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Calculus

Level · High school → introductory college · Calc I–III Chapters · 16 Explorers · 15 interactive Prerequisite · Algebra & functions

Foundations & Precalculus Review

Calculus is the study of how functions change and accumulate, so every later idea leans on fluency with functions themselves. This chapter gathers the algebra, function families, and notation the rest of the course quietly assumes.

● Definition 0.1 — Function, domain, range

A function $f$ assigns to each input $x$ in its domain exactly one output $f(x)$ . The set of resulting outputs is the range. We write $f:\mathbb{R}\to\mathbb{R}$ for a real-valued function of a real variable.

The function toolkit

A handful of parent functions, combined and transformed, generate almost everything in single-variable calculus.

Power

x^{n}

Exponential

a^{x},\ e^{x}

Logarithm

\log_a x,\ \ln x

Trigonometric

\sin x,\ \cos x,\ \tan x

● Definition 0.2 — Transformations

From a parent $f$ , the graph of $y=a\,f(x-h)+k$ is shifted right by $h$ , up by $k$ , and scaled vertically by $a$ (reflected across the $x$ -axis when $a\lt 0$ ).

⊙ Explorer · Transform a parent functioninteractive

vertical scale a: 10

horizontal shift h: 0

vertex (h, 0)(0.0, 0)

directionopens up

The faint grey curve is the parent $y=x^2$ ; the bold curve is $y=a(x-h)^2$ . Watch the vertex slide and flip.

Exponentials and logarithms

These are inverse operations, and their algebra appears constantly in growth models and integration. The key laws:

\log_a(xy)=\log_a x+\log_a y, \qquad \log_a\!\Big(\tfrac{x}{y}\Big)=\log_a x-\log_a y, \qquad \log_a(x^{r})=r\log_a x.

The natural base $e\approx 2.71828$ makes $\dfrac{d}{dx}e^{x}=e^{x}$ , which is why it dominates calculus.

Worked example 0.A — Finding a domain

Find the domain of $\displaystyle f(x)=\frac{\sqrt{x-2}}{x-5}$ .

The square root needs $x-2\ge 0$ , so $x\ge 2$ ; the denominator forbids $x=5$ . Combining,

\text{domain}=[2,5)\cup(5,\infty).

Worked example 0.B — Reading a transformation

Describe $y=-2(x-3)^{2}+1$ relative to $y=x^{2}$ .

It is shifted right $3$ and up $1$ , stretched vertically by $2$ , and reflected (since $a=-2\lt 0$ ). The vertex is $(3,1)$ and the parabola opens downward.

Parent parabola

y = x^2

alongside shifted and inverted parabola

y = -(x - 2)^2 + 3

with vertex at

(2, 3)

⚠ Common Traps & Exam Pitfalls

The 'Universal Distributive' Fallacy: Square roots, powers, and logs NEVER distribute across addition:
•

\sqrt{a^2 + b^2} \ne a + b

•

(a + b)^2 \ne a^2 + b^2

(remember the middle term

2ab

!)
•

\frac{1}{a + b} \ne \frac{1}{a} + \frac{1}{b}

•

\ln(a + b) \ne \ln a + \ln b

.

Domain Restrictions: In calculus, domains govern continuity and differentiability. Always inspect: (1) Even roots: radicand must be

\ge 0

; (2) Denominators: must be

\ne 0

; (3) Logarithms: argument must be strictly

> 0

Find the domain of

f(x) = \dfrac{\ln(x + 3)}{\sqrt{4 - x^2}}

For the numerator, the logarithm requires

x + 3 > 0 \implies x > -3

. For the denominator, the square root in the denominator requires

4 - x^2 > 0 \implies x^2 < 4 \implies -2 < x < 2

. Taking the intersection of

x > -3

and

-2 < x < 2

gives the domain

(-2, 2)

📝 Chapter Quiz

Questions Level

Limits & Continuity

Calculus rests on a single idea: what value does a function approach as its input creeps toward a target? It need not ever reach that value — what matters is the trend.

● Definition 1.1 — Limit (informal)

We write $\displaystyle \lim_{x \to c} f(x) = L$ if the outputs $f(x)$ can be made as close to $L$ as we like by taking $x$ sufficiently close to $c$ (but not equal to $c$ ).

The function may even be undefined exactly at $c$ . Consider $\displaystyle f(x)=\frac{x^2-1}{x-1}$ . At $x=1$ the formula gives $\tfrac{0}{0}$ , yet factoring reveals $f(x)=x+1$ everywhere else, so the curve heads straight for $2$ .

⊙ Explorer · Approaching a limitinteractive

Move x toward 1: -50

input x0.500

output f(x)1.500

limit as x→12.000

Slide from either side: $f(x)$ marches to $2$ , even though the point at $x=1$ is a hole.

One-sided limits

Sometimes the approach differs by direction. The left-hand limit $\displaystyle \lim_{x\to c^-} f(x)$ and right-hand limit $\displaystyle \lim_{x\to c^+} f(x)$ must agree for the two-sided limit to exist:

\lim_{x\to c} f(x)=L \quad\Longleftrightarrow\quad \lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x)=L.

● Definition 1.2 — Continuity

A function $f$ is continuous at $c$ when all three hold: (i) $f(c)$ is defined, (ii) $\displaystyle\lim_{x\to c} f(x)$ exists, and (iii) the two are equal, $\displaystyle\lim_{x\to c} f(x)=f(c)$ . Informally: you can draw it through $c$ without lifting the pen.

Two limits worth memorising

Trig squeeze

\displaystyle \lim_{x\to 0}\frac{\sin x}{x}=1

The number e

\displaystyle \lim_{x\to\infty}\!\Big(1+\tfrac1x\Big)^{x}=e

Worked example 1.A

Evaluate $\displaystyle \lim_{x\to 3}\frac{x^2-9}{x-3}$ .

Direct substitution gives $\tfrac00$ , so factor the numerator: $x^2-9=(x-3)(x+3)$ . Cancel the common factor (valid since $x\neq 3$ in the limit):

\lim_{x\to 3}\frac{(x-3)(x+3)}{x-3}=\lim_{x\to 3}(x+3)=6.

Plot of rational function

f(x) = \frac{x^2 - 9}{x - 3}

showing a removable discontinuity (hole) at

(3, 6)

⚠ Common Traps & Exam Pitfalls

Indeterminate Form 0/0: Obtaining

\frac{0}{0}

from direct substitution does NOT mean the limit is 0, 1, or does not exist. It is an indeterminate form signaling that an algebraic simplification (factoring, conjugate rationalization, trig identity, or L'Hôpital's Rule) must be performed.

Two-Sided Limit Existence: The overall limit

\lim_{x\to c} f(x) = L

exists IF AND ONLY IF both one-sided limits exist and agree:

\lim_{x\to c^-} f(x) = \lim_{x\to c^+} f(x) = L

. For piece-wise and absolute value functions, always verify both directions separately.

Continuity Checklist: A function

f

is continuous at

x = c

only if all 3 conditions hold: (1)

f(c)

is defined; (2)

\lim_{x\to c} f(x)

exists; (3)

\lim_{x\to c} f(x) = f(c)

Evaluate the one-sided limit

\displaystyle\lim_{x\to 2^-} \frac{|x - 2|}{x - 2}

When

x

approaches 2 from the left (

x < 2

x - 2 < 0

, so

|x - 2| = -(x - 2)

. Substituting gives

\lim_{x\to 2^-} \frac{-(x - 2)}{x - 2} = \lim_{x\to 2^-} (-1) = -1

. (Note: from the right, the limit is

+1

, so the two-sided limit does not exist).

📝 Chapter Quiz

Questions Level

Derivatives

The derivative is the first question of calculus made precise: the instantaneous rate of change of $f$ , equivalently the slope of the tangent line to its graph.

● Definition 2.1 — The derivative

The derivative of $f$ at $x$ is the limit of slopes of secant lines:

f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h},

when this limit exists. It is also written $\dfrac{dy}{dx}$ or $\dfrac{d}{dx}f(x)$ .

As $h\to 0$ , the secant through $\big(x,f(x)\big)$ and $\big(x+h,f(x+h)\big)$ pivots into the tangent. Watch it happen — drag the point of tangency and the slope readout below updates live.

⊙ Explorer · Tangent line & slopeinteractive

Function

f(x)

Point of tangency a = 62

Trace the derivative

f'(x)

in amber

height f(a)—

slope f′(a)—

The rules of differentiation

We rarely use the limit definition directly. These rules build every elementary derivative:

Power rule

\dfrac{d}{dx}x^{n}=n\,x^{\,n-1}

Constant multiple

\dfrac{d}{dx}\big(c\,f\big)=c\,f'

Sum

(f+g)'=f'+g'

Product

(fg)'=f'g+fg'

Quotient

\Big(\dfrac{f}{g}\Big)'=\dfrac{f'g-fg'}{g^{2}}

Chain

\dfrac{d}{dx}f\big(g(x)\big)=f'\!\big(g(x)\big)\,g'(x)

▲ Reference — Common derivatives

$\dfrac{d}{dx}\sin x=\cos x$ , $\quad \dfrac{d}{dx}\cos x=-\sin x$ , $\quad \dfrac{d}{dx}\tan x=\sec^2 x$ ,

$\dfrac{d}{dx}e^{x}=e^{x}$ , $\quad \dfrac{d}{dx}\ln x=\dfrac1x \quad (x>0).$

Worked example 2.A — Chain rule

Differentiate $y=\sin\!\big(3x^2\big)$ .

Outer function $\sin u$ with $u=3x^2$ . Then $\dfrac{dy}{du}=\cos u$ and $\dfrac{du}{dx}=6x$ . By the chain rule,

\frac{dy}{dx}=\cos\!\big(3x^2\big)\cdot 6x = 6x\cos\!\big(3x^2\big).

Worked example 2.B — Product rule

Differentiate $y=x^2 e^{x}$ .

With $f=x^2$ (so $f'=2x$ ) and $g=e^x$ (so $g'=e^x$ ):

y'=2x\,e^{x}+x^2 e^{x}=x e^{x}(x+2).

Tangent line

y = 2x - 1

to parabola

y = x^2

at point of tangency

P(1, 1)

, demonstrating gradient

m = f'(1) = 2

⚠ Common Traps & Exam Pitfalls

Product Rule Trap: The derivative of a product is NOT the product of derivatives:

(uv)' \ne u'v'

. Always use

(uv)' = u'v + uv'

.

Quotient Rule Order: Order matters in subtraction:

\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}

(the numerator is

u'v - uv'

, never reversed).

Chain Rule Inner Function: When differentiating composite functions such as

[g(x)]^n

e^{g(x)}

, or

\sin(g(x))

, never forget to multiply by the derivative of the inside:

g'(x)

. For example,

\frac{d}{dx}\cos(5x) = -5\sin(5x)

, not

-\sin(5x)

Find the equation of the tangent line to the curve

y = x^3 - 4x + 1

at the point where

x = 2

x = 2

y = 2^3 - 4(2) + 1 = 8 - 8 + 1 = 1

, so the point is

(2, 1)

. Differentiating gives

\frac{dy}{dx} = 3x^2 - 4

. At

x = 2

, the slope is

m = 3(4) - 4 = 8

. Using point-slope form:

y - 1 = 8(x - 2) \implies y = 8x - 15

📝 Chapter Quiz

Questions Level

Integrals

The integral is the second question of calculus: how much accumulates. Geometrically, the definite integral measures the signed area between a curve and the horizontal axis.

● Definition 3.1 — Definite integral (as a limit of sums)

Partition $[a,b]$ into $n$ strips of width $\Delta x=\frac{b-a}{n}$ . Sampling a height in each strip and summing the rectangle areas gives a Riemann sum; the integral is its limit:

\int_a^b f(x)\,dx=\lim_{n\to\infty}\sum_{i=1}^{n} f(x_i^\ast)\,\Delta x.

The explorer below builds that limit by hand. Increase the number of rectangles and watch the approximation tighten onto the true area — the error column shrinks toward zero.

⊙ Explorer · Riemann sums → areainteractive

Function

f(x)

[0,4]

Rectangles n = 6

Sample point

approx Sₙ—

exact ∫—

error—

Antiderivatives & the indefinite integral

Reversing differentiation, $F$ is an antiderivative of $f$ if $F'=f$ . Since constants vanish under differentiation, antiderivatives carry a $+C$ .

Power (n ≠ −1)

\displaystyle\int x^{n}\,dx=\frac{x^{n+1}}{n+1}+C

Reciprocal

\displaystyle\int \frac{1}{x}\,dx=\ln|x|+C

Exponential

\displaystyle\int e^{x}\,dx=e^{x}+C

Sine

\displaystyle\int \sin x\,dx=-\cos x+C

Cosine

\displaystyle\int \cos x\,dx=\sin x+C

Linearity

\displaystyle\int (af+bg)=a\!\int\! f+b\!\int\! g

Worked example 3.A

Find the area under $f(x)=x^2$ from $0$ to $2$ .

An antiderivative is $F(x)=\tfrac{x^3}{3}$ . Evaluating (using the Fundamental Theorem, next chapter):

\int_0^2 x^2\,dx=\Big[\tfrac{x^3}{3}\Big]_0^2=\tfrac{8}{3}-0=\tfrac{8}{3}\approx 2.667.

Parabola

y = x^2

over interval

[0, 2]

with shaded area representing the definite integral

\int_0^2 x^2\,dx = \frac{8}{3}

⚠ Common Traps & Exam Pitfalls

Forgetting the Constant + C: Every indefinite integral represents an entire family of antiderivatives and MUST end with

+ C

. Missing

+ C

loses points on free-response exams.

Power Rule Exception for 1/x: The power rule

\int x^n\,dx = \frac{x^{n+1}}{n+1} + C

is valid for all

n \ne -1

. When

n = -1

\int \frac{1}{x}\,dx = \ln|x| + C

(with absolute value bars!).

Trigonometric Integral Signs: Reversing signs between differentiation and integration is very common:

\int \sin x\,dx = -\cos x + C

(negative!), whereas

\int \cos x\,dx = +\sin x + C

.

Non-Distributivity Over Products: You CANNOT integrate term-by-term across multiplication:

\int f(x)g(x)\,dx \ne \left(\int f(x)\,dx\right)\left(\int g(x)\,dx\right)

Evaluate the indefinite integral

\displaystyle\int \left(4x^3 - \frac{2}{x} + 3\sin x\right)\,dx

Integrating term by term:
•

\int 4x^3\,dx = x^4

•

\int -\frac{2}{x}\,dx = -2\ln|x|

•

\int 3\sin x\,dx = -3\cos x

Combining and adding the constant of integration yields:

x^4 - 2\ln|x| - 3\cos x + C

📝 Chapter Quiz

Questions Level

The Fundamental Theorem

Here the two halves meet. The slope problem (indigo) and the area problem (amber) turn out to be inverses of one another — arguably the central result of all calculus.

▲ Theorem 4.1 — Fundamental Theorem of Calculus

Part I. If $\displaystyle A(x)=\int_a^x f(t)\,dt$ , then $A$ is differentiable and $A'(x)=f(x)$ . Differentiating an accumulated area recovers the original function.

Part II. If $F$ is any antiderivative of $f$ , then $\displaystyle\int_a^b f(x)\,dx=F(b)-F(a).$ This is what makes integration computable.

The twin explorer makes Part I visible. The amber area on the upper graph is exactly the indigo height of the accumulation graph below. Drag $x$ : the rate at which the lower curve rises equals the upper curve's height at that instant — that is, $A'(x)=f(x)$ .

⊙ Explorer · Accumulation A(x) = ∫ₐˣ finteractive

Upper limit x = 35

area A(x)—

slope A′(x)=f(x)—

Worked example 4.A — Putting it together

Evaluate $\displaystyle\int_0^{\pi} \sin x\,dx$ .

An antiderivative of $\sin x$ is $-\cos x$ . By Part II,

\int_0^{\pi}\sin x\,dx=\big[-\cos x\big]_0^{\pi}=(-\cos\pi)-(-\cos 0)=1+1=2.

▲ Fundamental Theorem of Calculus

Part I: If $F(x)=\int_a^x f(t)\,dt$ , then $F'(x)=f(x)$ .
Part II: $\int_a^b f(x)\,dx = F(b)-F(a)$ where $F'=f$ .

FTC Part I

\dfrac{d}{dx}\int_a^x f(t)\,dt = f(x)

FTC Part II

\int_a^b f(x)\,dx = F(b)-F(a)

Net change

\int_a^b F'(x)\,dx = F(b)-F(a)

Chain rule form

\dfrac{d}{dx}\int_a^{g(x)}\!f(t)\,dt = f(g(x))\,g'(x)

Worked example

Evaluate $\int_1^4 2x\,dx$ .

F(x)=x^2,\quad F(4)-F(1)=16-1=15.

Worked example

Find $\dfrac{d}{dx}\int_0^{x^2}\cos t\,dt$ .

\cos(x^2)\cdot 2x = 2x\cos(x^2).

Accumulation function

g(x) = \int_0^x f(t)\,dt

measuring net area under

f(t)

, illustrating the Fundamental Theorem of Calculus

g'(x) = f(x)

⚠ Common Traps & Exam Pitfalls

FTC Part 1 Chain Rule Missing: When differentiating an integral with a variable upper limit

u(x)

\frac{d}{dx}\int_a^{u(x)} f(t)\,dt = f(u(x)) \cdot u'(x)

Forgetting to multiply by

u'(x)

is the single most penalized error in AP Calculus!

Lower Limit is a Function: If the variable is in the lower limit, reverse limits and negate first:

\frac{d}{dx}\int_{u(x)}^b f(t)\,dt = -\frac{d}{dx}\int_b^{u(x)} f(t)\,dt = -f(u(x))u'(x)

Both Limits Variable: Split at a constant

c

\int_{v(x)}^{u(x)} = \int_c^{u(x)} - \int_c^{v(x)}

, yielding

f(u(x))u'(x) - f(v(x))v'(x)

Find the derivative

h'(x)

for the accumulation function

h(x) = \displaystyle\int_2^{x^3} \sqrt{1 + t^2}\,dt

By FTC Part 1 and the Chain Rule, let

u = x^3

so that

\frac{du}{dx} = 3x^2

. Then:

h'(x) = \sqrt{1 + (x^3)^2} \cdot \frac{d}{dx}(x^3) = 3x^2 \sqrt{1 + x^6}.

📝 Chapter Quiz

Questions Level

Applications of Derivatives

The sign of the derivative tells us where a function rises or falls; the sign of the second derivative tells us how it bends. Together they locate maxima, minima, and inflection points — the heart of optimization.

● Definition 5.1 — Critical point

A critical point of $f$ is a value $c$ where $f'(c)=0$ or $f'(c)$ does not exist. Local maxima and minima can occur only at critical points.

▲ Theorem 5.2 — Monotonicity & concavity

On an interval: if $f'>0$ then $f$ is increasing; if $f'<0$ then $f$ is decreasing. If $f''>0$ the graph is concave up; if $f''<0$ it is concave down. An inflection point is where concavity changes.

Second derivative test. At a critical point $c$ : if $f''(c)>0$ then $c$ is a local minimum; if $f''(c)<0$ then $c$ is a local maximum.

Optimization puts this to work: model the quantity, differentiate, set the derivative to zero, and confirm the extremum. The explorer animates the classic box problem — the volume rises to a single peak exactly where $V'(x)=0$ .

⊙ Explorer · Optimizing an open boxinteractive

Corner cut x = 25

volume V(x)—

maximum—

Cut squares of side $x$ from a $10\times 8$ sheet and fold up: $V(x)=x(10-2x)(8-2x)$ . The amber marker is the maximum.

Worked example 5.A — Local extrema

Find and classify the extrema of $f(x)=x^3-3x$ .

Critical points: $f'(x)=3x^2-3=3(x-1)(x+1)=0 \Rightarrow x=\pm 1$ . Use $f''(x)=6x$ :

f''(1)=6>0 \ (\text{min}),\qquad f''(-1)=-6<0 \ (\text{max}).

So a local minimum at $(1,-2)$ and a local maximum at $(-1,2)$ .

Worked example 5.B — Optimization

A rectangular pen is built against a wall using $100$ m of fence on the other three sides. Maximise the area.

Let the two perpendicular sides be $x$ ; the parallel side is $100-2x$ . Then $A(x)=x(100-2x)=100x-2x^2$ , so $A'(x)=100-4x=0 \Rightarrow x=25$ .

\text{Dimensions } 25\times 50, \qquad A_{\max}=1250\ \text{m}^2.

Related rates

When two quantities are tied together by an equation and both vary with time, differentiate the relation with respect to $t$ (chain rule) to connect their rates.

Worked example 5.C — Related rates

A spherical balloon is inflated so its radius grows at $\dfrac{dr}{dt}=2$ cm/s. How fast is the volume increasing when $r=5$ cm?

From $V=\tfrac{4}{3}\pi r^3$ , differentiate with respect to $t$ : $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$ . At $r=5$ :

\frac{dV}{dt}=4\pi(25)(2)=200\pi\ \text{cm}^3/\text{s}.

L'Hôpital's Rule

▲ Theorem 5.3 — L'Hôpital's Rule

If $\displaystyle\lim_{x\to a}\dfrac{f(x)}{g(x)}$ has indeterminate form $\tfrac00$ or $\tfrac{\infty}{\infty}$ , then

\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)},

provided the right-hand limit exists.

Worked example 5.D — L'Hôpital

Evaluate $\displaystyle\lim_{x\to\infty}\frac{\ln x}{x}$ (form $\tfrac{\infty}{\infty}$ ).

Differentiate top and bottom: $\dfrac{d}{dx}\ln x=\tfrac1x$ , $\dfrac{d}{dx}x=1$ :

\lim_{x\to\infty}\frac{\ln x}{x}=\lim_{x\to\infty}\frac{1/x}{1}=0.

▲ Key derivative applications

Derivatives measure rates of change. The first derivative test locates local extrema; the second derivative test distinguishes minima from maxima and reveals concavity.

Critical point

f'(c)=0

f'(c)

undefined

1st deriv. test

f'

changes

+\to-

: local max;

-\to+

: local min

2nd deriv. test

f''(c)>0

: min;

f''(c)<0

: max

L'Hôpital

\lim\dfrac{f}{g}\!=\!\lim\dfrac{f'}{g'}

when

\tfrac{0}{0}

\tfrac{\infty}{\infty}

Worked example — optimization

A farmer has 200 m of fencing to enclose a rectangle against a wall. Find dimensions maximizing area.

Let width $= x$ , length $= 200-2x$ . Area $A=x(200-2x)=200x-2x^2$ .

A'=200-4x=0 \implies x=50\text{ m},\; L=100\text{ m}.

Worked example — related rates

A ladder 10 m long slides down a wall. When the base is 6 m from the wall and moving out at 2 m/s, how fast is the top falling?

x^2+y^2=100 \implies 2x\tfrac{dx}{dt}+2y\tfrac{dy}{dt}=0.

At $x=6$ , $y=8$ : $12(2)+16\dfrac{dy}{dt}=0 \Rightarrow \dfrac{dy}{dt}=-1.5$ m/s.

Cubic polynomial

f(x) = x^3 - 3x

exhibiting local maximum at

(-1, 2)

and local minimum at

(1, -2)

, where

f'(x) = 0

⚠ Common Traps & Exam Pitfalls

Critical Points vs Extrema: A critical point

f'(c) = 0

or undefined is a candidate, NOT a guarantee of an extremum. For example,

f(x) = x^3

has

f'(0) = 0

, but changes sign neither way — it is an inflection point, not a max or min!

Absolute Extrema on Closed Intervals: By the Extreme Value Theorem, continuous

f

[a, b]

attains an absolute max and min. You MUST evaluate candidates at critical points AND at both endpoints

f(a)

and

f(b)

.

Related Rates Premature Substitution: Differentiate your geometric relation implicitly with respect to

t

FIRST. Never substitute instantaneous values (e.g.

x = 6

) before differentiating, or the derivative will erroneously treat variables as constants!

Find the absolute maximum and minimum values of

f(x) = 2x^3 - 3x^2 - 12x + 5

on the closed interval

[0, 3]

Find critical points:

f'(x) = 6x^2 - 6x - 12 = 6(x^2 - x - 2) = 6(x - 2)(x + 1) = 0

. The roots are

x = 2

and

x = -1

. Only

x = 2

lies in

[0, 3]

.
Evaluate at endpoints and interior critical point:
•

f(0) = 5

•

f(2) = 2(8) - 3(4) - 12(2) + 5 = 16 - 12 - 24 + 5 = -15

•

f(3) = 2(27) - 3(9) - 12(3) + 5 = 54 - 27 - 36 + 5 = -4

Absolute maximum is

5

(at

x = 0

); Absolute minimum is

-15

(at

x = 2

📝 Chapter Quiz

Questions Level

Techniques of Integration

Most integrals are not in a table. Two techniques reverse the differentiation rules you already know and unlock the majority of exam problems.

▲ Theorem 6.1 — Substitution (reverse chain rule)

If $u=g(x)$ then $du=g'(x)\,dx$ , and

\int f\big(g(x)\big)\,g'(x)\,dx=\int f(u)\,du.

Choose $u$ so that its derivative also appears (up to a constant) in the integrand.

▲ Theorem 6.2 — Integration by parts

From the product rule, $\displaystyle\int u\,dv = uv-\int v\,du.$ Pick $u$ by LIATE (logarithms, inverse-trig, algebraic, trig, exponential — earlier letters make better $u$ ).

Worked example 6.A — Substitution

Evaluate $\displaystyle\int 2x\cos\!\big(x^2\big)\,dx$ .

Let $u=x^2$ , so $du=2x\,dx$ . The integral becomes $\int \cos u\,du$ :

\int 2x\cos\!\big(x^2\big)\,dx=\sin\!\big(x^2\big)+C.

Worked example 6.B — By parts

Evaluate $\displaystyle\int x\,e^{x}\,dx$ .

Let $u=x,\ dv=e^{x}dx$ , so $du=dx,\ v=e^{x}$ . Then $uv-\int v\,du$ :

\int x\,e^{x}\,dx=x e^{x}-\int e^{x}\,dx=e^{x}(x-1)+C.

▲ Integration techniques

Beyond basic antiderivatives: substitution reverses the chain rule; integration by parts reverses the product rule; partial fractions decompose rational integrands.

Substitution

\int f(g(x))g'(x)\,dx = F(g(x))+C

By parts

\int u\,dv = uv - \int v\,du

Trig sub

\sqrt{a^2-x^2}

: set

x=a\sin\theta

Partial fractions

\dfrac{1}{(x-a)(x-b)}=\dfrac{A}{x-a}+\dfrac{B}{x-b}

Worked example — substitution

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

Let $u=x^2$ , $du=2x\,dx$ : $\int e^u\,du = e^{x^2}+C$ .

Worked example — by parts

Evaluate $\int x\cos x\,dx$ .

Let $u=x$ , $dv=\cos x\,dx$ . Then $du=dx$ , $v=\sin x$ :

x\sin x - \int\sin x\,dx = x\sin x + \cos x + C.

Graph of integrand

f(x) = x e^{-x}

, highlighting the weighting factor in integration by parts

\int x e^{-x}\,dx

⚠ Common Traps & Exam Pitfalls

Integration by Parts Sign Error: The formula is

\int u\,dv = uv - \int v\,du

. Forgetting that the second term has a MINUS sign (or mishandling double negatives when

v

is negative) is a chronic trap.

Choosing u via LIATE: Prioritize

u

in the order: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. Differentiating

u

must simplify the remaining integrand.

Partial Fractions Improper Rational Functions: The method of partial fractions ONLY applies when the numerator degree is strictly LESS than the denominator degree. If

\deg(P) \ge \deg(Q)

, you MUST perform polynomial long division first!

Definite Integrals with U-Sub: Remember to update limits of integration:

x = a \to u(a)

x = b \to u(b)

. Evaluating using original limits on a

u

-expression is an automatic deduction.

Evaluate the indefinite integral

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

using integration by parts.

Let

u = x

and

dv = e^{2x}\,dx

. Then

du = dx

and

v = \frac{1}{2}e^{2x}

.
Applying

\int u\,dv = uv - \int v\,du

\int x e^{2x}\,dx = \frac{1}{2}x e^{2x} - \int \frac{1}{2}e^{2x}\,dx = \frac{1}{2}x e^{2x} - \frac{1}{4}e^{2x} + C = \frac{1}{4}e^{2x}(2x - 1) + C.

📝 Chapter Quiz

Questions Level

Applications of Integration

Definite integrals measure accumulated quantities: the area trapped between two curves, the volume swept out by rotating a region, or the average height of a function.

▲ Theorem 7.1 — Area, volume, average

Area between curves: $\displaystyle A=\int_a^b \big[\,f(x)-g(x)\,\big]\,dx$ where $f$ is the upper curve.

Volume of revolution (disks): $\displaystyle V=\pi\int_a^b \big[R(x)\big]^2\,dx.$

Average value: $\displaystyle \bar f=\frac{1}{b-a}\int_a^b f(x)\,dx.$

Drag the line in the explorer: the amber region is the area enclosed between the parabola and the line, recomputed live between their intersection points.

⊙ Explorer · Area between two curvesinteractive

Slide the line c = 2

y = 4 − x² and y = x + 2

intersection—

enclosed area—

Worked example 7.A — Area between curves

Find the area between $y=x$ and $y=x^2$ on $[0,1]$ .

Here $x\geq x^2$ on $[0,1]$ , so integrate the difference:

\int_0^1 \big(x-x^2\big)\,dx=\Big[\tfrac{x^2}{2}-\tfrac{x^3}{3}\Big]_0^1=\tfrac12-\tfrac13=\tfrac16.

Worked example 7.B — Volume of revolution

The region under $y=\sqrt{x}$ on $[0,4]$ is revolved about the $x$ -axis. Find the volume.

Disk method with $R(x)=\sqrt{x}$ , so $R^2=x$ :

V=\pi\int_0^4 x\,dx=\pi\Big[\tfrac{x^2}{2}\Big]_0^4=8\pi.

▲ Geometric and physical applications

Definite integrals compute areas between curves, volumes of solids of revolution, arc lengths, and work done by variable forces.

Area between curves

\int_a^b[f(x)-g(x)]\,dx

Disk method

V=\pi\int_a^b[f(x)]^2\,dx

Shell method

V=2\pi\int_a^b x\,f(x)\,dx

Arc length

L=\int_a^b\sqrt{1+[f'(x)]^2}\,dx

Worked example — area

Find the area between $y=x^2$ and $y=x$ from $x=0$ to $x=1$ .

\int_0^1(x-x^2)\,dx = \left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1 = \frac{1}{6}.

Worked example — disk volume

Find the volume when $y=\sqrt{x}$ on $[0,4]$ is revolved about the $x$ -axis.

V=\pi\int_0^4 x\,dx = \pi\cdot 8 = 8\pi.

Region bounded between curves

y = x

and

y = x^2

[0, 1]

, where upper curve minus lower curve gives area

A = \frac{1}{6}

⚠ Common Traps & Exam Pitfalls

Top Minus Bottom Rule: When finding area between curves, always integrate

\int_a^b [y_{\text{top}} - y_{\text{bottom}}]\,dx

. If the curves intersect within the interval, you MUST split the integral at intersection points, or positive and negative areas will cancel each other out!

Washer Method Error: The washer formula for volume of revolution is

V = \pi \int [R(x)^2 - r(x)^2]\,dx

. It is NEVER

\pi \int [R(x) - r(x)]^2\,dx

(squaring the difference is wrong!).

Off-Axis Revolutions: When revolving around

y = k

instead of the

x

-axis, the radius is the distance to the axis of revolution:

R(x) = |f(x) - k|

Find the volume of the solid generated when the region bounded by

y = x^2

y = 0

, and

x = 2

is revolved about the

x

-axis.

Using the disk method with

R(x) = x^2

V = \pi \int_0^2 [R(x)]^2\,dx = \pi \int_0^2 x^4\,dx = \pi \left[\frac{x^5}{5}\right]_0^2 = \pi \left(\frac{32}{5} - 0\right) = \frac{32\pi}{5}.

📝 Chapter Quiz

Questions Level

Differential Equations

A differential equation relates a function to its own rate of change. Solving one means finding every function whose derivative behaves as prescribed — the language in which growth, cooling, and motion are written.

● Definition 8.1 — Differential equation

An equation involving an unknown function and its derivatives, such as $\dfrac{dy}{dx}=f(x,y)$ . A solution is a function $y(x)$ that satisfies it; an initial condition picks out one particular solution.

Separable equations. If $\dfrac{dy}{dx}=g(x)\,h(y)$ , gather each variable on its own side and integrate:

\int \frac{dy}{h(y)}=\int g(x)\,dx.

▲ Theorem 8.2 — Exponential model

The equation $\dfrac{dy}{dt}=ky$ has solution $y=y_0\,e^{kt}$ : exponential growth when $k>0$ and decay when $k<0$ .

A slope field draws the prescribed slope at a grid of points; solution curves simply flow along it. Pick an equation and slide the initial value — the amber curve threads through the field from $(0,\,y_0)$ .

⊙ Explorer · Slope field & solution curveinteractive

Equation

Initial value y(0) = 1

equationdy/dx = y

solution through (0, y₀)—

Worked example 8.A — Separable

Solve $\dfrac{dy}{dx}=2xy$ .

Separate and integrate: $\displaystyle\int\frac{dy}{y}=\int 2x\,dx$ , so $\ln|y|=x^2+C$ . Exponentiating,

y=C e^{\,x^2}.

Worked example 8.B — Initial value problem

Solve $\dfrac{dy}{dx}=3y$ with $y(0)=4$ .

By the exponential model $y=y_0 e^{kt}$ with $k=3$ , and $y(0)=4$ :

y=4e^{3x}.

▲ Ordinary differential equations

A differential equation relates a function to its derivatives. Separable ODEs are solved by separating variables and integrating both sides.

Separable ODE

\dfrac{dy}{dx}=g(x)h(y) \Rightarrow \int\dfrac{dy}{h(y)}=\int g(x)\,dx

Exponential growth

y'=ky \Rightarrow y=y_0 e^{kt}

Integrating factor

y'+P(x)y=Q(x)

\mu=e^{\int P\,dx}

Logistic model

\dfrac{dP}{dt}=kP\!\left(1-\dfrac{P}{M}\right)

Worked example

Solve $\dfrac{dy}{dx}=2xy$ , $y(0)=3$ .

\int\frac{dy}{y}=\int 2x\,dx \implies \ln|y|=x^2+C \implies y=3e^{x^2}.

Slope field for first-order differential equation

\frac{dy}{dx} = x - y

with particular solution curve passing through

(0, 1)

⚠ Common Traps & Exam Pitfalls

Separation of Variables Rules: Variables must separate completely into

g(y)\,dy = f(x)\,dx

. Equations with sums like

\frac{dy}{dx} = x + y

CANNOT be separated by subtraction! Treat

dy

and

dx

as algebraic differentials in numerators.

Position of the Constant + C: The constant

C

MUST be introduced at the exact moment of integration:

\ln|y| = kx + C

. Exponentiating gives

|y| = e^{kx+C} = e^C e^{kx} = A e^{kx}

. Adding

+ C

at the end to get

y = e^{kx} + C

is a fatal algebraic error.

Slope Field Zero Slopes: Points where

\frac{dy}{dx} = 0

produce horizontal tick marks. Points where the derivative is undefined (e.g. division by zero) produce no tick marks or vertical asymptotes.

Find the particular solution to

\dfrac{dy}{dx} = \dfrac{3x^2}{y}

given the initial condition

y(0) = 4

Separate variables:

y\,dy = 3x^2\,dx

.
Integrate both sides:

\int y\,dy = \int 3x^2\,dx \implies \frac{1}{2}y^2 = x^3 + C

.
Apply initial condition

y(0) = 4

\frac{1}{2}(4)^2 = 0^3 + C \implies 8 = C

.
Hence,

\frac{1}{2}y^2 = x^3 + 8 \implies y^2 = 2x^3 + 16

. Since

y(0) = 4 > 0

, we take the positive root:

y = \sqrt{2x^3 + 16}

📝 Chapter Quiz

Questions Level

Parametric & Polar

Not every curve is a function $y=f(x)$ . Parametric equations trace a path through time, and polar coordinates describe position by distance and angle — both unlock curves that loop, spiral, and cross themselves.

▲ Theorem 9.1 — Parametric calculus

For a curve $\big(x(t),\,y(t)\big)$ , the slope and arc length are

\frac{dy}{dx}=\frac{dy/dt}{dx/dt},\qquad L=\int_a^b\sqrt{\Big(\tfrac{dx}{dt}\Big)^2+\Big(\tfrac{dy}{dt}\Big)^2}\;dt.

▲ Theorem 9.2 — Polar coordinates & area

With $x=r\cos\theta,\ y=r\sin\theta$ , the area swept by a polar curve is

A=\frac12\int_\alpha^\beta r^2\,d\theta.

Drag the trace slider to watch each curve drawn by its parameter — the indigo portion is traced so far, the amber dot is the current point.

⊙ Explorer · Tracing parametric & polar curvesinteractive

Curve

Trace t = 42

current point—

Worked example 9.A — Parametric slope

For $x=t^3,\ y=t^2$ , find $\dfrac{dy}{dx}$ .

Differentiate each: $\dfrac{dx}{dt}=3t^2$ , $\dfrac{dy}{dt}=2t$ . Then

\frac{dy}{dx}=\frac{2t}{3t^2}=\frac{2}{3t}.

Worked example 9.B — Polar area

Find the area enclosed by the cardioid $r=1+\cos\theta$ .

Apply $A=\tfrac12\int_0^{2\pi} r^2\,d\theta$ with $r^2=1+2\cos\theta+\cos^2\theta$ :

A=\frac12\int_0^{2\pi}\big(1+2\cos\theta+\cos^2\theta\big)\,d\theta=\frac{3\pi}{2}.

▲ Parametric and polar calculus

Parametric curves express $x$ and $y$ in terms of a parameter $t$ . Polar coordinates use $(r,\theta)$ . Both require adapted formulas for slopes, arc lengths, and areas.

Parametric slope

\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}

Parametric arc length

L=\int_a^b\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt

Polar area

A=\dfrac{1}{2}\int_\alpha^\beta r^2\,d\theta

Polar to rect.

x=r\cos\theta,\; y=r\sin\theta

Worked example

Find $\dfrac{dy}{dx}$ for $x=t^2$ , $y=t^3$ .

\frac{dy}{dx}=\frac{3t^2}{2t}=\frac{3t}{2}.

Worked example — polar area

Area enclosed by $r=2\cos\theta$ , a circle of radius 1:

A=\frac{1}{2}\int_{-\pi/2}^{\pi/2}4\cos^2\theta\,d\theta = \pi.

Unit circle reference with terminal point

(\cos\theta, \sin\theta)

and angle

\theta = 45°

(

\frac{\pi}{4}

rad).

⚠ Common Traps & Exam Pitfalls

Second Derivative in Parametric: The second derivative is

\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(dy/dx)}{dx/dt}

. It is NEVER the quotient of second derivatives

\frac{d^2y/dt^2}{d^2x/dt^2}

! You must differentiate the first derivative with respect to

t

and divide again by

\frac{dx}{dt}

.

Missing 1/2 in Polar Area: Polar area is

A = \frac{1}{2} \int_\alpha^\beta r^2\,d\theta

. Forgetting the factor of

\frac{1}{2}

is the most frequent blunder on polar questions.

Polar Slope vs dr/dθ: In polar coordinates,

\frac{dr}{d\theta}

is the rate of change of radius with angle, NOT the slope of the curve! To find tangent slope

\frac{dy}{dx}

, convert via

x = r\cos\theta

y = r\sin\theta

and calculate

\frac{dy/d\theta}{dx/d\theta}

A particle moves along a curve defined parametrically by

x = t^2 + 1

and

y = t^3 - 3t

. Find the value of

\dfrac{d^2y}{dx^2}

t = 2

First find the components:

\frac{dx}{dt} = 2t

and

\frac{dy}{dt} = 3t^2 - 3

.
The first derivative is

\frac{dy}{dx} = \frac{3t^2 - 3}{2t} = \frac{3}{2}t - \frac{3}{2t}

.
Differentiating with respect to

t

\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{3}{2} + \frac{3}{2t^2}

.
Then

\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(dy/dx)}{dx/dt} = \frac{\frac{3}{2} + \frac{3}{2t^2}}{2t}

.
At

t = 2

: numerator is

\frac{3}{2} + \frac{3}{8} = \frac{15}{8}

; denominator is

2(2) = 4

.
Thus

\frac{d^2y}{dx^2} = \frac{15/8}{4} = \frac{15}{32}

📝 Chapter Quiz

Questions Level

Sequences & Series

A series adds infinitely many terms. Remarkably, many familiar functions can be written as “infinite polynomials” — and a short stretch of that polynomial already approximates the function beautifully near a point.

● Definition 10.1 — Series & convergence

An infinite series $\sum_{n=1}^{\infty} a_n$ converges if its partial sums $S_N=\sum_{n=1}^{N} a_n$ approach a finite limit as $N\to\infty$ ; otherwise it diverges.

▲ Theorem 10.2 — Geometric series

For $|r|<1$ , the geometric series converges:

\sum_{n=0}^{\infty} a\,r^{n}=\frac{a}{1-r}.

If $|r|\geq 1$ it diverges. The $p$ -series $\sum 1/n^{p}$ converges exactly when $p>1$ .

▲ Theorem 10.3 — Taylor & Maclaurin series

Near $x=0$ , a smooth function equals its Maclaurin series $\displaystyle f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^{n}$ . Three to know:

e^{x}=\sum \frac{x^{n}}{n!},\quad \sin x=\sum \frac{(-1)^{n}x^{2n+1}}{(2n+1)!},\quad \cos x=\sum \frac{(-1)^{n}x^{2n}}{(2n)!}.

The explorer builds the Maclaurin polynomial for $\sin x$ one term at a time. With a single term it is just the line $y=x$ ; add more and the amber curve wraps itself around $\sin x$ across an ever-wider interval.

⊙ Explorer · Taylor polynomials of sin xinteractive

Number of terms N = 2

functionsin x

approximationdegree 3

$N$ nonzero terms match $\sin x$ through degree $2N-1$ . Far from $0$ the polynomial eventually peels away.

Geometric series

\displaystyle\sum_{n=0}^\infty ar^n=\frac{a}{1-r},\;|r|<1

p-series test

\sum\frac{1}{n^p}

converges iff

p>1

Ratio test

L=\lim\left|\frac{a_{n+1}}{a_n}\right|

; converges if

L<1

, diverges if

L>1

Alternating series

Converges if

b_n\searrow 0

monotonically

Worked example 10.A — Infinite geometric series

Evaluate $\displaystyle\sum_{n=0}^{\infty}\Big(\tfrac{1}{2}\Big)^{n}$ .

Here $a=1,\ r=\tfrac12$ , and $|r|<1$ , so the series converges:

\frac{a}{1-r}=\frac{1}{1-\tfrac12}=2.

Worked example 10.B — Maclaurin approximation

Approximate $e^{0.1}$ with a third-degree Taylor polynomial.

Using $e^{x}\approx 1+x+\tfrac{x^2}{2}+\tfrac{x^3}{6}$ at $x=0.1$ :

1+0.1+0.005+0.000167\approx 1.1052.

Taylor polynomial approximations of

\sin x

: linear

T_1(x) = x

, cubic

T_3(x) = x - \frac{x^3}{6}

, and quintic

T_5(x) = x - \frac{x^3}{6} + \frac{x^5}{120}

⚠ Common Traps & Exam Pitfalls

The Divergence Test Fallacy: If

\lim_{n\to\infty} a_n = 0

, the series MAY converge or diverge. For example, the harmonic series

\sum \frac{1}{n}

diverges even though its terms shrink to zero! The nth-term test can ONLY establish divergence (when

\lim a_n \ne 0

), never convergence.

Taylor Series Center: If a Taylor series is centered at

x = a

, every term must be a power of

(x - a)

, not

x

P_n(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x - a)^k

.

Alternating Series Remainder Bound: For an alternating series satisfying Leibniz's conditions, the error in approximating the infinite sum by the

N

-th partial sum is strictly bounded by the absolute value of the first omitted term:

|S - S_N| \le a_{N+1}

Find the third-degree Maclaurin polynomial

P_3(x)

for

f(x) = \cos x

Compute derivatives at

x = 0

:
•

f(0) = \cos(0) = 1

•

f'(0) = -\sin(0) = 0

•

f''(0) = -\cos(0) = -1

•

f'''(0) = \sin(0) = 0

Applying the Maclaurin formula:

P_3(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 = 1 - \frac{x^2}{2}

. (Notice the cubic term has coefficient 0).

📝 Chapter Quiz

Questions Level

Convergence Tests & Power Series

A series $\sum a_n$ converges when its partial sums $S_N=\sum_{n=1}^{N}a_n$ approach a finite limit. Since we rarely sum infinitely many terms by hand, a battery of tests decides convergence from the shape of $a_n$ alone.

▲ Theorem 11.1 — The nth-term test

If $\displaystyle\lim_{n\to\infty}a_n\neq 0$ , then $\sum a_n$ diverges. Caution: $a_n\to 0$ does not guarantee convergence — the harmonic series $\sum \tfrac1n$ diverges even though its terms vanish.

▲ Theorem 11.2 — p-series & comparison

The p-series $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ converges if and only if $p\gt 1$ . For positive terms, if $0\le a_n\le b_n$ and $\sum b_n$ converges, so does $\sum a_n$ (and the contrapositive for divergence).

⊙ Explorer · Partial sums approaching a limitinteractive

terms N: 6

partial sum S_N0.9844

limit1.0000

remainder0.0156

The geometric series $\sum (\tfrac12)^n$ sums to $1$ . Each new term halves the remaining gap — visible convergence.

The ratio and root tests

For series with factorials or exponentials, compare a term to the one before it. Let $\displaystyle L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$ . Then the series converges absolutely if $L\lt 1$ , diverges if $L\gt 1$ , and the test is inconclusive if $L=1$ .

▲ Theorem 11.3 — Power series & radius of convergence

A power series $\displaystyle\sum_{n=0}^{\infty}c_n(x-a)^{n}$ converges for all $x$ with $|x-a|\lt R$ and diverges for $|x-a|\gt R$ , where $R$ is the radius of convergence. The endpoints $x=a\pm R$ must be tested separately.

Geometric

converges iff

|r|\lt 1

p-series

converges iff

p\gt 1

Ratio

L\lt 1\Rightarrow

converge

Alternating

terms

\downarrow 0 \Rightarrow

converge

Worked example 11.A — Ratio test

Test $\displaystyle\sum_{n=1}^{\infty}\frac{n}{2^{n}}$ for convergence.

Form the ratio of successive terms and take the limit:

L=\lim_{n\to\infty}\frac{n+1}{2^{n+1}}\cdot\frac{2^{n}}{n}=\lim_{n\to\infty}\frac{n+1}{2n}=\frac12\lt 1,

so the series converges absolutely.

Worked example 11.B — Radius of convergence

Find the radius of convergence of $\displaystyle\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$ .

The ratio of terms is $\left|\dfrac{x^{n+1}/(n+1)!}{x^{n}/n!}\right|=\dfrac{|x|}{n+1}\to 0$ for every $x$ . Since $L=0\lt 1$ always, $R=\infty$ — this series (which is $e^{x}$ ) converges everywhere.

Bar chart comparing term values

a_n = \frac{n}{2^n}

for

n = 1, \dots, 6

, displaying rapid geometric decay toward 0 that drives convergence.

⚠ Common Traps & Exam Pitfalls

Ratio Test Inconclusiveness (L = 1): If

L = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = 1

, the Ratio Test gives NO conclusion whatsoever. In particular, the Ratio Test ALWAYS gives

L = 1

for any p-series or rational fraction in

n

; use the Comparison Test or Integral Test instead!

Testing Endpoints for Interval of Convergence: Finding the radius of convergence

R

determines an open interval

(a - R, a + R)

. You MUST test each endpoint

x = a - R

and

x = a + R

separately, as convergence at boundaries varies (e.g. one end might be conditionally convergent via Alternating Series while the other diverges).

Absolute vs Conditional Convergence: A series

\sum a_n

is conditionally convergent if

\sum a_n

converges, but

\sum |a_n|

diverges (classic example:

\sum \frac{(-1)^{n+1}}{n}

Find the radius of convergence

R

of the power series

\displaystyle\sum_{n=1}^{\infty} \frac{(x - 3)^n}{n \cdot 5^n}

Apply the Ratio Test with

a_n = \frac{(x - 3)^n}{n \cdot 5^n}

L = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty} \left|\frac{(x - 3)^{n+1}}{(n+1)5^{n+1}} \cdot \frac{n \cdot 5^n}{(x - 3)^n}\right| = \lim_{n\to\infty} \frac{n}{n+1} \cdot \frac{|x - 3|}{5} = \frac{|x - 3|}{5}.

For absolute convergence, require

L < 1 \implies \frac{|x - 3|}{5} < 1 \implies |x - 3| < 5

. Thus the radius of convergence is

R = 5

📝 Chapter Quiz

Questions Level

Vectors & Vector-Valued Functions

Multivariable calculus begins with vectors — quantities carrying both magnitude and direction. Letting a vector depend on a parameter traces a curve through space, and differentiating it recovers velocity and acceleration.

● Definition 12.1 — Vector and magnitude

A vector $\mathbf{v}=\langle v_1,v_2,v_3\rangle$ is an ordered triple with length $\displaystyle|\mathbf{v}|=\sqrt{v_1^{2}+v_2^{2}+v_3^{2}}$ . Vectors add componentwise, and a unit vector is $\hat{\mathbf{v}}=\mathbf{v}/|\mathbf{v}|$ .

▲ Theorem 12.2 — Dot and cross products

The dot product $\mathbf{u}\cdot\mathbf{v}=|\mathbf{u}|\,|\mathbf{v}|\cos\theta$ measures alignment (zero when perpendicular). The cross product $\mathbf{u}\times\mathbf{v}$ has length $|\mathbf{u}|\,|\mathbf{v}|\sin\theta$ and points perpendicular to both.

⊙ Explorer · Vector additioninteractive

angle of v: 50

v components⟨1.67, 1.99⟩

|u + v|5.10

resultant angle35.5°

Blue is $\mathbf u=\langle3,1\rangle$ , amber is $\mathbf v$ , dark is the resultant $\mathbf u+\mathbf v$ — the parallelogram diagonal.

helix parameter t (Example 12.B): 0

r(t)⟨1.000, 0.000, 0.000⟩

|r′(t)| (speed)1.414

Vector-valued functions

A path in space is $\mathbf{r}(t)=\langle x(t),y(t),z(t)\rangle$ . Differentiating component-by-component gives the velocity $\mathbf{r}'(t)$ , whose length $|\mathbf{r}'(t)|$ is the speed; differentiating again gives acceleration.

Dot product

\mathbf{u}\cdot\mathbf{v}=u_1v_1+u_2v_2+u_3v_3

Unit vector

\hat{\mathbf{v}}=\mathbf{v}/|\mathbf{v}|

Angle

\cos\theta=\dfrac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}|\,|\mathbf{v}|}

Velocity

\mathbf{r}'(t)=\langle x',y',z'\rangle

Worked example 12.A — Angle between vectors

Find the angle between $\mathbf{u}=\langle 1,2,2\rangle$ and $\mathbf{v}=\langle 2,0,1\rangle$ .

Here $\mathbf{u}\cdot\mathbf{v}=2+0+2=4$ , $|\mathbf{u}|=3$ , $|\mathbf{v}|=\sqrt5$ , so

\cos\theta=\frac{4}{3\sqrt5}\approx 0.596 \quad\Rightarrow\quad \theta\approx 53.4^{\circ}.

Worked example 12.B — Velocity and speed

For the helix $\mathbf{r}(t)=\langle\cos t,\sin t,t\rangle$ , find the velocity and speed.

Differentiating each component, $\mathbf{r}'(t)=\langle-\sin t,\cos t,1\rangle$ . Its length is constant:

|\mathbf{r}'(t)|=\sqrt{\sin^{2}t+\cos^{2}t+1}=\sqrt2.

Right-angled vector component decomposition in 2D space, demonstrating

|\vec v| = \sqrt{v_x^2 + v_y^2}

and orthogonal directions.

⚠ Common Traps & Exam Pitfalls

Dot Product vs Cross Product Output: The dot product

\vec u \cdot \vec v = u_1v_1 + u_2v_2 + u_3v_3

produces a SCALAR (number). The cross product

\vec u \times \vec v

produces a VECTOR perpendicular to both original vectors.

Cross Product Anti-Commutativity: Order of factors reverses the sign:

\vec u \times \vec v = -(\vec v \times \vec u)

. Swapping vector order flips the resulting direction (Right-Hand Rule).

Unit Tangent Vector Normalization: The unit tangent vector

\vec T(t) = \frac{\vec r'(t)}{|\vec r'(t)|}

requires dividing velocity by speed. Forgetting to divide by magnitude leaves a velocity vector with non-unit length.

Determine whether the vectors

\vec u = \langle 2, -3, 4\rangle

and

\vec v = \langle 6, 8, 3\rangle

are orthogonal, parallel, or neither.

Compute the dot product:

\vec u \cdot \vec v = (2)(6) + (-3)(8) + (4)(3) = 12 - 24 + 12 = 0.

Since the dot product is zero and neither vector is the zero vector,

\vec u

and

\vec v

are strictly orthogonal (perpendicular).

📝 Chapter Quiz

Questions Level

Partial Derivatives & the Gradient

For a surface $z=f(x,y)$ , there is no single slope — it depends on the direction you walk. Partial derivatives measure the slope along each axis, and the gradient bundles them into the direction of steepest ascent.

● Definition 13.1 — Partial derivative

The partial derivative $\displaystyle f_x=\frac{\partial f}{\partial x}$ differentiates $f$ with respect to $x$ while holding $y$ constant; $f_y$ does the reverse. Geometrically, $f_x$ is the slope of the curve where the surface is sliced by a plane of fixed $y$ .

▲ Theorem 13.2 — Gradient & directional derivative

The gradient $\nabla f=\langle f_x,f_y\rangle$ points in the direction of greatest increase, with rate $|\nabla f|$ . The slope in a unit direction $\mathbf{u}$ is the directional derivative $D_{\mathbf{u}}f=\nabla f\cdot\mathbf{u}$ .

⊙ Explorer · Slicing a surfaceinteractive

slice at y = c: 0

f(1, c)2.800

f_x(1, c)-0.400

Fixing $y=c$ gives the curve $g(x)=f(x,c)$ . Its slope at $x=1$ is the partial $f_x(1,c)$ — and it shifts as $c$ moves.

Tangent planes and local extrema

Near a point, a smooth surface looks flat. The tangent plane at $(a,b)$ is

z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b).

Local maxima and minima occur where $\nabla f=\mathbf{0}$ ; the second-derivative test uses $D=f_{xx}f_{yy}-f_{xy}^{2}$ to classify them ( $D\gt 0$ and $f_{xx}\gt 0$ gives a minimum).

Partial

\dfrac{\partial f}{\partial x}

Gradient

\nabla f=\langle f_x,f_y\rangle

Directional

D_{\mathbf{u}}f=\nabla f\cdot\mathbf{u}

Second test

D=f_{xx}f_{yy}-f_{xy}^{2}

Worked example 13.A — Computing partials

For $f(x,y)=x^{2}y+\sin(xy)$ , find $f_x$ and $f_y$ .

Treating $y$ as constant for $f_x$ , then $x$ as constant for $f_y$ :

f_x=2xy+y\cos(xy), \qquad f_y=x^{2}+x\cos(xy).

Worked example 13.B — Gradient and steepest ascent

For $f(x,y)=x^{2}+y^{2}$ , find the gradient at $(1,2)$ and the direction of fastest increase.

Since $\nabla f=\langle 2x,2y\rangle$ , at $(1,2)$ we get $\nabla f=\langle 2,4\rangle$ . Fastest increase is along this vector, i.e. the unit direction

\hat{\mathbf{u}}=\frac{1}{\sqrt{20}}\langle 2,4\rangle, \qquad \text{rate } |\nabla f|=\sqrt{20}=2\sqrt5.

Level curves of paraboloid

z = x^2 + y^2

showing concentric circular contours and gradient vector

\nabla f = \langle 2, 4\rangle

orthogonal to the contour at

(1, 2)

⚠ Common Traps & Exam Pitfalls

Directional Derivative Requires a Unit Vector: The formula is

D_{\vec u}f = \nabla f \cdot \hat u

. If the problem gives an arbitrary direction

\vec v

, you MUST divide by its magnitude

|\vec v|

first. Forgetting to normalize scales the directional derivative by

|\vec v|

and produces an incorrect rate.

Gradient Geometric Meaning:

\nabla f(a, b)

points in the direction of maximum rate of increase, and its magnitude

|\nabla f|

is that maximum rate. Crucially, the gradient vector is ALWAYS orthogonal (perpendicular) to the level curve

f(x, y) = c

at that point.

Second Derivative Test Discriminant:

D = f_{xx}f_{yy} - (f_{xy})^2

. If

D < 0

, the point is a saddle point regardless of the sign of

f_{xx}

Find the directional derivative of

f(x, y) = 3x^2y - y^3

at the point

(1, 2)

in the direction of

\vec v = \langle 3, 4\rangle

First compute the gradient:

\nabla f = \langle 6xy, 3x^2 - 3y^2\rangle

.
At

(1, 2)

\nabla f(1, 2) = \langle 6(1)(2), 3(1)^2 - 3(2)^2\rangle = \langle 12, 3 - 12\rangle = \langle 12, -9\rangle

.
Normalize the direction vector

\vec v

|\vec v| = \sqrt{3^2 + 4^2} = 5

, so

\hat u = \langle \frac{3}{5}, \frac{4}{5}\rangle

.
Compute the dot product:

D_{\vec u}f(1, 2) = \nabla f \cdot \hat u = 12\left(\frac{3}{5}\right) + (-9)\left(\frac{4}{5}\right) = \frac{36 - 36}{5} = 0.

📝 Chapter Quiz

Questions Level

Multiple Integrals

A single integral accumulates over an interval; a double integral accumulates a function over a two-dimensional region — most concretely, the volume trapped under a surface $z=f(x,y)$ .

● Definition 14.1 — The double integral

Partition a region $R$ into small cells of area $\Delta A$ , sample $f$ in each, and sum. The double integral is the limit as the cells shrink:

\iint_{R} f\,dA=\lim_{\Delta A\to 0}\sum_{i,j} f(x_i,y_j)\,\Delta A.

▲ Theorem 14.2 — Fubini's theorem

Over a rectangle $R=[a,b]\times[c,d]$ , a double integral equals an iterated integral, and the order may be swapped:

\iint_{R} f\,dA=\int_a^b\!\!\int_c^d f(x,y)\,dy\,dx=\int_c^d\!\!\int_a^b f(x,y)\,dx\,dy.

⊙ Explorer · Double Riemann suminteractive

grid n × n: 4

cells16

approx ∬ f dA8.000

exact value8

Each cell is shaded by $f=x+y$ at its midpoint over $[0,2]^2$ . Refine the grid and the midpoint sum converges to $8$ .

Double integrals in polar coordinates

For circular regions, polar coordinates simplify everything — but the area element gains a factor of $r$ :

\iint_{R} f\,dA=\iint f(r\cos\theta,\,r\sin\theta)\,r\,dr\,d\theta.

That extra $r$ is the Jacobian of the polar change of variables, and forgetting it is the most common error in the topic.

Double integral

\iint_R f\,dA

Iterated

\int\!\!\int f\,dy\,dx

Area

\iint_R dA

Polar

r\,dr\,d\theta

Worked example 14.A — An iterated integral

Evaluate $\displaystyle\int_0^1\!\!\int_0^2 (x+2y)\,dx\,dy$ .

Integrate in $x$ first, treating $y$ as constant:

\int_0^1\Big[\tfrac{x^{2}}{2}+2xy\Big]_0^2 dy=\int_0^1(2+4y)\,dy=\big[2y+2y^{2}\big]_0^1=4.

Worked example 14.B — Area of a disk in polar

Confirm the area of the unit disk $x^{2}+y^{2}\le 1$ using a polar double integral.

\iint_{R} dA=\int_0^{2\pi}\!\!\int_0^1 r\,dr\,d\theta=\int_0^{2\pi}\tfrac12\,d\theta=\pi.

Rectangular region of integration

R = [0, 2] \times [0, 1]

in the

xy

-plane for double integral evaluation.

⚠ Common Traps & Exam Pitfalls

Polar Area Element (Jacobian r): In polar coordinates,

dA = r\,dr\,d\theta

. Omitting the extra factor of

r

is the single most common error in multivariable calculus!

Reversing Order of Integration: You CANNOT merely swap the integral signs and limits when the region is non-rectangular:

\int_0^1 \int_x^1 f(x,y)\,dy\,dx \ne \int_x^1 \int_0^1 f(x,y)\,dx\,dy

. You MUST sketch the 2D domain and reconstruct the bounding functions from horizontal vs vertical slices.

Coordinate Transformations Jacobian: When substituting

x = g(u, v)

y = h(u, v)

, always multiply the integrand by the absolute value of the Jacobian determinant

|J(u, v)| = |x_u y_v - x_v y_u|

Evaluate

\displaystyle\iint_R (2x + y)\,dA

where

R

is the rectangular region

[0, 3] \times [1, 2]

Set up iterated integral:

\int_1^2 \int_0^3 (2x + y)\,dx\,dy = \int_1^2 \left[ x^2 + xy \right]_0^3 dy = \int_1^2 (9 + 3y)\,dy

Integrate with respect to

y

\left[ 9y + \frac{3y^2}{2} \right]_1^2 = \left(18 + \frac{12}{2}\right) - \left(9 + \frac{3}{2}\right) = 24 - 10.5 = 13.5 = \frac{27}{2}.

📝 Chapter Quiz

Questions Level

Vector Calculus

The capstone of Calc III integrates vector fields — assignments of a vector to every point — along curves and across surfaces. Three great theorems (Green, Stokes, Divergence) tie these integrals back to the derivatives $\nabla\cdot\mathbf{F}$ and $\nabla\times\mathbf{F}$ .

● Definition 15.1 — Line integral & work

The line integral $\displaystyle\int_C \mathbf{F}\cdot d\mathbf{r}$ measures the work a field $\mathbf{F}$ does along a curve $C$ . A field is conservative when $\mathbf{F}=\nabla f$ for some $f$ ; then the integral is path-independent and equals $f(\text{end})-f(\text{start})$ .

▲ Theorem 15.2 — Green's theorem

For a positively oriented simple closed curve $C$ bounding a region $R$ in the plane,

\oint_C \big(P\,dx+Q\,dy\big)=\iint_{R}\Big(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\Big)\,dA.

⊙ Explorer · Vector field — divergence & curlinteractive

blend t (source → rotation): 100

divergence ∇·F0.00

curl ∇×F2.00

field typerotational

At $t=1$ the field is $\langle -y,x\rangle$ (pure rotation, curl $2$ ); at $t=0$ it is $\langle x,y\rangle$ (pure source, divergence $2$ ).

Stokes' and the Divergence theorems

Green's theorem generalizes in two directions. Stokes' theorem relates circulation around a boundary curve to the curl over a surface; the Divergence theorem relates outward flux through a closed surface to the divergence inside:

\oint_C \mathbf{F}\cdot d\mathbf{r}=\iint_S (\nabla\times\mathbf{F})\cdot d\mathbf{S}, \qquad \iint_S \mathbf{F}\cdot d\mathbf{S}=\iiint_E (\nabla\cdot\mathbf{F})\,dV.

Line integral

\int_C \mathbf{F}\cdot d\mathbf{r}

Green

\oint=\iint(\,Q_x-P_y)

Divergence

\nabla\cdot\mathbf{F}

Curl

\nabla\times\mathbf{F}

Worked example 15.A — Green's theorem

Evaluate $\displaystyle\oint_C (-y\,dx+x\,dy)$ around the unit circle, counterclockwise.

Here $P=-y,\ Q=x$ , so $Q_x-P_y=1-(-1)=2$ . By Green's theorem,

\oint_C(-y\,dx+x\,dy)=\iint_{R}2\,dA=2(\text{area of disk})=2\pi.

Worked example 15.B — The Divergence theorem

Find the outward flux of $\mathbf{F}=\langle x,y,z\rangle$ through the unit sphere.

Since $\nabla\cdot\mathbf{F}=1+1+1=3$ , the Divergence theorem turns the surface flux into a volume integral over the unit ball $E$ :

\iint_S \mathbf{F}\cdot d\mathbf{S}=\iiint_E 3\,dV=3\cdot\tfrac{4}{3}\pi=4\pi.

Positively oriented (counterclockwise) closed boundary curve

C

enclosing planar region

R

for Green's Theorem.

⚠ Common Traps & Exam Pitfalls

Green's Theorem Orientation: Green's Theorem

\oint_C (P\,dx + Q\,dy) = \iint_R (Q_x - P_y)\,dA

strictly requires a positively oriented (counterclockwise) boundary curve

C

keeping the region on the left. If traversing clockwise, you MUST insert a negative sign!

Conservative Vector Fields Checklist: A vector field

\vec F = \nabla f

has

\text{curl}(\vec F) = \vec 0

. On simply connected domains: (1) Path independence

\int_{C_1} = \int_{C_2}

; (2) Zero work around any closed loop:

\oint_C \vec F \cdot d\vec r = 0

; (3) Fundamental Theorem for Line Integrals:

\int_C \nabla f \cdot d\vec r = f(B) - f(A)

.

Divergence vs Curl: Divergence

\nabla \cdot \vec F

yields a scalar measuring net outflow; Curl

\nabla \times \vec F

yields a vector measuring rotational circulation.

Use Green's Theorem to evaluate

\displaystyle\oint_C (3y\,dx + 5x\,dy)

where

C

is the counterclockwise circle

x^2 + y^2 = 4

Identify components:

P = 3y

and

Q = 5x

.
Compute partial derivatives:

\frac{\partial Q}{\partial x} = 5

and

\frac{\partial P}{\partial y} = 3

.
Apply Green's Theorem:

\oint_C (3y\,dx + 5x\,dy) = \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)\,dA = \iint_R (5 - 3)\,dA = 2 \iint_R dA.

The region

R

is a disk of radius

r = 2

, so its area is

\pi(2^2) = 4\pi

.
Therefore,

2 \iint_R dA = 2(4\pi) = 8\pi

📝 Chapter Quiz

Questions Level

Practice set

Fifty mixed problems spanning all ten chapters, ordered roughly from limits to series and tagged by difficulty. Click any problem to reveal a complete worked solution.

Evaluate $\displaystyle\lim_{x\to 2}\frac{x^2-4}{x-2}$ .

Basic

Factor: $x^2-4=(x-2)(x+2)$ . Cancel $(x-2)$ :

\lim_{x\to 2}(x+2)=4.

Differentiate $f(x)=5x^4-3x^2+7$ .

Basic

Apply the power and sum rules term by term:

f'(x)=20x^3-6x.

Differentiate $y=\dfrac{x}{x^2+1}$ .

Intermediate

Quotient rule with $f=x,\ g=x^2+1$ , so $f'=1,\ g'=2x$ :

y'=\frac{(1)(x^2+1)-(x)(2x)}{(x^2+1)^2}=\frac{1-x^2}{(x^2+1)^2}.

Differentiate $y=e^{2x}\cos x$ .

Intermediate

Product rule; the first factor needs the chain rule, $\tfrac{d}{dx}e^{2x}=2e^{2x}$ :

y'=2e^{2x}\cos x - e^{2x}\sin x = e^{2x}\big(2\cos x-\sin x\big).

Evaluate $\displaystyle\int_1^{3}\big(2x+1\big)\,dx$ .

Basic

Antiderivative $x^2+x$ . By the Fundamental Theorem:

\big[x^2+x\big]_1^{3}=(9+3)-(1+1)=12-2=10.

Find the area under $f(x)=\sin x$ over $\big[0,\tfrac{\pi}{2}\big]$ .

Intermediate

Use $\int \sin x\,dx=-\cos x$ :

\int_0^{\pi/2}\sin x\,dx=\big[-\cos x\big]_0^{\pi/2}=0-(-1)=1.

Differentiate $y=\big(3x^2+1\big)^{5}$ .

Advanced

Chain rule: outer power $u^5$ gives $5u^4$ ; inner derivative $\tfrac{d}{dx}(3x^2+1)=6x$ :

y'=5\big(3x^2+1\big)^{4}\cdot 6x=30x\big(3x^2+1\big)^{4}.

A particle's position is $s(t)=t^3-6t^2+9t$ . Find the times when it is momentarily at rest.

Advanced

Velocity is the derivative; rest means $v(t)=0$ :

v(t)=s'(t)=3t^2-12t+9=3(t-1)(t-3).

So the particle is at rest at $t=1$ and $t=3$ .

Find and classify the local extrema of $f(x)=2x^3-9x^2+12x$ .

Advanced

$f'(x)=6x^2-18x+12=6(x-1)(x-2)$ , so critical points at $x=1,2$ . With $f''(x)=12x-18$ :

f''(1)=-6<0\ (\text{max}),\quad f''(2)=6>0\ (\text{min}).

Local maximum $(1,5)$ , local minimum $(2,4)$ .

P10

Evaluate $\displaystyle\int x\sin\!\big(x^2\big)\,dx$ .

Intermediate

Let $u=x^2$ , so $du=2x\,dx$ and $x\,dx=\tfrac12\,du$ :

\int x\sin\!\big(x^2\big)\,dx=\tfrac12\int\sin u\,du=-\tfrac12\cos\!\big(x^2\big)+C.

P11

Find the area enclosed between $y=4-x^2$ and $y=x+2$ .

Advanced

Intersections: $4-x^2=x+2\Rightarrow x^2+x-2=0\Rightarrow x=-2,1$ . The parabola is on top, so

\int_{-2}^{1}\big[(4-x^2)-(x+2)\big]\,dx=\int_{-2}^{1}\big(2-x-x^2\big)\,dx=\frac{9}{2}.

P12

Evaluate $\displaystyle\sum_{n=1}^{\infty} 3\Big(\tfrac{1}{4}\Big)^{n}$ .

Intermediate

The first term (at $n=1$ ) is $a=\tfrac34$ and the ratio is $r=\tfrac14$ :

\sum_{n=1}^{\infty} 3\Big(\tfrac14\Big)^{n}=\frac{a}{1-r}=\frac{3/4}{3/4}=1.

P13

Solve the separable equation $\dfrac{dy}{dx}=\dfrac{x}{y}$ .

Intermediate

Separate: $y\,dy=x\,dx$ . Integrate both sides:

\tfrac{y^2}{2}=\tfrac{x^2}{2}+C\ \Longrightarrow\ y^2-x^2=C.

P14

For $x=t^2,\ y=t^3$ , find $\dfrac{dy}{dx}$ .

Intermediate

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

P15

Evaluate $\displaystyle\lim_{x\to 0}\frac{1-\cos x}{x^2}$ .

Advanced

Form $\tfrac00$ ; apply L'Hôpital twice:

\lim_{x\to 0}\frac{1-\cos x}{x^2}=\lim_{x\to 0}\frac{\sin x}{2x}=\lim_{x\to 0}\frac{\cos x}{2}=\frac12.

P16

Evaluate $\displaystyle\lim_{x\to 2}\big(3x^2-x+1\big)$ .

Basic

The polynomial is continuous, so substitute directly:

3(2)^2-2+1=11.

P17

Evaluate $\displaystyle\lim_{x\to 0}\frac{\sin 3x}{x}$ .

Basic

Use $\lim_{u\to 0}\tfrac{\sin u}{u}=1$ :

\lim_{x\to 0}\frac{\sin 3x}{x}=3\lim_{x\to 0}\frac{\sin 3x}{3x}=3.

P18

Evaluate $\displaystyle\lim_{x\to\infty}\frac{2x^2-3x+1}{5x^2+x}$ .

Intermediate

Divide by $x^2$ ; lower-order terms vanish, leaving the ratio of leading coefficients:

\frac{2}{5}.

P19

Evaluate $\displaystyle\lim_{x\to 0}\frac{\sqrt{x+4}-2}{x}$ .

Intermediate

Multiply by the conjugate $\sqrt{x+4}+2$ :

\lim_{x\to 0}\frac{(x+4)-4}{x\big(\sqrt{x+4}+2\big)}=\lim_{x\to 0}\frac{1}{\sqrt{x+4}+2}=\frac14.

P20

Find $k$ so that $f(x)=\begin{cases} x^2,&x\le 1\\ kx,&x>1\end{cases}$ is continuous at $x=1$ .

Intermediate

The two pieces must agree at $x=1$ : $1^2=k(1)$ .

k=1.

P21

Differentiate $f(x)=7x^3-4x+9$ .

Basic

f'(x)=21x^2-4.

P22

Differentiate $y=\cos 5x$ .

Basic

Chain rule with inner $5x$ :

y'=-5\sin 5x.

P23

Differentiate $y=e^{2x}$ .

Basic

y'=2e^{2x}.

P24

Differentiate $y=\ln(3x)$ .

Basic

Chain rule: $\tfrac{1}{3x}\cdot 3$ .

y'=\frac1x.

P25

Differentiate $y=x^3\ln x$ .

Intermediate

Product rule: $3x^2\ln x + x^3\cdot\tfrac1x$ .

y'=x^2\big(3\ln x+1\big).

P26

Differentiate $y=\dfrac{x}{x+1}$ .

Intermediate

Quotient rule:

y'=\frac{(1)(x+1)-x(1)}{(x+1)^2}=\frac{1}{(x+1)^2}.

P27

Differentiate $y=\sqrt{x^2+1}$ .

Intermediate

Chain rule on $(x^2+1)^{1/2}$ :

y'=\frac{x}{\sqrt{x^2+1}}.

P28

Differentiate $y=\tan 3x$ .

Intermediate

y'=3\sec^2 3x.

P29

Find $\dfrac{dy}{dx}$ if $x^2+y^2=25$ .

Advanced

Differentiate implicitly: $2x+2y\,y'=0$ .

\frac{dy}{dx}=-\frac{x}{y}.

P30

Find $\dfrac{dy}{dx}$ if $xy+y^2=6$ .

Advanced

Differentiate implicitly (product rule on $xy$ ): $y+x\,y'+2y\,y'=0$ .

\frac{dy}{dx}=-\frac{y}{x+2y}.

P31

Find the tangent line to $y=x^2$ at $x=3$ .

Intermediate

Point $(3,9)$ ; slope $y'=2x=6$ . Then $y-9=6(x-3)$ :

y=6x-9.

P32

On what intervals is $f(x)=x^3-3x$ increasing?

Intermediate

$f'(x)=3x^2-3>0 \Rightarrow x^2>1$ :

x<-1 \ \text{or}\ x>1.

P33

A rectangular field is enclosed with 60 m of fencing on all four sides. What dimensions maximise the area?

Advanced

With $l+w=30$ , area $A=l(30-l)$ , so $A'(l)=30-2l=0\Rightarrow l=15$ .

15\times 15,\qquad A_{\max}=225\ \text{m}^2.

P34

A 10 m ladder leans on a wall; its base slides away at 1 m/s. How fast is the top sliding down when the base is 6 m from the wall?

Advanced

From $x^2+y^2=100$ , at $x=6$ we get $y=8$ . Differentiate: $2x\,x'+2y\,y'=0$ :

y'=-\frac{x\,x'}{y}=-\frac{6(1)}{8}=-0.75\ \text{m/s}.

The top slides down at $0.75$ m/s.

P35

Evaluate $\displaystyle\lim_{x\to 0}\frac{e^{x}-1}{x}$ .

Intermediate

Form $\tfrac00$ ; by L'Hôpital, $\lim e^{x}/1$ :

=1.

P36

Evaluate $\displaystyle\int \big(6x^2-4x+3\big)\,dx$ .

Basic

2x^3-2x^2+3x+C.

P37

Evaluate $\displaystyle\int_0^{1} e^{x}\,dx$ .

Basic

\big[e^{x}\big]_0^{1}=e-1.

P38

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

Basic

\big[\ln x\big]_1^{e}=\ln e-\ln 1=1.

P39

Evaluate $\displaystyle\int_0^{\pi/2}\cos x\,dx$ .

Intermediate

\big[\sin x\big]_0^{\pi/2}=1-0=1.

P40

Evaluate $\displaystyle\int_0^{4}\sqrt{x}\,dx$ .

Intermediate

Write $\sqrt{x}=x^{1/2}$ :

\Big[\tfrac{2}{3}x^{3/2}\Big]_0^{4}=\tfrac{2}{3}(8)=\frac{16}{3}.

P41

Evaluate $\displaystyle\int 2x\,e^{x^2}\,dx$ .

Intermediate

Let $u=x^2$ , $du=2x\,dx$ :

\int e^{u}\,du=e^{x^2}+C.

P42

Evaluate $\displaystyle\int x\cos x\,dx$ .

Intermediate

By parts with $u=x,\ dv=\cos x\,dx$ :

x\sin x-\int\sin x\,dx=x\sin x+\cos x+C.

P43

Evaluate $\displaystyle\int \ln x\,dx$ .

Advanced

By parts with $u=\ln x,\ dv=dx$ :

x\ln x-\int x\cdot\tfrac1x\,dx=x\ln x-x+C.

P44

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

Intermediate

The numerator is the derivative of the denominator, so let $u=x^2+x$ :

\int\frac{du}{u}=\ln\big|x^2+x\big|+C.

P45

Find the area under $y=4-x^2$ from $x=-2$ to $x=2$ .

Intermediate

\int_{-2}^{2}\big(4-x^2\big)\,dx=\Big[4x-\tfrac{x^3}{3}\Big]_{-2}^{2}=\frac{32}{3}.

P46

The region under $y=x$ from $0$ to $2$ is revolved about the $x$ -axis. Find the volume.

Advanced

Disk method with $R(x)=x$ :

\pi\int_0^{2}x^2\,dx=\pi\Big[\tfrac{x^3}{3}\Big]_0^{2}=\frac{8\pi}{3}.

P47

Find the average value of $f(x)=x^2$ on $[0,3]$ .

Intermediate

\bar f=\frac{1}{3}\int_0^{3}x^2\,dx=\frac{1}{3}\cdot 9=3.

P48

Solve $\dfrac{dy}{dx}=2y$ with $y(0)=5$ .

Intermediate

Exponential model $y=y_0e^{kt}$ with $k=2$ , $y_0=5$ :

y=5e^{2x}.

P49

For $x=t^2,\ y=2t$ , find $\dfrac{dy}{dx}$ .

Intermediate

\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2}{2t}=\frac{1}{t}.

P50

Evaluate $\displaystyle\sum_{n=0}^{\infty}\Big(\tfrac{2}{3}\Big)^{n}$ .

Intermediate

Geometric with $a=1,\ r=\tfrac23$ and $|r|<1$ :

\frac{1}{1-\tfrac23}=3.

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Calculus notes & practice

The full chapters above plus every worked solution, formatted as a clean print document.

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Problem-set generator

Randomised practice sheets with answer keys, ready to print per class.

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Official · Past Papers

Past Papers & Official Tests

Officially released exams and mark schemes from the examining body.

AP Calculus AB — Past Exam Questions Free-response questions & scoring guidelines AP Calculus BC — Past Exam Questions Free-response questions & scoring guidelines

Worksheet Library

24 original ClipSAT worksheets with full answer keys, covering Calculus I-III, organized by unit — free to download and print.

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Calculus

Foundations & Precalculus Review

The function toolkit

Exponentials and logarithms

Limits & Continuity

One-sided limits

Two limits worth memorising

Derivatives

The rules of differentiation

Integrals

Antiderivatives & the indefinite integral

The Fundamental Theorem

Applications of Derivatives

Related rates

L'Hôpital's Rule

Techniques of Integration

Applications of Integration

Differential Equations

Parametric & Polar

Sequences & Series

Convergence Tests & Power Series

The ratio and root tests

Vectors & Vector-Valued Functions

Vector-valued functions

Partial Derivatives & the Gradient

Tangent planes and local extrema

Multiple Integrals

Double integrals in polar coordinates

Vector Calculus

Stokes' and the Divergence theorems

Practice set

Test generator

Downloads

Calculus notes & practice

Native Word packet

Problem-set generator

Past Papers & Official Tests

Worksheet Library

Mock Exam