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Welcome to Calculus II

Calculus I built the derivative and the integral. Calculus II puts the integral to work, solves equations about change, and learns to write functions as infinite sums.

1. Where Calculus I left you

Calculus I gave you two operations on functions. Differentiation turns a function into its rate of change; integration turns a rate into an accumulated total. The Fundamental Theorem of Calculus showed that they undo each other, and you learned the first two techniques for running differentiation backwards: substitution (the chain rule in reverse) and integration by parts (the product rule in reverse).

Calculus II takes those tools in three new directions. First it finishes the integration toolkit and uses it to measure things โ€” areas, volumes, lengths, work, centres of mass. Then it turns to equations whose unknown is a function, the differential equations that describe cooling coffee, growing populations and draining tanks. Finally it asks a surprising question: can a function like $e^x$ or $\sin x$ be written as an infinite polynomial? The answer โ€” series โ€” is how your calculator actually computes $\sin(0.5)$, and it closes the course with curves traced out by a moving point.

๐Ÿ“ฆ Calculus I ended withโ€ฆ

$$\int u\,dv=uv-\int v\,du$$ $$\int_a^bf(x)\,dx=F(b)-F(a)$$

๐Ÿš€ Calculus II begins withโ€ฆ

$$\int\sin^2x\,dx,\quad\int\sqrt{1-x^2}\,dx,\quad\int_1^\infty\frac{dx}{x^2}.$$

2. A story that needs all of Calculus II

A doctor prescribes a 100 mg tablet to be taken every 8 hours. The body removes the drug at a rate proportional to the amount present, so between doses the amount $A(t)$ obeys the differential equation $\dfrac{dA}{dt}=-kA$ (Chapter 7), whose solution is the exponential decay $A=A_0e^{-kt}$. For a drug with a half-life of 6 hours, a fraction $r=2^{-8/6}\approx0.397$ of each dose is still in the body when the next one is taken.

So just after the $n$-th tablet the amount is $100\left(1+r+r^2+\cdots+r^{n-1}\right)$ mg โ€” a partial sum of a geometric series (Chapter 8). Does it grow without bound, putting the patient at risk, or does it level off? And how much drug does the body "see" in total from a single tablet? That is the area under $100e^{-kt}$ from $0$ to $\infty$ โ€” an improper integral (Chapter 5). Pharmacologists call it the AUC, and it is printed on every drug approval document. Run the lab and watch all three ideas at once.

Interactive 1 โ€” One Tablet Every $\tau$ Hours

Each dose adds $100$ mg; between doses the amount decays exponentially. The peaks climb, but they climb by less each time and settle at the dashed line โ€” the sum of an infinite geometric series, $\dfrac{100}{1-r}$. Try a long half-life with a short interval and see how much higher the drug builds up.

8
time (h)โ€”
doses taken $n$โ€”
latest peak $=100\frac{1-r^n}{1-r}$ (mg)โ€”
limit $\frac{100}{1-r}$ (mg)โ€”
$r$ (fraction left at next dose)โ€”
โœ๏ธ Example 1 โ€” The steady state is a series
With a half-life of 6 hours and a 100 mg tablet every 8 hours, find the level the peaks approach, and the total exposure $\displaystyle\int_0^\infty100e^{-kt}\,dt$ from one tablet.
Decay rate Half-life 6 h means $e^{-6k}=\tfrac12$, so $k=\dfrac{\ln2}{6}\approx0.1155$ per hour, and the fraction left after 8 h is $r=e^{-8k}=2^{-4/3}\approx0.3969$.
Series The $n$-th peak is $100\left(1+r+\cdots+r^{n-1}\right)=100\dfrac{1-r^n}{1-r}$. Since $0\lt r\lt1$, $r^n\to0$, so the peaks approach $$\frac{100}{1-r}\approx\frac{100}{0.6031}\approx165.8\ \text{mg}.$$ The troughs approach $165.8\,r\approx65.8$ mg.
Improper integral $$\int_0^\infty100e^{-kt}\,dt=\lim_{b\to\infty}\Big[-\tfrac{100}{k}e^{-kt}\Big]_0^b=\frac{100}{k}=\frac{600}{\ln2}\approx865.6\ \text{mgยทh}.$$
Meaning Three chapters in one problem: a differential equation gave the decay, a geometric series gave the steady state, and an improper integral gave the total exposure. You will meet each of them properly in Chapters 7, 8 and 5.

3. The six chapters at a glance

Calculus II continues the chapter numbering of Calculus I, so it starts at Chapter 5.

5 ยท Techniquesโ†’6 ยท Applicationsโ†’7 ยท Differential equationsโ†’8 ยท Seriesโ†’9 ยท Power seriesโ†’10 ยท Parametric & polar
Chapter 5

Techniques of Integration

Trigonometric integrals and substitutions, partial fractions, numerical methods and improper integrals.

$\int\sqrt{a^2-x^2}\,dx\ :\ x=a\sin\theta$
Open chapter โ†’ Chapter 6

Applications of Integration

Slice, approximate, add up: areas, volumes of solids, arc length, surface area, work and centres of mass.

$V=\int_a^b\pi\,[f(x)]^2\,dx$
Open chapter โ†’ Chapter 7

Differential Equations

Equations for an unknown function: direction fields, Euler's method, separable equations and logistic growth.

$\dfrac{dy}{dx}=ky\ \Rightarrow\ y=Ce^{kx}$
Open chapter โ†’ Chapter 8

Sequences and Series

When does adding infinitely many numbers give a finite answer? Convergence tests and how to choose one.

$\sum_{n=0}^{\infty}ar^n=\dfrac{a}{1-r}\ (|r|\lt1)$
Open chapter โ†’ Chapter 9

Power Series

Functions as infinite polynomials: radius of convergence, Taylor and Maclaurin series, approximation with error bounds.

$e^x=\sum_{n=0}^{\infty}\dfrac{x^n}{n!}$
Open chapter โ†’ Chapter 10

Parametric and Polar

Curves traced by a moving point, polar coordinates, and calculus โ€” slopes, areas, lengths โ€” on these curves.

$A=\tfrac12\int_\alpha^\beta r^2\,d\theta$
Open chapter โ†’
๐Ÿ—บ๏ธ How the chapters connect
  1. Techniques (Ch. 5) feed everything after. Every volume, length and work problem in Chapter 6 ends in an integral, and solving a separable differential equation in Chapter 7 means integrating both sides โ€” often with partial fractions.
  2. Applications (Ch. 6) use one idea over and over: slice the object into thin pieces, approximate each piece simply, add up, and let the pieces shrink โ€” the same Riemann-sum idea that defined the integral in Calculus I.
  3. Differential equations (Ch. 7) turn a statement about rates into a function. When no formula exists, Euler's method approximates the solution step by step.
  4. Series (Ch. 8) make "adding infinitely many numbers" precise, using limits of partial sums; the integral test links them straight back to improper integrals from 5.5.
  5. Power series (Ch. 9) let $x$ vary in the series. Taylor series turn $e^x$, $\sin x$ and $\ln(1+x)$ into polynomials with infinitely many terms โ€” and give a way to integrate $e^{-x^2}$ that Chapter 5 could not.
  6. Parametric and polar curves (Ch. 10) revisit derivatives, areas and arc length for curves that are not graphs of functions โ€” orbits, cycloids, cardioids.

Threads to watch for: every new object is defined by a limit (improper integrals, sums of series, Taylor polynomials); approximation always comes with an error estimate; and the same pattern โ€” approximate, add, take a limit โ€” appears in every chapter.

โœ๏ธ Example 2 โ€” An integral that Calculus I cannot evaluate
Find $\displaystyle\int_0^1\sqrt{1-x^2}\,dx$, and explain why the methods of Calculus I do not produce an antiderivative.
Geometry $y=\sqrt{1-x^2}$ is the upper half of the unit circle, so the integral is the area of a quarter of the unit disc: $\dfrac{\pi}{4}\approx0.7854$.
Calculus I fails Substitution $u=1-x^2$ needs a factor $x$ (since $du=-2x\,dx$), and there is none. Integration by parts with $u=\sqrt{1-x^2}$, $dv=dx$ gives $x\sqrt{1-x^2}+\displaystyle\int\frac{x^2}{\sqrt{1-x^2}}\,dx$, which is no easier.
Calculus II succeeds In 5.2 the substitution $x=\sin\theta$ turns the integral into $\displaystyle\int_0^{\pi/2}\cos^2\theta\,d\theta=\frac\pi4$, and it gives the antiderivative $\tfrac12\left(\sin^{-1}x+x\sqrt{1-x^2}\right)$ for every $x$ in $[-1,1]$, not just this one area.
๐Ÿงญ How to use this course
  • Read with a pen. Try every worked example before opening its solution, then compare step by step.
  • Play with the labs. Each section has two interactives; change the settings until you can predict what will happen before you press โ–ถ Play.
  • Do the exercises in order. They run from routine to challenging, and every part has a full solution. The small check-labs let you test an antiderivative by differentiating it.
  • Use the quiz as a gate. Five questions end every section; if you score below 4, reread before moving on.
  • Bilingual terms. Dotted words show their Arabic name and meaning; the ุน button shows Arabic terms throughout. The Glossary and Formula Sheet collect everything in one place.

4. Prerequisite diagnostic โ€” are you ready?

Eight quick questions on the Calculus I skills this course uses most. Answer all of them; for each one you miss you will get a link to the Calculus I section to review.

Q1. $\displaystyle\int2x\cos\left(x^2\right)dx=$
Q2. $\displaystyle\int_0^1xe^{x^2}dx=$
Q3. $\displaystyle\int xe^x\,dx=$
Q4. $\displaystyle\int\ln x\,dx=$
Q5. $\dfrac{d}{dx}\displaystyle\int_0^x\sqrt{1+t^3}\,dt=$
Q6. $\displaystyle\int_1^4\frac{dx}{\sqrt x}=$
Q7. $\displaystyle\lim_{x\to\infty}\frac{3x^2+1}{5x^2-x}=$
Q8. $\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^2}=$

When you are ready, start with Chapter 5 โ€” Techniques of Integration.

โ† Calculus II โ€” ContentsChapter 5 โ€” Techniques of Integration โ†’