Mathematics

A. Analytic geometry and trigonometry

Analytic Geometry and Trigonometry

Lines, conics, polar conversion, and the triangle laws and identities that every geometry, kinematics, and phasor problem on the FE quietly leans on.

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B. Differential equations

First- and Second-Order Differential Equations

Solve the linear ODEs the FE actually asks: separable and first-order linear, plus second-order constant-coefficient over-, critically, and under-damped responses.

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C. Numerical methods

Numerical Methods

Root-finding by bisection and Newton-Raphson and numerical integration by the trapezoidal and Simpson rules, with the error and convergence sense the FE rewards.

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D. Linear algebra

Linear Algebra

Matrix algebra, determinants, inverses, and the three ways the FE solves a linear system — Cramer's rule, Gaussian elimination, and the inverse — plus an eigenvalue primer.

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E. Single-variable calculus

Single-Variable Calculus

Limits, derivatives, optimization, definite and indefinite integrals, the Fundamental Theorem, area, and L'Hopital — the calculus the FE asks across nearly every topic.

Calculus is the single most pervasive idea on the FE: a velocity is the derivative of position, a flow rate the derivative of volume, an accumulated mass the integral of a rate, an area the integral of a height. The exam almost never asks for calculus in the abstract — it hands you a function and wants a maximum, a slope, an area, or the value of a stubborn limit. The Mathematics chapter of the NCEES FE Reference Handbook prints the full Derivatives and Indefinite Integrals tables, so your edge is fluency: knowing which operation the question is really asking for and executing it without a slip. This concept walks the whole arc, from limits to the Fundamental Theorem.

Limits and continuity

A limit asks what value

f(x)

heads toward as

x

approaches

a

, independent of

f(a)

itself. For a continuous function you simply substitute,

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

. The work begins when substitution gives an indeterminate form like

0/0

or

\infty/\infty

: then you factor, rationalize, divide by the highest power, or call in L'Hopital's rule. Continuity at

a

— the limit exists and equals

f(a)

— is the prerequisite for differentiability.

\lim_{x\to a} f(x)=L\quad\text{means } f(x)\to L\text{ as } x\to a

The derivative and its rules

The derivative is the instantaneous rate of change — the slope of the tangent line — defined as the limit of a difference quotient. In practice you apply the table rules: the power rule for polynomials, the product and quotient rules for combinations, and the chain rule for compositions. The chain rule is the one examinees forget: differentiate the outer function, then multiply by the derivative of the inner. Define inner and outer explicitly so you never drop a factor.

f'(x)=\lim_{\Delta x\to 0}\dfrac{f(x+\Delta x)-f(x)}{\Delta x}

Maxima, minima, and inflection

To optimize, set the first derivative to zero to locate critical points, then classify each with the second derivative:

f''(a)>0

is a local minimum (concave up),

f''(a)<0

a local maximum (concave down). A point of inflection occurs where

f''

changes sign, typically where

f''(a)=0

. On a closed interval, also test the endpoints — the absolute extremum can live at a boundary even when no interior critical point wins. This is the engine behind every "largest area," "least material," or "minimum cost" problem.

f'(a)=0;\quad f''(a)>0\Rightarrow\text{min},\quad f''(a)<0\Rightarrow\text{max}

Indefinite integrals (antiderivatives)

The indefinite integral reverses differentiation and carries an arbitrary constant

C

, since any constant differentiates away. When the table does not hand you the antiderivative directly, the handbook's three workhorses are substitution (reverse chain rule), integration by parts for products, and partial fractions for rational functions. Integration by parts is

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

; choose

u

to simplify when differentiated (the LIATE guideline).

\int u\,dv = uv-\int v\,du,\qquad \int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+C\ (n\ne -1)

The definite integral and the Fundamental Theorem

The definite integral is the signed area under a curve, built as the limit of a sum of thin rectangles. The Fundamental Theorem of Calculus evaluates it as the net change of any antiderivative across the limits — this is the bridge that turns the hard limit-of-a-sum into a two-step lookup-and-subtract. Its companion statement says differentiation and integration undo each other, so

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

.

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

Area between curves

The area enclosed between two curves on

[a,b]

is the integral of the upper minus the lower function. Find the intersection points first — they set the limits — and integrate the difference; if the curves cross within the interval, split the integral so "upper minus lower" stays correct on each piece. This same construction gives the area swept by a velocity curve (a distance) or a power curve (an energy).

A=\int_a^b\big[\,f_{\text{upper}}(x)-f_{\text{lower}}(x)\,\big]\,dx

L'Hopital's rule

When a limit produces the indeterminate form

0/0

or

\infty/\infty

, L'Hopital's rule replaces it with the limit of the ratio of derivatives, repeated until the form is determinate. Confirm the form is genuinely indeterminate before applying it — using L'Hopital on a determinate or

2/0

limit gives wrong answers. Other indeterminate forms (

0\cdot\infty

,

\infty-\infty

,

1^{\infty}

) must first be algebraically recast into a quotient.

\lim_{x\to a}\dfrac{f(x)}{g(x)}=\lim_{x\to a}\dfrac{f'(x)}{g'(x)}\quad\left(\tfrac{0}{0}\text{ or }\tfrac{\infty}{\infty}\right)

Exam strategy

Identify the operation first: "slope," "rate," or "velocity" means derivative; "area," "total," or "accumulated" means integral; a stuck

0/0

means factor or L'Hopital; "largest/smallest" means set

f'=0

and apply the second-derivative test. Then go straight to the handbook's Derivatives or Indefinite Integrals table rather than deriving from scratch. For optimization never skip the second-derivative test or the endpoints; carry the

+C

on indefinite integrals; and keep every trig argument in radians, the convention the handbook's tables assume.

Key equations

Derivative (definition)

f'(x)=\lim_{\Delta x\to 0}\dfrac{f(x+\Delta x)-f(x)}{\Delta x}

Instantaneous rate / tangent slope; units of $f'$ are units of $f$ per unit of $x$ .

Power rule

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

Differentiates any monomial; the backbone of polynomial differentiation.

Product and quotient rules

(uv)'=u'v+uv',\qquad\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^2}

For products and quotients of functions $u(x),v(x)$ ; mind the numerator order in the quotient.

Chain rule

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

Differentiate compositions: outer derivative times inner derivative.

Critical point + 2nd-derivative test

f'(a)=0,\quad f''(a)>0\Rightarrow\text{min},\ f''(a)<0\Rightarrow\text{max}

Locate then classify extrema; inflection where $f''$ changes sign.

Power rule for integrals

\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+C\quad(n\ne -1)

Antiderivative of a power; for $n=-1$ , $\int x^{-1}dx=\ln|x|+C$ .

Integration by parts

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

For products; pick $u$ to simplify on differentiation (LIATE guideline).

Fundamental Theorem of Calculus

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

Evaluate a definite integral (signed area) via any antiderivative $F$ .

Area between curves

A=\int_a^b\big[f_{\text{upper}}(x)-f_{\text{lower}}(x)\big]dx

Limits are the intersection points; split where the curves cross.

L'Hopital's rule

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

Valid only for $0/0$ or $\infty/\infty$ ; repeat until determinate.

Common derivatives

\dfrac{d}{dx}\sin x=\cos x,\ \dfrac{d}{dx}e^{x}=e^{x},\ \dfrac{d}{dx}\ln x=\dfrac{1}{x}

Table staples; trig arguments are in radians.

Worked examples

Optimization: maximum enclosed area

Problem. A rancher has $400\,\text{m}$ of fence to enclose a rectangular pen along a straight river, fencing only the three non-river sides. What dimensions maximize the enclosed area, and what is that area?

Solution. Let

x

be each side perpendicular to the river and

y

the side parallel to it. Fence used:

2x+y=400

, so

y=400-2x

. Area:

A(x)=x\,y=x(400-2x)=400x-2x^2

. Maximize:

A'(x)=400-4x=0\Rightarrow x=100\,\text{m}

, and

A''=-4<0

confirms a maximum. Then

y=400-200=200\,\text{m}

and

A=100\times 200=20{,}000\,\text{m}^2

. Sanity check:

2(100)+200=400\,\text{m}

of fence used exactly;

A''<0

confirms a max. Answer:

x=\mathbf{100\,\text{m}}

,

y=\mathbf{200\,\text{m}}

,

A=\mathbf{2.00\times 10^4\,\text{m}^2}

.

A'(x)=400-4x=0\ \Rightarrow\ x=100\,\text{m}

Area between two curves

Problem. Find the area enclosed between $y=2x$ and $y=x^2$ .

Solution. Intersections:

2x=x^2\Rightarrow x^2-2x=0\Rightarrow x=0,\,2

. On

(0,2)

, the line

2x

lies above the parabola

x^2

.

A=\displaystyle\int_0^2 (2x-x^2)\,dx=\left[x^2-\dfrac{x^3}{3}\right]_0^2=4-\dfrac{8}{3}=\dfrac{4}{3}

. So

A=1.33

. Sanity check: at

x=1

the gap is

2(1)-1^2=1

, and the region spans width 2, so an area near

1.3

is reasonable. Answer:

A=\mathbf{1.33}

(i.e.

4/3

).

A=\int_0^2(2x-x^2)\,dx=4-\tfrac{8}{3}=\tfrac{4}{3}

Indeterminate limit by L'Hopital

Problem. Evaluate $\displaystyle\lim_{x\to 0}\dfrac{e^{x}-1-x}{x^{2}}$ .

Solution. Direct substitution gives

0/0

— indeterminate, so L'Hopital applies. First application:

\dfrac{d}{dx}(e^x-1-x)=e^x-1

and

\dfrac{d}{dx}(x^2)=2x

, giving

\lim_{x\to 0}\dfrac{e^x-1}{2x}

, still

0/0

. Second application:

\dfrac{e^x}{2}

, which at

x=0

is

\dfrac{1}{2}=0.500

. Sanity check: the Maclaurin series gives

e^x-1-x=\tfrac{x^2}{2}+\cdots

, so the ratio

\to \tfrac12

— agrees. Answer:

\mathbf{0.500}

.

\lim_{x\to 0}\dfrac{e^{x}-1-x}{x^{2}}=\lim_{x\to 0}\dfrac{e^{x}}{2}=\tfrac{1}{2}

Velocity from position (derivative application)

Problem. A particle moves with position $s(t)=t^3-6t^2+9t$ meters ( $t$ in seconds). Find the times when it is momentarily at rest and its position at each.

Solution. Velocity is the derivative:

v(t)=s'(t)=3t^2-12t+9=3(t^2-4t+3)=3(t-1)(t-3)

. At rest when

v=0

:

t=1\,\text{s}

and

t=3\,\text{s}

. Positions:

s(1)=1-6+9=4\,\text{m}

;

s(3)=27-54+27=0\,\text{m}

. Sanity check: acceleration

a(t)=6t-12

is negative at

t=1

(a position maximum,

s=4

) and positive at

t=3

(a minimum,

s=0

), consistent with motion out and back. Answer: at rest at

t=\mathbf{1\,\text{s}}

(

s=4\,\text{m}

) and

t=\mathbf{3\,\text{s}}

(

s=0\,\text{m}

).

v(t)=3(t-1)(t-3)=0\ \Rightarrow\ t=1,\,3\,\text{s}

Common pitfalls

•Forgetting the chain rule's inner factor: $\frac{d}{dx}\sin(3x)=3\cos(3x)$ , not $\cos(3x)$ . Always multiply by the derivative of the inside.
•Skipping the second-derivative test: $f'(a)=0$ alone does not tell max from min, and a closed interval can hide its extremum at an endpoint.
•Dropping the constant of integration: indefinite integrals carry $+C$ , and some answer choices are written specifically to catch its omission.
•Applying L'Hopital to a determinate form: confirm $0/0$ or $\infty/\infty$ first; on a $2/0$ or a finite limit the rule gives a wrong value.
•Radians vs. degrees: the handbook's derivative and integral tables assume radians. $\frac{d}{dx}\sin x=\cos x$ holds only in radian measure.
•Area sign error: $\int f\,dx$ is signed area — below the axis it is negative. For a geometric area between curves, integrate upper minus lower and split where they cross.
•Wrong integration limits for area: the limits are the intersection points of the curves, not the bounds stated for some other part of the problem — solve $f_{\text{upper}}=f_{\text{lower}}$ first.

References

NCEES FE Reference Handbook — Mathematics (Differential Calculus, Integral Calculus, Derivatives and Indefinite Integrals tables)
Stewart, Calculus: Early Transcendentals — limits, optimization, FTC