asymptotes, f′f' sign table (increasing/decreasing, local extrema), f′′f'' sign table (concavity, inflection points). Without a sign change there is no extremum or inflection point.

Optimization: objective function + constraint, reduce to one variable, write the interval, f′=0f'=0, show that it is a max/min.

Linear approximation: L(x)=f(a)+f′(a)(x−a)L(x)=f(a)+f'(a)(x-a), with aa a nearby point whose value is known. Error: dQ=Q′ dxdQ=Q'\,dx, relative error dQQ\dfrac{dQ}{Q}.

Block 4: integrals (weeks 11-14)

Riemann sum: Δx=b−an\Delta x=\dfrac{b-a}{n}, xi=a+iΔxx_i=a+i\Delta x, sigma formulas, limit. The definite integral is signed area.

Fundamental Theorem: ddx∫ag(x)f(t) dt=f(g(x))g′(x)\dfrac{d}{dx}\displaystyle\int_a^{g(x)}f(t)\,dt=f(g(x))g'(x) and ∫abf=F(b)−F(a)\displaystyle\int_a^bf=F(b)-F(a).

When you see an integral, try this in order:

1. Directly from the table?       (first expand, divide, simplify)
2. Inner function + its derivative?           -> u-substitution
3. Product of two different types of function? -> integration by parts (LIATE)
4. Polynomial / polynomial?                   -> degree check, partial fractions
5. a^2 - x^2, a^2 + x^2, x^2 - a^2 present?   -> trigonometric substitution
6. ax^2 + bx + c present?                     -> complete the square first

+C+C in an indefinite integral, an absolute value in a ln⁡\ln answer, converting the limits when substituting in a definite integral. These three are where points are lost most often.

Time and score management in the exam

  • Finish the easy blocks first. Limit and direct derivative questions are two-to-three-minute jobs; starting with them both secures points and warms up the mind.
  • In word problems, write the figure and the variable definitions on paper. Even if the result comes out wrong, a correctly set up equation and a correct derivative earn partial credit.
  • Check every antiderivative by differentiating. It takes 20 seconds and catches most sign errors.
  • Check the units and re-read the question. Is “maximum value” asked, or “where”? Velocity or magnitude of velocity? If it says “exact value”, leave ln⁡2\ln 2, π\pi, 3\sqrt3 instead of a decimal.
  • Leave a question you are stuck on and come back. Losing fifteen minutes on a single integral means losing two easy questions.

There is also a quick list for testing yourself: can you recite each of the following from memory? lim⁡sin⁡xx\lim\frac{\sin x}{x}, the product and quotient rules, the six trigonometric derivatives, the derivatives of exe^x, ln⁡x\ln x, arctan⁡x\arctan x, arcsin⁡x\arcsin x, the closed interval method, the formula for L(x)L(x), the two parts of the Fundamental Theorem, the formula for ∫u dv\int u\,dv, the three trigonometric substitutions. Re-read the week of any item you can’t recite.

Common mistakes (whole semester)

Wrong thinkingCorrect
00=0\tfrac00=0Indeterminate; simplify
x2=x\sqrt{x^2}=x always∣x∣\lvert x\rvert
(sin⁡5x)′=cos⁡5x(\sin5x)'=\cos5x5cos⁡5x5\cos5x
(fg)′=f′g′(fg)'=f'g'f′g+fg′f'g+fg'
ddxy2=2y\dfrac{d}{dx}y^2=2y2yy′2yy'
Substituting the number first in related ratesDerivative first
f′(c)=0f'(c)=0 means an extremumThe sign must change
Forgetting the endpointsEndpoints are required on a closed interval
∫dxx=ln⁡x\int\dfrac{dx}{x}=\ln xln⁡∣x∣+C\ln\lvert x\rvert+C
∫fg=∫f∫g\int fg=\int f\int gIntegration by parts
Numerator degree larger, go straight to partial fractionsDivide first
Leaving θ\theta in the answer of a trig substitutionGo back to xx with the triangle

Test yourself

Exam-style questions. Solve them on paper first, then reveal the answer.

Question 1easy

Evaluate lim⁡x→−1x2+3x+2x2−1\displaystyle\lim_{x\to-1}\frac{x^2+3x+2}{x^2-1}.

Show answer
-1/2

00\tfrac00. Factor: (x+1)(x+2)(x+1)(x−1)=x+2x−1\dfrac{(x+1)(x+2)}{(x+1)(x-1)}=\dfrac{x+2}{x-1}. At x=−1x=-1: 1−2=−12\dfrac{1}{-2}=-\dfrac12.

Question 2medium

Evaluate lim⁡x→∞(9x2+2x−3x)\displaystyle\lim_{x\to\infty}\left(\sqrt{9x^2+2x}-3x\right).

Show answer
1/3

Multiply by the conjugate: 2x9x2+2x+3x\dfrac{2x}{\sqrt{9x^2+2x}+3x}. Divide by xx: 29+2/x+3→26=13\dfrac{2}{\sqrt{9+2/x}+3}\to\dfrac{2}{6}=\dfrac13.

Question 3easy

Find kk so that f(x)={sin⁡3xx,x≠0k,x=0f(x)=\begin{cases}\dfrac{\sin 3x}{x}, & x\ne0\\ k, & x=0\end{cases} is continuous at x=0x=0.

Show answer
k = 3

Continuity: f(0)=lim⁡x→0f(x)f(0)=\lim_{x\to0}f(x). lim⁡x→0sin⁡3xx=3⋅lim⁡sin⁡3x3x=3\lim_{x\to0}\dfrac{\sin3x}{x}=3\cdot\lim\dfrac{\sin3x}{3x}=3. So k=3k=3.

Question 4medium

Differentiate y=ex2cos⁡3xy=e^{x^2}\cos3x.

Show answer
y′=ex2(2xcos⁡3x−3sin⁡3x)y'=e^{x^2}(2x\cos3x-3\sin3x)

Product rule, chain rule in each piece: (ex2)′=2xex2(e^{x^2})'=2xe^{x^2}, (cos⁡3x)′=−3sin⁡3x(\cos3x)'=-3\sin3x. y′=2xex2cos⁡3x−3ex2sin⁡3xy'=2xe^{x^2}\cos3x-3e^{x^2}\sin3x.

Question 5medium

Find dydx\dfrac{dy}{dx} at (0,1)(0,1) for the curve ey+xy=ee^y+xy=e.

Show answer
-1/e

Check: e1+0=ee^1+0=e, the point is on the curve. Implicit derivative: eyy′+y+xy′=0⇒y′=−yey+xe^yy'+y+xy'=0 \Rightarrow y'=-\dfrac{y}{e^y+x}. At (0,1)(0,1): −1e-\dfrac{1}{e}.

Question 6easy

The side of a square increases at 22 cm/s. How fast is its area increasing when the side is 55 cm?

Show answer
20 cm²/s

A=s2⇒dAdt=2sdsdt=2⋅5⋅2=20A=s^2 \Rightarrow \dfrac{dA}{dt}=2s\dfrac{ds}{dt}=2\cdot5\cdot2=20 cm²/s.

Question 7medium

Find the absolute maximum and minimum values of f(x)=x3−3xf(x)=x^3-3x on [−2,3][-2,3].

Show answer
max 18 (at x = 3), min -2 (at x = -2 and x = 1)

f′=3x2−3=0⇒x=±1f'=3x^2-3=0 \Rightarrow x=\pm1. Values: f(−2)=−8+6=−2f(-2)=-8+6=-2, f(−1)=−1+3=2f(-1)=-1+3=2, f(1)=1−3=−2f(1)=1-3=-2, f(3)=27−9=18f(3)=27-9=18. Max 1818 (at an endpoint), min −2-2.

Question 8easy

Among all rectangles with perimeter 3636 cm, find the one with the largest area.

Show answer
9 cm × 9 cm square, area 81 cm²

2x+2y=36⇒y=18−x2x+2y=36 \Rightarrow y=18-x. A=x(18−x)A=x(18-x), A′=18−2x=0⇒x=9A'=18-2x=0 \Rightarrow x=9, y=9y=9. A′′=−2<0A''=-2<0: max. A=81A=81 cm².

Question 9easy

Use a linear approximation to estimate 4.1\sqrt{4.1}.

Show answer
≈ 2.025

f(x)=xf(x)=\sqrt x, a=4a=4: f(4)=2f(4)=2, f′(4)=14f'(4)=\tfrac14. L(4.1)=2+14(0.1)=2.025L(4.1)=2+\tfrac14(0.1)=2.025. The true value is 2.024852.02485.

Question 10medium

Evaluate ∫01xex2 dx\displaystyle\int_0^1xe^{x^2}\,dx.

Show answer
(e - 1)/2 ≈ 0.859

u=x2u=x^2, du=2x dxdu=2x\,dx, limits 0→00\to0, 1→11\to1. 12∫01eu du=12(e−1)\tfrac12\int_0^1e^u\,du=\tfrac12(e-1).

Question 11medium

Find ∫xln⁡x dx\displaystyle\int x\ln x\,dx.

Show answer
x2ln⁡x2−x24+C\dfrac{x^2\ln x}{2}-\dfrac{x^2}{4}+C

LIATE: u=ln⁡xu=\ln x, dv=x dxdv=x\,dx, v=x22v=\tfrac{x^2}{2}. ∫=x22ln⁡x−∫x22⋅1x dx=x2ln⁡x2−x24+C\int=\tfrac{x^2}{2}\ln x-\int\tfrac{x^2}{2}\cdot\tfrac1x\,dx=\tfrac{x^2\ln x}{2}-\tfrac{x^2}{4}+C.

Question 12hard

Find ∫dxx2−x−6\displaystyle\int\frac{dx}{x^2-x-6}.

Show answer
15ln⁡∣x−3x+2∣+C\dfrac15\ln\left|\dfrac{x-3}{x+2}\right|+C

x2−x−6=(x−3)(x+2)x^2-x-6=(x-3)(x+2). 1(x−3)(x+2)=Ax−3+Bx+2\dfrac{1}{(x-3)(x+2)}=\dfrac{A}{x-3}+\dfrac{B}{x+2}, 1=A(x+2)+B(x−3)1=A(x+2)+B(x-3). x=3x=3: A=15A=\tfrac15. x=−2x=-2: B=−15B=-\tfrac15. 15ln⁡∣x−3∣−15ln⁡∣x+2∣+C\tfrac15\ln|x-3|-\tfrac15\ln|x+2|+C.

Verification: All answers were checked with node (verify/MAT1051/14.mjs, 'Final review' section): limits with values near the point, derivatives with a central difference, implicit derivatives with a local Newton solution, extrema by scanning the interval, integrals with Simpson's rule and by comparing F' = f.

Review card

LIMIT: substitute -> if 0/0 factor/conjugate/sin u/u -> if c/0 check right-left sign x -> ±inf: degree rule; sqrt(x^2) = |x|; inf - inf: conjugate CONTINUOUS: lim = f(a); piecewise: left = right = f(a); IVT for existence of a root DERIVATIVE: (fg)' = f'g + fg' (f/g)' = (f'g - fg')/g^2 chain: f'(g)·g' sin->cos cos->-sin tan->sec^2 e^u->e^u u' ln u->u'/u arctan u->u'/(1+u^2) arcsin u->u'/√(1-u^2) Implicit: y' in every term with y; x^x: take ln APPLICATIONS: related rates -> derivative then numbers; absolute extremum -> critical points + endpoints Curve: sign of f' for increasing/decreasing, sign of f'' for concavity; the sign must CHANGE Optimization: objective + constraint -> one variable -> f' = 0 -> verify max/min L(x) = f(a) + f'(a)(x - a) dy = f'(x)dx INTEGRAL: FTC d/dx ∫_a^g(x) f = f(g)g'; ∫_a^b f = F(b) - F(a) Order of methods: table -> u-sub -> parts (LIATE) -> partial fractions -> trig substitution