Calculus Volume 2 PDF Download – Strang & Herman (OpenStax)

Students taking Calculus II can download the complete textbook “Calculus Volume 2” by Gilbert Strang and Edwin “Jed” Herman, free as a PDF from OpenStax. It is a traditional, example-and-exercise-driven calculus text — every section works through several fully-solved examples before a large practice-problem set, and every chapter closes with a Key Terms, Key Equations, and Key Concepts review before its Chapter Review Exercises.

This is Volume 2 of OpenStax’s three-volume calculus series, covering the full second semester: integration and its applications, techniques of integration, differential equations, sequences and series, power series, and parametric equations and polar coordinates. It is designed to stand alone, so Calculus Volume 1 is not required — Multivariable Calculus continues in Volume 3, published separately.

Book Overview

Course	Calculus II (single-variable, second semester)
Degree Programs	BS Mathematics, BSCS, Engineering — general first-year/second-semester requirement
Level	University — second semester (second-year calculus)
Edition	OpenStax edition — published March 30, 2016
Author	Gilbert Strang, Edwin “Jed” Herman
Structure	7 chapters, from a self-contained integration review through parametric equations and polar coordinates
Exercises	Every section ends with a substantial practice-problem set, and every chapter closes with Key Terms, Key Equations, Key Concepts, and a full Chapter Review Exercises set
Language	English
License	Creative Commons Attribution-NonCommercial-ShareAlike 4.0 (CC BY-NC-SA 4.0) — Model: Link-only
Format	Free PDF and web/HTML reader; also available as a low-cost print edition through third-party printers

Chapter List

Chapter 1: Integration

Difficulty: Medium · Calculus II · Key topics: approximating areas, the definite integral, the Fundamental Theorem of Calculus, integration formulas and the net change theorem, substitution, integrals involving exponential and logarithmic functions, integrals resulting in inverse trigonometric functions

Volume 2 opens by rebuilding the integral from the ground up, so this book stands on its own even for a reader who hasn’t seen Volume 1. Chapter 1 approximates the area under a curve with a sum of rectangles, takes the limit as the number of rectangles grows without bound to define the definite integral precisely, and then proves the Fundamental Theorem of Calculus — the fact that this area-based definition and the antiderivative are secretly the same idea, which is what makes definite integrals computable without ever summing infinitely many rectangles by hand. It closes with substitution and the antiderivatives of exponential, logarithmic, and inverse trigonometric functions.

Key Points:

• A Riemann sum approximates area under a curve using rectangles: ∑f(x_i)·Δx, with left, right, or midpoint sample points.
• The definite integral ∫_a^b f(x)dx is defined as the limit of Riemann sums as the number of rectangles → ∞.
• FTC Part 1: if F is any antiderivative of f, then ∫_a^b f(x)dx = F(b)−F(a) — no limit of sums is needed once an antiderivative is known.
• FTC Part 2: the accumulation function A(x)=∫_a^x f(t)dt satisfies A′(x)=f(x), so integration and differentiation genuinely undo each other.
• u-substitution reverses the chain rule for integrals: choosing u=g(x) and rewriting the whole integral in terms of u.
• ∫(1/x)dx = ln|x|+C and ∫e^xdx = e^x+C are the two integral facts the chapter’s exponential/log work depends on.

Practice Tip: If your course starts directly with Calculus II, don’t skip this chapter as “review” — work through it once at full speed. Every technique in Chapters 2 and 3 assumes total fluency with the Fundamental Theorem and basic substitution, not just a vague memory of them.

Common Mistake: Forgetting that a definite integral’s value can be negative when the curve dips below the x-axis. ∫_a^bf(x)dx measures signed area — a curve entirely below the x-axis on [a,b] gives a negative integral, not zero or an error.

Important Questions:

• Use the Fundamental Theorem of Calculus to evaluate ∫₁⁴ 2x dx. An antiderivative of 2x is x², so the integral equals 4²−1² = 16−1 = 15.
• Evaluate ∫ 2x·cos(x²) dx using substitution. Let u=x², du=2x dx, giving ∫cos(u)du = sin(u)+C = sin(x²)+C.

Chapter 2: Applications of Integration

Difficulty: Hard · Calculus II · Key topics: areas between curves, determining volumes by slicing, volumes of revolution (cylindrical shells), arc length and surface area, physical applications, moments and centers of mass, integrals of exponential functions and logarithms, exponential growth and decay, calculus of the hyperbolic functions

Chapter 2 shows how far the definite integral’s core idea — summing infinitely many infinitesimal pieces — extends beyond area. Area between two curves and arc length come first, followed by two different methods (slicing into disks/washers, and cylindrical shells) for finding the volume of a solid of revolution. The chapter then applies the same accumulation idea to physical quantities: work done by a variable force, and moments and centers of mass for an object with varying density. It closes by defining the exponential and logarithmic functions properly in terms of integrals, and introducing the hyperbolic functions as a new function family built from e^x.

Key Points:

• Area between curves: ∫_a^b[f(x)−g(x)]dx, where f(x)≥g(x) on [a,b].
• The disk/washer method and the shell method both compute volumes of revolution by integrating cross-sectional area or shell surface area — pick whichever gives a simpler integral for a given axis of rotation.
• Arc length of a curve y=f(x) on [a,b] is ∫_a^b√(1+[f′(x)]²)dx — a direct consequence of summing infinitesimal straight-line segments.
• Work done by a variable force F(x) moving an object from a to b is W=∫_a^bF(x)dx.
• The center of mass of a region balances the region’s weighted position, computed by dividing a moment integral by a mass (or area) integral.
• The hyperbolic functions sinh x and cosh x are built from e^x and e^−x and satisfy identities that mirror ordinary trig functions, but with different signs in places (e.g. cosh²x − sinh²x = 1).

Practice Tip: For any volume-of-revolution problem, sketch the 2D region AND the axis of rotation before choosing disks/washers or shells — the picture alone usually tells you which method gives an integral in terms of a single, easy variable.

Common Mistake: Using the disk method when the region being revolved doesn’t touch the axis of rotation. A gap between the region and the axis means you need the washer method — subtracting an inner radius — not a single disk radius.

Important Questions:

• Set up the integral for the area between y=x² and y=x on [0,1]. Since x≥x² on [0,1], the area is ∫₀¹(x−x²)dx = [x²/2 − x³/3]₀¹ = 1/6.
• A variable force F(x)=3x newtons moves an object from x=0 to x=4 meters. Find the work done. W=∫₀⁴3x dx = [3x²/2]₀⁴ = 24 joules.

Chapter 3: Techniques of Integration

Difficulty: Hard · Calculus II · Key topics: integration by parts, trigonometric integrals, trigonometric substitution, partial fractions, other strategies for integration, numerical integration, improper integrals

Chapter 3 is where Calculus II earns its reputation — it’s a full toolkit of techniques for finding antiderivatives that substitution alone can’t reach. Integration by parts handles products of unlike function types; trigonometric integrals and trigonometric substitution handle powers of sine/cosine and expressions with √(a²−x²)-type radicals; and partial fractions breaks a rational function into simpler pieces before integrating each one. The chapter closes practically: a strategy section for choosing the right technique, numerical integration for when no elementary antiderivative exists, and improper integrals for unbounded intervals or unbounded functions.

Key Points:

• Integration by parts ∫u dv = uv − ∫v du reverses the product rule; choosing u via LIATES (Log, Inverse trig, Algebraic, Trig, Exponential) usually picks the right factor.
• Trigonometric integrals of sin^mx cosⁿx are solved by splitting off one factor and using a Pythagorean identity on the rest, with the odd-power factor chosen for substitution.
• Trigonometric substitution replaces √(a²−x²), √(a²+x²), or √(x²−a²) with x=a sinθ, a tanθ, or a secθ respectively, turning an algebraic radical into a trig identity.
• Partial fractions rewrites a rational function P(x)/Q(x) as a sum of simpler fractions with linear or irreducible-quadratic denominators, each of which integrates by elementary methods.
• Numerical integration (the Trapezoidal Rule, Simpson’s Rule) approximates a definite integral directly from function values when no elementary antiderivative exists — error-bound formulas quantify how accurate the approximation is.
• An improper integral (an infinite interval, or an integrand blowing up somewhere on the interval) is defined as a limit of a proper integral; the improper integral converges only if that limit exists and is finite.

Practice Tip: Before reaching for a named technique, always try the simplest thing first: can u-substitution alone finish it? A large fraction of Chapter 3’s practice problems look like they need partial fractions or trig substitution but actually collapse to ordinary substitution once you look for a hidden derivative pair.

Common Mistake: Forgetting to check whether a rational function’s degree needs polynomial long division before partial fractions applies. Partial fractions only decomposes a proper fraction (numerator degree less than denominator degree) — skipping the long-division step on an improper fraction produces a wrong decomposition entirely.

Important Questions:

• Use integration by parts to evaluate ∫ x·e^x dx. Let u=x, dv=e^xdx, so du=dx, v=e^x. Then ∫x·e^xdx = x·e^x − ∫e^xdx = x·e^x − e^x + C.
• Determine whether ∫₁^∞ (1/x²) dx converges, and find its value if so. ∫₁^t(1/x²)dx = [−1/x]₁^t = 1−1/t, and as t→∞ this approaches 1, so the improper integral converges to 1.

Chapter 4: Introduction to Differential Equations

Difficulty: Medium · Calculus II · Key topics: basics of differential equations, direction fields and numerical methods, separable equations, the logistic equation, first-order linear equations

Chapter 4 turns calculus toward modeling: many real phenomena — population growth, radioactive decay, predator-prey dynamics, spring-mass systems — are naturally described by an equation relating a function to its own rate of change, rather than by the function itself. The chapter defines what a differential equation and its solution are, visualizes solution behavior with direction fields before solving anything algebraically, and works through the two solution methods a first course needs most: separable equations (where variables can be split and integrated independently) and first-order linear equations (solved with an integrating factor). The logistic equation gets its own section as the standard model for growth that levels off at a carrying capacity, rather than growing without bound.

Key Points:

• A differential equation relates a function to one or more of its derivatives; a solution is a function (not a number) that makes the equation true.
• A direction field plots short line segments showing dy/dx at a grid of points, letting you sketch a solution curve’s shape before solving the equation algebraically.
• A separable equation dy/dx = f(x)g(y) is solved by rewriting it as dy/g(y) = f(x)dx and integrating both sides independently.
• The logistic equation dP/dt = kP(1−P/M) models growth that starts nearly exponential but levels off as P approaches the carrying capacity M — unlike unrestricted exponential growth.
• A first-order linear equation y′ + p(x)y = q(x) is solved using an integrating factor μ(x)=e^∫p(x)dx, which turns the left side into the derivative of a single product.
• Euler’s Method numerically approximates a solution curve step-by-step using the direction field’s slopes, useful when a differential equation has no elementary closed-form solution.

Practice Tip: Before solving any differential equation algebraically, sketch its direction field (or at least a few slope segments by hand) and predict roughly what the solution curve should look like. If your algebraic answer’s shape doesn’t match that prediction, you likely made a sign or integration error.

Common Mistake: Losing the constant of integration, or forgetting to solve for it using the initial condition, when solving a separable equation. Every differential equation has infinitely many solutions differing by a constant — an initial condition (e.g. y(0)=3) is what picks out the one particular solution a word problem is actually asking for.

Important Questions:

• Solve the separable equation dy/dx = xy, given y(0)=2. Separate: dy/y = x dx, integrate: ln|y| = x²/2 + C, so y = Ae^x²/2. Using y(0)=2 gives A=2, so y = 2e^x²/2.
• A population grows according to the logistic equation with carrying capacity M=1000. What happens to the population as t→∞, regardless of the starting population (as long as it’s positive)? The population approaches the carrying capacity M=1000 — this is the defining long-term behavior of the logistic model, in contrast to unrestricted exponential growth, which would grow without bound.

Chapter 5: Sequences and Series

Difficulty: Hard · Calculus II · Key topics: sequences, infinite series, the divergence and integral tests, comparison tests, alternating series, ratio and root tests

Chapter 5 asks a question that sounds like it shouldn’t have a sensible answer: can adding infinitely many numbers together ever produce a finite total? Using the Koch snowflake as a motivating example — a shape with infinite perimeter but finite area — the chapter builds sequences and then infinite series from them, defining convergence as the limit of the sequence of partial sums. From there it develops a toolbox of convergence tests (divergence, integral, comparison, alternating series, ratio, and root), since very few series can be summed in closed form and most calculus applications only need to know whether a series converges at all, not its exact sum.

Key Points:

• A sequence converges to L if its terms get arbitrarily close to L as n→∞; an infinite series ∑a_n converges if its sequence of partial sums S_n converges.
• The Divergence Test is a quick first check: if lim_n→∞a_n≠0, the series diverges immediately — but a_n→0 alone never proves convergence (the harmonic series ∑1/n is the standard counterexample).
• The Integral Test compares a series to the improper integral of a related continuous, positive, decreasing function — both converge or both diverge together.
• Comparison Tests (direct and limit comparison) determine a series’ behavior by comparing its terms to a series already known to converge or diverge, typically a p-series or geometric series.
• The Alternating Series Test guarantees convergence for a series ∑(−1)ⁿb_n when b_n is decreasing and b_n→0, and bounds the error by the first omitted term.
• The Ratio Test and Root Test examine lim|a_n+1/a_n| or lim(|a_n|)^1/n: a limit less than 1 means convergence, greater than 1 means divergence, and exactly 1 is inconclusive — a different test is needed.

Practice Tip: Build a mental flowchart before starting a convergence-test problem: check the Divergence Test first (it’s nearly free), then look at the series’ shape — factorials or n-th powers point to the Ratio Test, alternating signs point to the Alternating Series Test, and a resemblance to 1/n^p points to a Comparison Test against a p-series.

Common Mistake: Concluding a series converges just because its terms shrink toward zero. The harmonic series ∑1/n has terms approaching 0 but still diverges (its partial sums grow without bound, just very slowly) — a_n→0 is necessary for convergence but never sufficient on its own.

Important Questions:

• Does the series ∑_n=1^∞ 1/n² converge or diverge? Justify with a named test. It converges, by the Integral Test (or as a p-series with p=2>1): ∫₁^∞(1/x²)dx converges to 1, so the series converges as well.
• Apply the Ratio Test to ∑_n=1^∞ n!/nⁿ and state the conclusion. The ratio a_n+1/a_n simplifies to (n/(n+1))ⁿ, whose limit as n→∞ is 1/e ≈ 0.368 < 1, so by the Ratio Test the series converges.

Chapter 6: Power Series

Difficulty: Hard · Calculus II · Key topics: power series and functions, properties of power series, Taylor and Maclaurin series, working with Taylor series

Chapter 6 specializes the series ideas of Chapter 5 to power series — series of the form ∑c_nxⁿ — which behave like “infinite polynomials” and can represent genuinely complicated functions on an interval of convergence. The chapter develops how to find that interval, shows that power series can be differentiated and integrated term by term within it, and introduces Taylor and Maclaurin series as the way to build a power series representation for any sufficiently smooth function directly from its derivatives at a point. The closing section applies Taylor series practically: approximating function values, evaluating otherwise-impossible limits, and estimating definite integrals that have no elementary antiderivative.

Key Points:

• A power series ∑c_n(x−a)ⁿ converges on an interval centered at a, found using the Ratio Test to determine a radius of convergence R, then checking the two endpoints separately.
• Within its interval of convergence, a power series can be differentiated or integrated term by term, and the result is a power series for the derivative or antiderivative of the original function.
• A Taylor series for f centered at a is ∑[f⁽ⁿ⁾(a)/n!](x−a)ⁿ; a Maclaurin series is simply the special case a=0.
• Common Maclaurin series worth memorizing: e^x=∑xⁿ/n!, sin x=∑(−1)ⁿx²ⁿ⁺¹/(2n+1)!, and 1/(1−x)=∑xⁿ for |x|<1.
• The remainder term (Taylor’s Theorem) bounds the error when a Taylor series is truncated to a finite number of terms — this is what makes a truncated series a usable approximation, not just a curiosity.
• Known Maclaurin series can be combined (substituted, multiplied, differentiated, or integrated) to build new series without recomputing derivatives from scratch — e.g. integrating the series for 1/(1+x²) directly gives the series for arctan x.

Practice Tip: Memorize the four or five standard Maclaurin series (e^x, sin x, cos x, 1/(1−x), ln(1+x)) cold, the way you memorized derivative rules in Volume 1 — nearly every problem in this chapter is faster solved by substituting into or combining a known series than by computing derivatives term by term from the Taylor formula.

Common Mistake: Assuming a Taylor series converges to f(x) everywhere the series itself converges. Convergence of the series and equality with f(x) are two separate facts (governed by the remainder term going to zero) — there exist smooth functions whose Taylor series converges but to the wrong value, though this is rare for the standard functions this chapter covers.

Important Questions:

• Find the radius of convergence of ∑_n=0^∞ xⁿ/n!. By the Ratio Test, lim|a_n+1/a_n| = lim|x|/(n+1) = 0 for every x, so the series converges for all x — the radius of convergence is R=∞. (This is exactly the Maclaurin series for e^x.)
• Use the Maclaurin series for sin x to approximate sin(0.1) using just the first two nonzero terms. sin x ≈ x − x³/6, so sin(0.1) ≈ 0.1 − (0.001)/6 ≈ 0.0998333, matching the true value of sin(0.1) to five decimal places.

Chapter 7: Parametric Equations and Polar Coordinates

Difficulty: Medium · Calculus II · Key topics: parametric equations, calculus of parametric curves, polar coordinates, area and arc length in polar coordinates, conic sections

The closing chapter steps outside the x-y function framework entirely. Parametric equations describe a curve, or the position of a moving particle, using a third variable (usually t) that both x and y depend on — useful for curves like circles or cycloids that aren’t the graph of any single function y=f(x). The chapter extends derivatives, arc length, and surface area to parametric curves, then introduces polar coordinates as a second alternative coordinate system particularly suited to circular and spiral shapes that would be needlessly complicated in rectangular (x,y) form. It closes with area and arc length formulas adapted to polar coordinates, and a review of the conic sections (ellipse, parabola, hyperbola) from a unified focus-directrix viewpoint.

Key Points:

• A parametric curve is given by x=f(t), y=g(t); it can trace loops, cusps, and self-intersections that no single function y=h(x) could represent.
• dy/dx for a parametric curve is (dy/dt)/(dx/dt), not dy/dt alone — the chain rule is what connects the two parametric derivatives to the curve’s actual slope.
• Polar coordinates (r,θ) locate a point by distance from the origin and angle from the positive x-axis; the conversions x=r cosθ, y=r sinθ translate between polar and rectangular forms.
• Area enclosed by a polar curve is (1/2)∫r²dθ, not ∫r dθ — a common point of confusion since it looks unlike the rectangular area formula.
• Arc length in polar coordinates is ∫√(r²+(dr/dθ)²)dθ, derived the same way as parametric arc length once r(θ) is treated as a parametric curve in disguise.
• All three conic sections (ellipse, parabola, hyperbola) can be written with a single unified polar equation using eccentricity e: e<1 gives an ellipse, e=1 a parabola, e>1 a hyperbola.

Practice Tip: When converting a polar equation to rectangular form (or back), multiply both sides by r first if the equation isn’t already in terms of r² and r cosθ/r sinθ — e.g. r=2cosθ becomes r²=2r cosθ, which converts cleanly to x²+y²=2x. Trying to substitute before this multiplication step usually leads nowhere.

Common Mistake: Using ∫r dθ instead of (1/2)∫r² dθ for polar area. This mirrors a genuinely different derivation (summing infinitesimal circular sectors, not rectangles) from rectangular area, and mixing the two formulas up is the single most common error in this chapter.

Important Questions:

• Eliminate the parameter from x=t², y=2t (for all real t) to find the equation of the curve in x and y. From y=2t, t=y/2; substituting into x=t² gives x=(y/2)²=y²/4, i.e. y²=4x — a sideways parabola.
• Find the area enclosed by the polar curve r=2 (a circle of radius 2) using the polar area formula, and confirm it against the familiar circle-area formula. Area = (1/2)∫₀^2π 2² dθ = (1/2)(4)(2π) = 4π, which matches πr² = π(2)² = 4π exactly, as expected for a full circle of radius 2.

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Download Calculus Volume 2 PDF (Free)

This book is free from its official source, OpenStax. Click below to download the complete PDF — a free web-based reader edition (with per-section, linkable pages) is also available on the OpenStax site if you’d rather read online.

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How to Study This Book

Read the seven chapters in order — each one leans directly on the last. Chapter 1 (Integration) rebuilds the definite integral and the Fundamental Theorem of Calculus from scratch, so this book is a complete, self-contained starting point even if you haven’t read Volume 1 recently — but don’t skip it as pure review, since Chapters 2 and 3 assume total fluency with it, not just a vague memory.

Chapters 2 and 3 (Applications of Integration, Techniques of Integration) are where the course gets genuinely harder. Chapter 3 in particular is a full toolkit — integration by parts, trig integrals, trig substitution, partial fractions — and most courses expect you to be able to look at an integral and immediately recognize which technique applies, which only comes from working many problems, not from reading the chapter once.

Chapter 4 (Differential Equations) is a self-contained modeling unit — some Calculus II syllabi cover it in full, others only touch separable equations before moving to series. Check your syllabus before assuming you need every section.

Chapters 5 and 6 (Sequences and Series, Power Series) are usually the most conceptually demanding part of a Calculus II course. Build the convergence-test flowchart described in Chapter 5’s Practice Tip early, and don’t move into Chapter 6’s Taylor series until the convergence tests from Chapter 5 are comfortable — every interval-of-convergence problem in Chapter 6 depends on them.

Chapter 7 (Parametric Equations and Polar Coordinates) is largely independent of Chapters 4–6 and can be read right after Chapter 2 if your course sequences it earlier — check your syllabus, since ordering varies by institution.

This is Volume 2 of a three-volume OpenStax series. Chapters 1–2 intentionally overlap with the end of Calculus Volume 1 (Integration and Applications of Integration) so this book stands alone; if your course continues into Multivariable Calculus (vectors, partial derivatives, multiple integrals), that content is in Calculus Volume 3, not this book.

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Used In These Programs

This book is used for the Calculus II course in: BS Mathematics, BSCS, and Engineering programs. Browse all Calculus books or all Mathematics books.

Who Should Read This

Calculus Volume 2 is written for a student who has completed a first-semester (Calculus I) course and is moving into Calculus II — it assumes solid fluency with limits, derivatives, and basic integration, but rebuilds the integral and the Fundamental Theorem of Calculus from the ground up in Chapter 1, so it does not require Calculus Volume 1 specifically. Its example-and-exercise-driven style, with large practice sets at the end of every section, suits a student who learns by working through many worked examples and problems. It is also a strong fit for self-study, since the free web reader links every term to its definition and every chapter closes with a Key Terms/Key Equations/Key Concepts review before the practice problems.

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Applicable Universities

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

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Calculus Volume 2 continues OpenStax’s three-volume calculus series into the second semester, and its large, traditional practice-problem sets make it a strong match for students who learn best by working through many worked examples. Browse more Mathematics books for the rest of your semester.

Calculus Volume 2, by Gilbert Strang and Edwin “Jed” Herman. OpenStax, Rice University. Free under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 licence. Access for free at https://openstax.org/details/books/calculus-volume-2