Calculus by Gilbert Strang PDF Download – MIT OpenCourseWare (3rd Edition) BS Math

Gilbert Strang’s Calculus is the open textbook edition of MIT’s own first-year calculus sequence — the same course text Strang teaches from in MIT’s 18.01 (single-variable) and 18.02 (multivariable) lectures. Free through MIT OpenCourseWare (RES.18-001) in its current 3rd edition, it is Strang’s own single-author MIT text from 1991 — not the same book as the “Calculus Volume 1/2/3” series already on this site, which is a separate OpenStax collaboration between Strang and Edwin Herman.

For a BS Mathematics or Engineering student, this book is unusual among free calculus texts: instead of stopping at the end of a Calculus II course, it carries straight on into vectors, partial derivatives, multiple integrals and vector calculus, so an entire two- or three-semester calculus sequence lives in one volume. Its explanations lean on physical intuition — distance and velocity, slope and area — before formal proof, which makes it a strong choice for board-exam-style revision as well as first-time learning.

Book Overview

Chapter List

Chapter 0: Highlights of Calculus

Difficulty: Easy · Key topics: distance/speed and height/slope, the power rule from limits, the second derivative, linear approximation, building e^x from dy/dx = y, series definition of e^x, video/practice summaries

Chapter 0 is the newest chapter, added to give the book a fast, visual on-ramp that pairs with Strang’s own “Highlights of Calculus” video series. It opens with the same paired idea as the rest of the book — distance/speed and height/slope — showing that slope = Δy/Δx and asking what that ratio means at a single instant. It then pushes into the derivative of y = x² and the general power rule dy/dx = n·xⁿ⁻¹ by direct limit computation, defines the second derivative as “the slope of the slope,” and introduces linear approximation with the tangent line. The chapter’s centerpiece constructs e^x directly by solving dy/dx = y with y(0) = 1, building it as the infinite series y = 1 + x + x²/2! + … rather than defining it as an inverse logarithm. It closes with video-lecture summaries and worked practice spanning max/min, product/quotient/chain rules, logarithms, growth, and Newton’s method.

Key Points:

• Slope Δy/Δx is the same idea as speed = distance/time — calculus asks for that ratio at one instant, not over an interval.
• Power rule from first principles: for y = xⁿ, dy/dx = n·xⁿ⁻¹, derived directly from the limit of Δy/Δx as Δx→0.
• The second derivative d²y/dx² is “the slope of the slope” — it measures how fast the rate of change is itself changing.
• e^x is the unique function solving dy/dx = y with y(0) = 1, and equals the series 1 + x + x²/2! + x³/3! + …
• Linear approximation: y ≈ y₀ + (dy/dx)(x − x₀) — the tangent line is the simplest local model of any curve.

Memory Tip: Read “the derivative of e^x is e^x” as the defining property, not a fact to memorize separately — every other exponential rule in the book (growth/decay, compound interest, b^x = e^x ln b) falls out of that one differential equation.

Common Mistake: Treating dy/dx = y as “obvious” rather than as a definition. Students often assume e^x is just “2.718 raised to a power” and miss that e is special precisely because it is the one base whose derivative equals itself.

Important Questions:

• What is the geometric meaning of the second derivative? It measures concavity — how quickly the slope itself is changing, i.e. whether the curve bends upward (concave up) or downward (concave down).
• Using y = 1 + x + x²/2! + x³/3! + …, verify that dy/dx = y term by term. Differentiating each term xⁿ/n! gives xⁿ⁻¹/(n−1)!, which reproduces the same series shifted by one term, so the whole series is its own derivative.

Chapter 1: Introduction to Calculus

Difficulty: Easy · Key topics: average vs. instantaneous velocity, the derivative from limits, circular motion, trigonometry review, point-plotting/graphing

Chapter 1 opens the classic sequence with the falling-body example used throughout the book: a distance function d = ½gt² produces the velocity v = gt as its derivative, establishing that the slope of a distance graph is velocity. Section 1.2 shows how far plain algebra and averages can go before any limit is needed, and 1.3 takes the genuine limiting step, shrinking Δt to 0 to define the velocity at one exact instant. Section 1.4 applies the same derivative idea to circular motion, connecting angular velocity ω to linear velocity through v = ωr, and 1.5 reviews the trigonometry that circular motion depends on. Section 1.6, “A Thousand Points of Light,” closes the chapter with a point-plotting idea — approximating a curve’s shape from many individual points, a preview of how computers draw graphs today.

Key Points:

• Slope of a distance-vs-time graph = velocity; the derivative is the general version of this idea for any function.
• For a falling body, d = ½gt² (g ≈ 32 ft/s²) gives instantaneous velocity v = gt as its derivative.
• Average velocity = Δd/Δt over an interval; instantaneous velocity is the limit of that ratio as Δt → 0.
• Circular motion links angular velocity ω (radians per second) to linear speed through v = ωr.
• Radians, not degrees, are calculus’ natural angle unit — every trig derivative formula later in the book assumes radians.

Memory Tip: Keep the “distance/velocity” pair as your mental model for every new derivative you meet later — asking “if this were a distance graph, what would the slope mean?” turns an abstract rate of change into something concrete.

Common Mistake: Confusing average velocity over an interval with instantaneous velocity at a point. They only agree in the limit as the interval shrinks to zero; for any finite interval they are generally different numbers.

Important Questions:

• A ball is dropped and falls d = 16t² feet in t seconds. Find its instantaneous velocity at t = 2 seconds. v = dd/dt = 32t, so at t = 2, v = 64 ft/s.
• Explain why radians, rather than degrees, are used throughout calculus. Radian measure makes the derivative formulas for sine and cosine come out cleanly (d/dx sin x = cos x); using degrees would introduce an extra conversion constant into every trigonometric derivative.

Chapter 2: Derivatives

Difficulty: Easy · Key topics: limit definition, power rule, tangent line, trigonometric derivatives, product/quotient rule, limits, continuity

Chapter 2 formalizes the derivative that Chapter 1 built from examples, stating the general limit definition f′(x) = lim_h→0 [f(x+h)−f(x)]/h. It proves the power rule dy/dx = n·xⁿ⁻¹ for any power function and extends it to sums and constant multiples so whole polynomials differentiate term by term. It then connects the derivative back to geometry, defining the tangent line as the best straight-line approximation at a point, and derives the two core trigonometric derivatives, d/dx sin x = cos x and d/dx cos x = −sin x, before introducing the product and quotient rules for combining functions. It closes by giving limits a rigorous footing and defining continuity, including the fact that differentiability implies continuity but not the reverse.

Key Points:

• Limit definition: f′(x) = lim_h→0 [f(x+h) − f(x)]/h.
• Power rule: d/dx[xⁿ] = n·xⁿ⁻¹, extended term by term to any polynomial.
• Product rule: (fg)′ = f′g + fg′. Quotient rule: (f/g)′ = (f′g − fg′)/g².
• d/dx sin x = cos x and d/dx cos x = −sin x — the base case every later trig derivative builds on.
• Differentiable ⇒ continuous, but continuous does not imply differentiable — e.g. f(x) = |x| is continuous but has no derivative at x = 0.

Memory Tip: “Low d-high minus high d-low, over low squared” is the classic line for the quotient rule (f/g)′ = (g·f′ − f·g′)/g² — say it until the subtraction order stops tripping you up.

Common Mistake: Applying the product rule to a quotient, or vice versa — writing (f/g)′ as f′/g′ instead of the correct (f′g−fg′)/g². Mixing the two rules up is one of the most common early differentiation errors.

Important Questions:

• Use the limit definition to find the derivative of f(x) = x². f′(x) = lim_h→0 [(x+h)² − x²]/h = lim_h→0 (2xh+h²)/h = 2x.
• Differentiate f(x) = x²·sin x using the product rule. f′(x) = 2x·sin x + x²·cos x.

Chapter 3: Applications of the Derivative

Difficulty: Medium · Key topics: linear approximation, max/min problems, second derivative test, curve sketching, conic sections, iteration/fixed points, Newton’s method, Mean Value Theorem, L’Hôpital’s Rule

Chapter 3 turns the derivative into a problem-solving tool. It uses the tangent line for linear approximation, f(x) ≈ f(a) + f′(a)(x−a), then applies derivatives to max/min problems by finding critical points where f′(x) = 0. The second derivative test follows — f″(x) > 0 signals a local minimum, f″(x) < 0 a local maximum — and this combines with increasing/decreasing behavior to sketch accurate graphs. A section on conic sections studies parabolas, ellipses, and hyperbolas geometrically, before the chapter develops iteration and Newton’s method, x_n+1 = x_n − f(x_n)/f′(x_n), for solving f(x) = 0 numerically through successive tangent-line steps. It closes with the Mean Value Theorem and L’Hôpital’s Rule for resolving 0/0 indeterminate limits.

Key Points:

• Linear approximation: f(x) ≈ f(a) + f′(a)(x−a) for x near a.
• Critical points occur where f′(x) = 0 (or is undefined); the second derivative test classifies them as a local max (f″<0) or local min (f″>0).
• Newton’s method: x_n+1 = x_n − f(x_n)/f′(x_n) — each step follows the tangent line to its x-intercept, converging rapidly to a root.
• The Mean Value Theorem guarantees a point where the instantaneous rate equals the average rate over an interval.
• L’Hôpital’s Rule: for a 0/0 or ∞/∞ limit, lim f(x)/g(x) = lim f′(x)/g′(x), when that second limit exists.

Memory Tip: For any optimization word problem, follow the same three steps every time — write the quantity to optimize as a function of one variable, set its derivative to zero, then check the second derivative (or the endpoints) to confirm max vs. min. The algebra changes; the three steps never do.

Common Mistake: Finding where f′(x) = 0 and stopping there, without checking whether that point is a maximum, a minimum, or neither — and without checking the endpoints of a closed interval, which can hold the true global max or min even where f′≠0.

Important Questions:

• Find the critical points of f(x) = x³ − 3x and classify each using the second derivative test. f′(x) = 3x²−3 = 0 gives x = ±1; f″(x) = 6x, so f″(1) = 6>0 (local min at x = 1) and f″(−1) = −6<0 (local max at x = −1).
• Use Newton’s method with x₀ = 1 to take one step toward a root of f(x) = x² − 2. x₁ = 1 − (1²−2)/(2·1) = 1.5, already close to √2 ≈ 1.41421.

Chapter 4: Derivatives by the Chain Rule

Difficulty: Medium · Key topics: chain rule, implicit differentiation, related rates, inverse functions, inverse trig derivatives

Chapter 4 solves the problem of differentiating one function nested inside another. It states the chain rule, dz/dx = (dz/dy)(dy/dx), almost as canceling dy between two fractions, working through examples like z = (x⁵)⁴ by separating outer and inner functions. It then uses the chain rule for implicit differentiation, finding dy/dx from equations not solved for y (such as circles), and extends it to related-rates problems where several quantities change together over time. A section on inverse functions shows their derivatives are reciprocal, dx/dy = 1/(dy/dx), which is exactly what makes the derivative of ln x work out to 1/x, and the chapter closes by applying that same idea to the inverse trig functions, deriving formulas such as d/dx[sin⁻¹x] = 1/√(1−x²).

Key Points:

• Chain rule: d/dx[f(g(x))] = f′(g(x))·g′(x) — differentiate the outside function, then multiply by the derivative of the inside.
• Implicit differentiation differentiates both sides of an equation in x and y with respect to x, treating y as a function of x.
• Related rates connect the rates of change of two or more linked quantities via the chain rule, usually after differentiating a geometric equation with respect to time.
• Inverse function derivative: if g = f⁻¹, then g′(x) = 1/f′(g(x)).
• d/dx[sin⁻¹x] = 1/√(1−x²) is the standard example of turning an inverse-function derivative into a closed formula.

Memory Tip: Say the chain rule as “outside function derivative, times inside function derivative” every time — for (x⁵)⁴, that’s “4·(x⁵)³” (outside) “times 5x⁴” (inside), not the other way around.

Common Mistake: Forgetting to multiply by the derivative of the inner function — differentiating (x⁵)⁴ as 4(x⁵)³ and stopping there, instead of also multiplying by the derivative of x⁵, which is 5x⁴.

Important Questions:

• Differentiate y = (3x²+1)⁵ using the chain rule. dy/dx = 5(3x²+1)⁴·(6x) = 30x(3x²+1)⁴.
• A 10‑ft ladder leans against a wall; its base slides away at 2 ft/s. How fast is the top sliding down when the base is 6 ft from the wall? With x²+y²=100, differentiating gives 2x(dx/dt)+2y(dy/dt)=0; at x=6, y=8, so dy/dt = −(x/y)(dx/dt) = −(6/8)(2) = −1.5 ft/s.

Chapter 5: Integrals

Difficulty: Medium · Key topics: integration as accumulation, antiderivatives, sums vs. integrals, substitution, definite vs. indefinite integrals, average value, Fundamental Theorem of Calculus, numerical integration

Chapter 5 introduces integration as the reverse of differentiation — “adding up small pieces” to recover a total from a rate of change. It develops antiderivatives directly, recognizing and reversing known derivative formulas, then compares discrete sums to continuous integrals to motivate why summation becomes integration in the limit. Indefinite integrals and the substitution method follow, for handling composite functions, and the chapter distinguishes the definite integral, which computes an exact accumulated quantity between two bounds, from the indefinite integral’s arbitrary constant. After developing integral properties and average value, it states the Fundamental Theorem of Calculus — the integral of a rate of change equals the total change — and closes with numerical integration for functions with no simple antiderivative.

Key Points:

• An antiderivative F of f satisfies F′(x) = f(x); the symbol ∫f(x)dx represents the whole family of antiderivatives.
• Fundamental Theorem of Calculus: ∫_a^b f′(x)dx = f(b) − f(a) — the integral of a rate of change equals the net change.
• Substitution reverses the chain rule: set u = g(x), rewrite the whole integral in terms of u, and don’t forget du.
• A definite integral (with bounds) gives one number; an indefinite integral (no bounds) gives a family of functions differing by a constant C.
• Average value of f on [a,b]: (1/(b−a))∫_a^b f(x)dx — the “height” of a rectangle with the same area as the region under the curve.

Memory Tip: Read the Fundamental Theorem of Calculus as “total change equals the integral of the rate of change” — it is the same distance/velocity relationship from Chapter 1, run in reverse: velocity integrated over time gives back distance traveled.

Common Mistake: Dropping the “+C” on an indefinite integral, or losing it partway through a multi-step substitution. It only disappears automatically on a definite integral, because the constant cancels in f(b)−f(a).

Important Questions:

• Evaluate ∫(3x² + 2x)dx. x³ + x² + C, by reversing the power rule on each term.
• Evaluate ∫₀² 2x dx and state what it represents geometrically. [x²]₀² = 4 − 0 = 4, the area under the line y=2x from x=0 to x=2 (a triangle of base 2 and height 4).

Chapter 6: Exponentials and Logarithms

Difficulty: Medium · Key topics: e^x from dy/dx=y, growth and decay, natural logarithm, logarithm rules, separable differential equations/logistic equation, exponential vs. power functions, hyperbolic functions

Chapter 6 develops the exponential and logarithm functions Chapter 0 previewed. It builds e^x as the unique function solving dy/dx = y, the defining property that makes it “equal to its own derivative,” then applies that property to growth and decay models — population growth, radioactive decay, and continuous compound interest, where interest is added at every instant rather than in discrete steps. The natural logarithm ln x is introduced as the inverse of e^x, with rules such as ln(yz) = ln y + ln z, and a section on separable differential equations includes the logistic equation, which models growth that slows as a population approaches a carrying capacity. The chapter compares exponential growth to polynomial growth, showing e^x eventually outgrows any power xⁿ, and closes with the hyperbolic functions, built from combinations of e^x and e^−x.

Key Points:

• e^x is the unique function with dy/dx = e^x — its own defining property, not a separately derived fact.
• Continuous compounding models interest added “at every instant” rather than at fixed periods, leading directly to the exponential function.
• ln x is the inverse of e^x; logarithm rules: ln(yz) = ln y+ln z, ln(y/z) = ln y−ln z, ln(yⁿ) = n ln y.
• A separable equation dy/dx = ky has general solution y = Ce^kt — the universal model for unconstrained growth or decay.
• The logistic equation dy/dx = ky(1 − y/L) models growth that slows and levels off as y approaches the carrying capacity L.

Memory Tip: Any time a rate of change is proportional to the current amount (population growth, radioactive decay, compound interest), the underlying equation is dy/dx = ky and the solution is automatically y = Ce^kt — recognizing the pattern saves re-deriving it every time.

Common Mistake: Mixing up ln(x+y) with ln x + ln y — logarithms distribute over multiplication and division, not addition. ln(x+y) generally cannot be simplified at all.

Important Questions:

• A population grows according to dy/dt = 0.03y with y(0) = 1000. Find y(t). Separating and integrating gives y(t) = 1000e^0.03t.
• Explain why exponential functions eventually grow faster than any polynomial function. For large x, each additional factor of x in xⁿ is fixed, but e^x‘s own growth rate keeps growing proportionally to e^x itself, so no fixed-degree polynomial can keep pace with e^x in the long run.

Chapter 7: Techniques of Integration

Difficulty: Hard · Key topics: integration by parts, trigonometric integrals, trigonometric substitution, partial fractions, improper integrals

Chapter 7 builds a systematic toolkit for integrals that don’t yield to direct antiderivative recognition. It develops integration by parts, ∫u dv = uv − ∫v du, as the reverse of the product rule, where the choice of u and dv determines whether the new integral is actually easier. Trigonometric integrals use identities like sin²x+cos²x=1 to rewrite powers of sine, cosine, and tangent into integrable forms, and trigonometric substitution replaces expressions such as √(x²+a²) with a trig function of a new variable to remove the square root. Partial fractions break a rational function like (cx+d)/[(x−a)(x−b)] into simpler pieces that integrate individually, and the chapter closes with improper integrals — integrals with an infinite limit or an unbounded integrand — evaluated as a limit to determine whether the accumulated area converges.

Key Points:

• Integration by parts: ∫u dv = uv − ∫v du — choose u to get simpler when differentiated, and dv to stay easy to integrate.
• sin²x + cos²x = 1 is the identity behind every trigonometric-integral technique in the chapter.
• Trig substitution removes a square root: √(a²−x²) → x=a sin θ, √(x²+a²) → x=a tan θ, √(x²−a²) → x=a sec θ.
• Partial fractions rewrite a rational function as a sum of simpler fractions, one per factor of the denominator, before integrating term by term.
• An improper integral converges if its defining limit exists and is finite, and diverges otherwise — the same convergence idea used for infinite series in Chapter 10.

Memory Tip: For integration by parts, use “LIATE” (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential) as a rough priority order for choosing u — the type listed first usually makes the better choice.

Common Mistake: Choosing u and dv so the new integral ∫v du ends up harder than the original — usually the result of picking u as the part that’s easy, not hard, to differentiate.

Important Questions:

• Evaluate ∫x·ln x dx using integration by parts. With u=ln x, dv=x dx: (x²/2)ln x − ∫(x²/2)(1/x)dx = (x²/2)ln x − x²/4 + C.
• Evaluate the improper integral ∫₁^∞ 1/x² dx. Rewriting as lim_b→∞ ∫₁^b x⁻²dx = lim_b→∞[1 − 1/b] = 1, so the integral converges to 1.

Chapter 8: Applications of the Integral

Difficulty: Medium · Key topics: areas and volumes by slicing, arc length, surface area of revolution, probability, mass/center of mass, work/force/energy

Chapter 8 shows how far the definite integral reaches once it isn’t restricted to area alone. Volumes are computed by slicing a solid perpendicular to an axis and integrating the slice areas, and arc length uses ds = √(1+(dy/dx)²)dx to measure the length of a curve; surface area of revolution extends the same slicing idea to a surface swept out when a curve rotates around an axis. A probability section connects density functions to probability, where the integral of a density function over an interval gives the probability of an outcome in that range. Mass-and-moments integrates a varying density to find center of mass, and force/work/energy integrates a varying force over displacement to compute work, linking calculus directly to physics.

Key Points:

• Volume by slicing: integrate the cross-sectional area A(x) along an axis, V = ∫A(x)dx.
• Arc length: ds = √(1+(dy/dx)²)dx, integrated over the curve’s domain.
• Work done by a variable force: W = ∫F(x)dx — e.g. stretching a spring or pumping water out of a tank.
• A probability density function’s integral over an interval gives the probability of an outcome in that interval; over the whole domain it must equal 1.
• Center of mass integrates position weighted by density, balancing the “moments” of mass on either side.

Memory Tip: For any volume, arc-length, work, or center-of-mass problem, draw and label one representative slice first — a disk, a thin strip of arc, or a thin layer of mass — before writing any integral.

Common Mistake: Setting up a volume-of-revolution integral with the disk method when the region doesn’t touch the axis of rotation. A gap between region and axis needs the washer method (subtracting an inner radius), not a plain disk.

Important Questions:

• Set up the integral for the arc length of y = x^3/2 from x=0 to x=4. ds = √(1+(dy/dx)²)dx with dy/dx = (3/2)x^1/2, so length = ∫₀⁴ √(1+(9/4)x) dx.
• A variable force F(x) = 3x newtons acts on an object as it moves from x=0 to x=2 meters. Find the work done. W = ∫₀² 3x dx = [3x²/2]₀² = 6 joules.

Chapter 9: Polar Coordinates and Complex Numbers

Difficulty: Medium · Key topics: polar coordinates, polar graphs, slope/length/area in polar form, complex numbers, the complex plane, Euler’s formula

Chapter 9 introduces two coordinate systems that describe points differently from the Cartesian x-y plane. Polar coordinates define a point by radius r and angle θ, giving the conversion x = r cos θ, y = r sin θ, and polar graphs trace curves that look very different from their Cartesian counterparts. A section adapts the derivative and integral to polar curves directly, so tangent slopes, arc lengths, and enclosed areas can be computed without first converting to x-y form. Complex numbers are introduced through the complex plane, where a+bi is plotted as a point and multiplication corresponds to rotating and scaling. The chapter’s centerpiece is Euler’s formula, e^iθ = cos θ + i sin θ, which unifies exponentials, trigonometry, and complex numbers into one identity.

Key Points:

• Polar-to-Cartesian conversion: x = r cos θ, y = r sin θ.
• Circular motion in Cartesian form: x = r cos ωt, y = r sin ωt, connecting back to Chapter 1’s circular motion.
• Euler’s formula: e^iθ = cos θ + i sin θ — links the exponential function directly to sine and cosine.
• Multiplying two complex numbers rotates and scales one by the other in the complex plane — multiplication becomes geometry.
• Area enclosed by a polar curve r(θ): A = ½∫r²dθ, the polar analogue of the Cartesian area integral.

Memory Tip: Think of Euler’s formula as “sine and cosine on the unit circle, written as a single exponential” — e^iθ traces the unit circle as θ increases, cos θ as its real part and sin θ as its imaginary part.

Common Mistake: Forgetting that a single point in polar coordinates has infinitely many representations — (r,θ), (r,θ+2π), (−r,θ+π) — so an equation that looks different in polar form may describe the exact same curve.

Important Questions:

• Convert the point (r,θ) = (4, π/3) to Cartesian coordinates. x = 4cos(π/3) = 2, y = 4sin(π/3) = 2√3, so the point is (2, 2√3).
• Use Euler’s formula to show that e^iπ + 1 = 0. Setting θ=π in e^iθ=cosθ+i sinθ gives e^iπ=cosπ+i sinπ = −1+0i = −1, so e^iπ+1 = 0.

Chapter 10: Infinite Series

Difficulty: Hard · Key topics: geometric series, convergence tests (positive series), convergence tests (all series), Taylor series, power series

Chapter 10 studies what happens when infinitely many terms are added together. The geometric series is the founding example: a series with ratio r converges to a/(1−r) exactly when |r| < 1, otherwise it diverges. Convergence tests for positive-term series — comparison and ratio tests — extend to series with mixed signs, covering alternating series and the stronger notion of absolute convergence. Taylor series show how a function’s derivatives at a single point determine an entire infinite series representing it, with e^x, sin x, and cos x as running examples, and power series study functions defined directly as series in powers of x, including their interval of convergence and use in solving differential equations.

Key Points:

• Geometric series with ratio r: Σarⁿ converges to a/(1−r) when |r|<1, and diverges otherwise.
• The ratio test examines lim|a_n+1/a_n| to determine convergence for series where comparison is awkward.
• Alternating series can converge even when the series of absolute values diverges — conditional convergence.
• Taylor series: f(x) = Σ [f⁽ⁿ⁾(a)/n!]·(x−a)ⁿ, built entirely from a function’s derivatives at the single point a.
• Every power series has a radius of convergence: an interval around its center where the infinite sum equals a finite value.

Memory Tip: Every convergence test answers one question — “does the tail of this series shrink fast enough?” — comparison tests check it against a known series, and the ratio test checks how quickly consecutive terms shrink relative to each other.

Common Mistake: Assuming that because a series’ terms shrink toward zero, the series itself must converge. The harmonic series Σ1/n has terms shrinking to 0 but still diverges — “terms → 0” is necessary but never sufficient for convergence.

Important Questions:

• Determine whether Σ(1/3)ⁿ (n=0 to ∞) converges, and find its sum if it does. |r|=1/3<1, so it converges to a/(1−r) = 1/(1−1/3) = 3/2.
• Find the first three nonzero terms of the Taylor series for sin x centered at a = 0. Using f⁽ⁿ⁾(0) for sine’s derivatives, sin x = x − x³/3! + x⁵/5! − …

Chapter 11: Vectors and Matrices

Difficulty: Medium · Key topics: dot product, planes/projections, cross product, determinants, matrices and linear equations, eigenvalues/eigenvectors

Chapter 11 introduces the linear-algebra machinery multivariable calculus depends on. Vectors and dot products define a vector as a quantity with magnitude and direction, and the dot product u·v = u₁v₁+u₂v₂+u₃v₃ measures alignment and computes projections. Planes and projections extend this to three dimensions, defining a plane through a normal vector and ax+by+cz=d, then decomposing a vector into components along a given direction. Cross products and determinants introduce the cross product, producing a vector perpendicular to both inputs, and determinants, whose value represents a scaling of area or volume. Matrices and linear equations set up and solve systems Ax=b, and the closing section introduces eigenvalues and eigenvectors, describing how a matrix stretches or rotates space along special directions.

Key Points:

• Dot product: u·v = u₁v₁+u₂v₂+u₃v₃ — zero exactly when the two vectors are perpendicular.
• A plane is defined by a normal vector and a point: a(x−x₀)+b(y−y₀)+c(z−z₀) = 0.
• The cross product u×v produces a vector perpendicular to both u and v, with magnitude equal to the area of the parallelogram they span.
• A determinant’s value represents how a linear transformation scales area (2D) or volume (3D), and whether it flips orientation.
• Eigenvalues and eigenvectors of matrix A satisfy Av = λv — directions the matrix only stretches or shrinks, without rotating.

Memory Tip: The dot product answers “how much do these vectors point the same way?” while the cross product answers “what new direction is perpendicular to both?” — one gives a number (alignment), the other a vector (a new direction).

Common Mistake: Confusing the dot product (a number, measuring alignment) with the cross product (a vector, only defined in three dimensions) — trying to “cross” two-dimensional vectors the same way.

Important Questions:

• Compute the dot product of u=(1,2,3) and v=(4,−1,2), and state whether they are perpendicular. u·v = 1(4)+2(−1)+3(2) = 4−2+6 = 8, so they are not perpendicular (that would require a dot product of 0).
• Find the equation of the plane through (1,0,2) with normal vector (2,−1,3). 2(x−1)−1(y−0)+3(z−2) = 0, which simplifies to 2x−y+3z = 8.

Chapter 12: Motion Along a Curve

Difficulty: Medium · Key topics: position/velocity/acceleration vectors, plane motion (projectile, cycloid), curvature and normal vector, polar coordinates and planetary motion (Kepler’s laws)

Chapter 12 uses vectors to describe objects moving along curved paths rather than a straight line. Position is defined as r(t) = (x(t),y(t),z(t)), with velocity found by differentiating position and acceleration by differentiating velocity again. Plane motion works through concrete trajectories, including projectile motion under gravity and the cycloid traced by a point on a rolling circle, showing the velocity vector is always tangent to the path. Curvature and the normal vector quantify how sharply a path bends, using derivatives of the position vector, and the chapter’s centerpiece — polar coordinates and planetary motion — derives Kepler’s laws of orbital motion from the inverse-square law of gravity: elliptical orbits, equal areas swept in equal times, and the relationship between orbital period and distance.

Key Points:

• Position vector r(t)=(x(t),y(t),z(t)); velocity v(t)=r′(t); acceleration a(t)=v′(t)=r″(t).
• The velocity vector is always tangent to the path of motion, whatever the curve’s shape.
• Curvature measures how sharply a path bends — a straight line has curvature 0, a tight circle has high curvature.
• Kepler’s first law: planets move in elliptical orbits with the sun at one focus.
• Kepler’s second law: a line from the sun to a planet sweeps out equal areas in equal times — a direct consequence of conserved angular momentum.

Memory Tip: Keep “position → velocity → acceleration” straight by treating each arrow as “take one derivative” — the same relationship as distance → velocity → acceleration from Chapter 1, now written with vectors.

Common Mistake: Assuming speed is constant just because the path is a circle. In non-uniform circular motion, or an elliptical orbit under Kepler’s laws, both the direction and magnitude of the velocity vector change continuously.

Important Questions:

• An object has position r(t)=(cos t, sin t). Find its velocity and acceleration vectors. v(t)=(−sin t, cos t), a(t)=(−cos t, −sin t) — note a(t) = −r(t), pointing back toward the origin, characteristic of circular motion.
• State Kepler’s second law and explain what conserved quantity it reflects. A line from the sun to an orbiting planet sweeps out equal areas in equal times; this reflects conservation of angular momentum in a system where gravity acts only along the line connecting the two bodies.

Chapter 13: Partial Derivatives

Difficulty: Hard · Key topics: functions of two variables, level curves, partial derivatives, tangent planes/linear approximation, gradient/directional derivatives, multivariable chain rule, max/min/saddle points, Lagrange multipliers

Chapter 13 extends the derivative to functions of more than one variable. Surfaces and level curves introduce z = f(x,y) as a surface in three dimensions, with level curves working like elevation lines on a topographic map. Partial derivatives ∂f/∂x and ∂f/∂y measure the rate of change in one direction while every other variable is held fixed, and tangent planes extend linear approximation to two variables, f(x,y) ≈ f(a,b)+(∂f/∂x)(x−a)+(∂f/∂y)(y−b). The gradient ∇f = (∂f/∂x,∂f/∂y) points in the direction of steepest increase, the chain rule extends to composite multivariable functions, and the chapter closes with critical points — classified by a second-derivative test into max, min, or saddle — and Lagrange multipliers, solving constrained optimization with ∇f = λ∇g.

Key Points:

• Partial derivative ∂f/∂x measures f’s rate of change in the x-direction alone, holding every other variable fixed.
• The gradient ∇f = (∂f/∂x,∂f/∂y) points in the direction of steepest increase, and its magnitude is the steepest slope.
• Linear approximation: f(x,y) ≈ f(a,b) + (∂f/∂x)(x−a) + (∂f/∂y)(y−b).
• Critical points satisfy ∂f/∂x=0 and ∂f/∂y=0; the test D = f_xxf_yy−(f_xy)² classifies them as max, min, or saddle.
• Lagrange multipliers solve “maximize f subject to g=constant” by solving ∇f = λ∇g together with the constraint.

Memory Tip: Think of the gradient as a compass needle for the surface — it always points “uphill” in the steepest direction, and its length tells you how steep that climb is.

Common Mistake: Setting only ∂f/∂x=0 (or only ∂f/∂y=0) and stopping there. Both partial derivatives must equal zero simultaneously to locate a genuine critical point of a two-variable function.

Important Questions:

• Find the critical point of f(x,y) = x²+y²−4x+2y and classify it. ∂f/∂x=2x−4=0 and ∂f/∂y=2y+2=0 give (2,−1); with f_xx=2, f_yy=2, f_xy=0, D=4>0 and f_xx>0, so it’s a local minimum.
• What does the gradient vector ∇f at a point tell you about the surface z=f(x,y) there? Its direction is the direction of steepest increase of f from that point, and its magnitude is the value of that steepest slope; moving perpendicular to the gradient keeps f approximately constant.

Chapter 14: Multiple Integrals

Difficulty: Hard · Key topics: double integrals, changing coordinates/Jacobian, triple integrals, cylindrical and spherical coordinates

Chapter 14 extends integration to functions of two and three variables. A double integral ∫∫f(x,y)dA over a planar region is computed as an iterated integral — integrating with respect to one variable while treating the other as constant, then integrating the result with respect to the second — and this computes both volumes under a surface and areas of a region. The change-of-coordinates section introduces the Jacobian determinant, accounting for how area scales when switching coordinate systems, making circularly symmetric regions much easier to integrate over. Triple integrals extend the same iterated-integral idea to three dimensions, and cylindrical (r,θ,z) and spherical (ρ,φ,θ) coordinates present two non-Cartesian systems suited to axis-symmetric and point-symmetric regions respectively, each with its own Jacobian factor for the volume element.

Key Points:

• A double integral ∫∫f(x,y)dA is computed as an iterated integral: integrate with respect to one variable first, then the other.
• The Jacobian determinant accounts for how area or volume elements scale under a change of coordinates.
• Cylindrical coordinates (r,θ,z) suit regions symmetric about an axis; the volume element is dV = r dr dθ dz.
• Spherical coordinates (ρ,φ,θ) suit regions symmetric about a point; the volume element is dV = ρ²sin φ dρ dφ dθ.
• Matching the coordinate system to a region’s symmetry is usually what turns an intractable integral into a manageable one.

Memory Tip: Before setting up any multiple integral, sketch the region and ask “does this look circular/spherical or rectangular?” — that answer picks the coordinate system that will make the integral tractable.

Common Mistake: Forgetting the extra factor r (cylindrical) or ρ²sin φ (spherical) when converting from Cartesian coordinates — omitting the Jacobian factor is one of the most common errors in multiple integration.

Important Questions:

• Set up (don’t evaluate) the double integral for the volume under z=x+y over 0≤x≤2, 0≤y≤3. V = ∫₀² ∫₀³ (x+y) dy dx.
• Why does converting to polar/cylindrical coordinates require an extra factor of r in the integrand? Because a small area element in polar coordinates is r dr dθ, not dr dθ — the factor r accounts for area elements at a larger radius sweeping out more actual area for the same angular width dθ.

Chapter 15: Vector Calculus

Difficulty: Hard · Key topics: vector fields, line integrals, Green’s Theorem, surface integrals, Divergence Theorem, Stokes’ Theorem

Chapter 15 closes the sequence with tools for analyzing vector fields — assignments of a vector to every point in space, such as a fluid’s velocity field. Line integrals ∫F·dr measure the accumulated component of a vector field along a curve, computing the work done by a force along a path. Green’s Theorem connects a line integral around a closed curve in the plane to a double integral of the field’s curl over the enclosed region, trading a boundary calculation for an interior one. Surface integrals generalize this to two-dimensional surfaces, computing flux via ∫∫F·n dS, and the Divergence Theorem relates the flux through a closed surface to the integral of the field’s divergence throughout the enclosed volume. Stokes’ Theorem generalizes Green’s Theorem to three dimensions, relating a surface integral of curl to a line integral around its boundary curve.

Key Points:

• A line integral ∫F·dr sums the component of vector field F along a curve — physically, the work done by a force along a path.
• Green’s Theorem: ∮F·dr = ∫∫(curl F) dA — a closed-curve line integral equals a double integral of curl over the enclosed region.
• Flux through a surface: ∫∫F·n dS, measuring how much of a vector field passes through it.
• Divergence Theorem: flux of F through a closed surface equals the integral of div F over the enclosed volume.
• Stokes’ Theorem: ∫∫(curl F)·n dS = ∮F·dr — the 3D generalization of Green’s Theorem.

Memory Tip: Think of the Fundamental Theorem of Calculus, Green’s, Divergence, and Stokes’ Theorems as one family — each converts an integral over a boundary into an integral over the whole region (or the reverse), just in successively higher dimensions.

Common Mistake: Applying Green’s Theorem to a curve that isn’t closed, or a region that isn’t simply connected, without adjusting for it — both conditions are required for the basic version of the theorem.

Important Questions:

• What physical quantity does ∫F·dr represent when F is a force field? The work done by the force field on an object moving along the given path.
• State the Divergence Theorem in words. The total flux of a vector field out through a closed surface equals the integral of the field’s divergence over the volume it encloses — summing all the “local outflow” throughout the region gives the same answer as measuring everything crossing the boundary.

Chapter 16: Mathematics after Calculus

Difficulty: Easy · Key topics: differential equations, discrete mathematics, probability and statistics, linear algebra — a preview chapter

Chapter 16 closes the book by looking ahead rather than introducing new techniques to master. It previews differential equations as a natural next step, extending the single-variable groundwork from Chapters 6–7 to systems with several interacting variables, and gestures toward discrete mathematics — combinatorics, graph theory — as a genuinely different style of mathematics built on finite structures rather than limits and continuity. It connects integration to probability and statistics, where the techniques from Chapters 5, 7, and 8 become the tool for computing probabilities and analyzing data, and points to linear algebra as the formal framework behind the vectors and matrices already introduced in Chapter 11. This chapter is deliberately short and has no exercises of its own — its role is orientation, not instruction.

Key Points:

• Differential equations generalize the single-variable equations of Chapters 6–7 to systems with several interacting variables.
• Discrete mathematics studies finite structures (combinatorics, graph theory) rather than the continuous limits calculus is built on.
• Probability and statistics reuse the integration techniques from Chapters 5, 7, and 8 to compute probabilities from density functions.
• Linear algebra formalizes and generalizes the vectors and matrices from Chapter 11 to any number of dimensions.
• This chapter previews where calculus leads next — it is not itself an exam-tested chapter.

Important Questions:

• According to this chapter, which later subject formalizes the vector and matrix ideas from Chapter 11? Linear algebra.
• Why does the book connect integral calculus to probability and statistics here? Computing a probability from a continuous density function is itself a definite integral, so the same integration techniques from Chapters 5, 7, and 8 carry over directly into probability and statistics.

Download Calculus (Strang) PDF (Free)

This book is free from its official source. Click below to open the complete book as a single PDF, hosted directly by MIT OpenCourseWare — the same page also offers each chapter as its own smaller PDF and links to Strang’s companion “Highlights of Calculus” video series.

How to Study This Book

Start with Chapter 0 only as an optional fast refresher if you already know a little calculus and want the big picture before the full course — it is not a prerequisite for Chapter 1, and it pairs directly with Strang’s free “Highlights of Calculus” videos if you prefer to watch before you read.

Chapters 1 through 4 form a complete Calculus I course — limits, the derivative, and its first applications — and match MIT’s own 18.01 opening unit. Read them in strict order: Chapter 3’s optimization and Newton’s-method work leans directly on the derivative rules built in Chapter 2, and Chapter 4’s chain rule is used constantly from Chapter 5 onward.

Chapters 5 through 10 form the Calculus II half: integration (5), exponentials and logarithms (6), integration techniques (7), applications of the integral (8), polar coordinates and complex numbers (9), and infinite series (10). Work through Chapter 5’s substitution and integration-by-parts before Chapter 7, since Chapter 7 assumes both are already comfortable, and read Chapter 10 (series) last in this half — it leans on the exponential and trig derivatives from Chapters 2 and 6.

Chapters 11 through 15 are the multivariable half of the book (MIT’s 18.02) and assume Chapters 1–10 are solid, especially the chain rule from Chapter 4. Read Chapter 11 (vectors and matrices) first, since Chapters 12–15 all use vector notation without re-explaining it; Chapter 13’s partial derivatives and Chapter 14’s multiple integrals build on each other directly, and Chapter 15 (vector calculus) is the hardest chapter in the book and assumes every earlier multivariable chapter is already solid.

Chapter 16 is a short preview chapter with no exercises — read it once for orientation, not for exam preparation.

If your course covers only single-variable calculus (a typical BSCS or first-year Engineering Calculus I/II requirement), Chapters 1–10 alone are a complete, self-contained book; Chapters 11–15 (multivariable calculus / Calculus III) can be picked up separately when your program requires them.

Used In These Programs

This book is used for the full Calculus I, II & III sequence in: BS Mathematics and Engineering programs. Browse all Calculus books or all Mathematics books.

Who Should Read This

This book is written for a BS Mathematics or Engineering student who wants a single, free text covering the entire calculus sequence — single-variable through multivariable — rather than switching books partway through. It assumes only a solid pre-calculus and trigonometry background (Chapter 1.5 reviews the trig you’ll need), not any prior calculus, so it suits a first-time learner as well as a student revising for finals. Its explanations lean on physical intuition (distance/velocity, slope/area) before formal proof, which is a different style from the guided-activity approach of Active Calculus also on this site — students who prefer working through preview activities before reading may want to pair the two, using Active Calculus for guided practice and this book for its wider multivariable coverage and its companion video series.

Applicable Universities

This book is useful for students at Pakistani universities offering BS Mathematics or Engineering programs, including Punjab University, Virtual University, COMSATS, FAST, UET, NUST, and other HEC-recognized institutions.

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Related Books

• Active Calculus – Matthew Boelkins (2nd Edition)
• Calculus Volume 1 – Strang & Herman (OpenStax)
• Calculus Volume 2 – Strang & Herman (OpenStax)
• Calculus Volume 3 – Strang & Herman (OpenStax)

Calculus by Gilbert Strang is the most complete free calculus text in this collection’s University Books section, carrying a single-variable-through-multivariable sequence in one volume for BS Mathematics and Engineering students. Browse more Mathematics books for the rest of your semester.

Calculus, 3rd Edition (RES.18-001), by Gilbert Strang, Massachusetts Institute of Technology. Free under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International (CC BY-NC-SA 4.0) licence via MIT OpenCourseWare. The original 1991 print edition is published by Wellesley-Cambridge Press. Official source: https://ocw.mit.edu/courses/res-18-001-calculus-fall-2023/