If a car drives at a steady 60 mph for 2 hours it covers \(60 \times 2 = 120\) miles. Draw that on a graph with time across and speed up, and the journey is a horizontal line at height 60 running from \(t = 0\) to \(t = 2\). The distance, 120, is the area of the rectangle under that line: height (speed) times width (time).

Now let the speed change in jumps. Say 30 mph for the first hour, 60 mph for the next 1.5 hours, then 20 mph for the last half hour. Each piece is a rectangle, and the total distance is the sum of their areas:

Real cars do not change speed in jumps. Suppose the speed in mph after \(t\) hours is

It starts at 10 mph, peaks at 55 mph at \(t = 1.5\), and falls back to 10 mph. How far did the car go?

There is no single speed to multiply by 3 hours. But the rectangle picture suggests what to do. On a short enough time interval the speed hardly changes, so "speed times time" is nearly right for that sliver. Chop the journey into many slivers, treat the speed as constant on each, add up the little distances, and make the slivers thinner and thinner. That is the whole idea of the integral, and it is the same limiting move you met with derivatives, applied to a sum instead of a ratio.

Notice what has happened. We started with a physical question (distance from speed) and ended with a geometric one (the area under a curve). These are the same problem, and the integral is the tool that answers both. By the end of the section on the Fundamental Theorem we will find that this car covered exactly 120 miles.

Notice too how this reverses the previous chapter. There, position \(s(t)\) was given and we differentiated to get speed \(v(t) = s'(t)\). Here the speed is given and we want the change in position. Integration will turn out to be the inverse of differentiation, which is why it is sometimes called antidifferentiation.

Take the simplest curved example: the area under \(y = x^2\) between \(x = 0\) and \(x = 2\). No school formula gives this region’s area: it is not a triangle, rectangle or circle segment. So we approximate it.

Cut the interval \({[0, 2]}\) into \(n\) equal pieces. Above each piece, build a rectangle whose height is the curve’s height somewhere in that piece. Add the rectangle areas. The choice of "somewhere" gives three standard versions:

Because \(x^2\) is increasing on this interval, the left sum leaves gaps under the curve (an underestimate) and the right sum overshoots (an overestimate). The midpoint sum overshoots on half of each piece and undershoots on the other half, so the errors largely cancel.

Approximate the area under \(f(x) = x^2\) on \({[0, 2]}\) using \(n = 4\). Try it: Draw the three sums; Watch the rectangles.

Step 1: the width. \(\Delta x = (2 - 0)/4 = 0.5\).

Step 2: the partition points. \(x_0 = 0,\ x_1 = 0.5,\ x_2 = 1,\ x_3 = 1.5,\ x_4 = 2.\)

Step 3: left sum. Heights at \(0, 0.5, 1, 1.5\):

Step 4: right sum. Heights at \(0.5, 1, 1.5, 2\):

Step 5: midpoint sum. Midpoints are \(0.25, 0.75, 1.25, 1.75\):

The true area, as we will soon prove, is \(8/3 \approx 2.6667\). The midpoint sum is already within 0.042 of it with only four rectangles; the left and right sums are off by about 0.92 and 1.08.

All three columns close in on \(2.6666\ldots\). Look at the rates: each time \(n\) is multiplied by 4, the left and right errors shrink by roughly a factor of 4, while the midpoint error shrinks by roughly 16. The midpoint rule is not just a little better; it improves faster. We will see why in Section 19.12.

For this one function we can compute the sum exactly for every \(n\) and watch the limit happen. Use right endpoints: \(\Delta x = 2/n\) and \(x_i = 2i/n\), so \(f(x_i) = 4i^2/n^2\).

Check with \(n = 4\): \(\tfrac43 \times 1.25 \times 2.25 = 3.75\), matching the hand calculation. As \(n \to \infty\), both \(1/n\) terms vanish and

That is a genuine derivation of the area under a parabola. It also shows why we need something better: this needed a special formula for \(\sum i^2\), and a function like \(\sin x\) or \(e^{-x^2}\) has no such convenient formula. The Fundamental Theorem will remove the need for these sums entirely.

Riemann sums are not just a stepping stone to the integral; they are how totals are computed from measurements in practice.

The definite integral of \(f\) from \(a\) to \(b\) is the limit of Riemann sums as the rectangles become infinitely thin:

For continuous \(f\) (and many others) the limit exists and is the same whichever sample points you use, which is why left, right and midpoint sums all converged to the same 8/3.

The notation is a frozen picture of the Riemann sum.

So \(\int_0^2 x^2\,dx\) reads "the integral from 0 to 2 of \(x^2\) d-x" and means "add up \(x^2\) times a sliver of width, for all \(x\) from 0 to 2".

If \(f(x) < 0\), the "rectangle height" \(f(x_i^*)\) is negative, so that rectangle contributes negative area. The definite integral therefore counts area above the axis as positive and area below as negative. For example, \(\sin x\) is positive on \((0, \pi)\) and negative on \((\pi, 2\pi)\) with identical shapes, so

This is the right behaviour for many applications: if \(v(t)\) is a velocity (which can be negative when you reverse), \(\int v\,dt\) is your net displacement, and driving 5 miles forward then 5 back gives 0.

These follow directly from the rectangle picture, and you will use them constantly.

The first two together are called linearity. The last one is a definition that keeps the third rule true in all orders.

Computing \(\int_0^2 x^2\,dx\) from the limit took a special summation formula and some algebra. For \(\int_0^\pi \sin x\,dx\) the same route is miserable. We need a shortcut, and there is a spectacular one.

Fix a starting point \(a\) and let the right-hand end vary. Define the accumulation function

the area under \(f\) from \(a\) up to \(x\). (We use \(t\) inside so that \(x\) can be the moving endpoint; the dummy variable’s name does not matter.)

Ask a derivative question about it: how fast does \(A\) grow as \(x\) moves right? Picture yourself painting the region under the curve with a roller moving left to right. The rate at which paint goes on is the height of the strip under the roller right now. Tall curve, area grows quickly; curve near zero, area barely grows; curve negative, the signed area shrinks.

Move the endpoint from \(x\) to \(x + h\). The extra area is a thin strip of width \(h\) and height about \(f(x)\):

Divide by \(h\):

The left side is the difference quotient from Chapter 17. As \(h \to 0\) the strip gets thinner, the top of the strip flattens out (because \(f\) is continuous, its height across the strip approaches \(f(x)\)), and the approximation becomes exact. So

To make the "approximately" honest: on the strip, \(f\) lies between its minimum \(m_h\) and maximum \(M_h\), so the strip’s area lies between \(m_h h\) and \(M_h h\), and the difference quotient lies between \(m_h\) and \(M_h\). Continuity forces both to approach \(f(x)\) as \(h \to 0\), so the quotient is squeezed to \(f(x)\).

Part 1 says \(A\) is a function whose derivative is \(f\). Any such function is called an antiderivative of \(f\).

Suppose you have found some antiderivative \(F\) of \(f\) by any means, perhaps by recognising a derivative backwards. Both \(F\) and \(A\) have derivative \(f\), so their difference has derivative zero, and a function with zero derivative everywhere on an interval is constant (its graph has slope 0 everywhere, so it is flat). Hence

Find \(C\) by putting \(x = a\): the area from \(a\) to \(a\) is zero, so \(0 = A(a) = F(a) + C\), giving \(C = -F(a)\). Therefore \(A(x) = F(x) - F(a)\), and at \(x = b\):

This is astonishing when you first take it in. An infinite limit of sums (an area) is computed by evaluating some other function at two points. All the detail of the curve in between is already encoded in \(F\).

Example 1: the parabola again. An antiderivative of \(x^2\) is \(x^3/3\). Try it: Integrate x² from 0 to 2.

One line, versus the page of summation algebra earlier, and the same answer.

Example 2: one arch of the sine curve. Since \(\frac{d}{dx}(-\cos x) = \sin x\), an antiderivative of \(\sin x\) is \(-\cos x\). Try it: Paint one arch of sin.

The area under one arch of the sine wave is exactly 2. Nothing about the curve suggested a whole number; the theorem delivers it effortlessly.

Example 3: the car. Back to \(v(t) = 20t(3-t) + 10 = -20t^2 + 60t + 10\). Find an antiderivative term by term (each term: raise the power by one, divide by the new power, which we justify properly in the next section): Try it: The car: 120 miles.

Check by differentiating: \(S'(t) = -20t^2 + 60t + 10 = v(t)\). Good. Now evaluate:

The opening question is answered: the car travelled 120 miles. And notice what \(S(t)\) is: a position function, the distance travelled by time \(t\). The FTC is the formal statement that "integrating speed gives distance" and "differentiating distance gives speed" are inverse operations.

Part 1 is not only theory. It lets you differentiate functions defined by integrals, even when the integral has no formula:

If the upper limit is a function of \(x\), combine with the chain rule. With \(G(u) = \int_0^u \cos(t^2)\,dt\) and upper limit \(u = x^3\):

The FTC turns every definite integral into a search: find a function whose derivative is the integrand. So we need a toolkit for running derivative rules in reverse.

Why must \(+C\) be there? Because \(x^2\), \(x^2 + 1\) and \(x^2 - 5\) all have derivative \(2x\). Graphically, they are vertical shifts of the same curve and have identical slopes at every \(x\). Knowing only the slope, you cannot know the height. For a car: knowing your speed at every moment tells you how far you went, but not where you started. To pin down \(C\) you need one extra fact, an initial condition such as "at \(t = 0\) the car was at mile marker 40".

In a definite integral \(C\) always cancels, \((F(b) + C) - (F(a) + C) = F(b) - F(a)\), which is why we drop it there.

Each entry is a derivative fact read from right to left. You can check any of them by differentiating the right-hand side.

And the linearity rules carry over from derivatives: \(\int (f + g)\,dx = \int f\,dx + \int g\,dx\) and \(\int c f\,dx = c\int f\,dx\).

Example 1. Evaluate \(\int_1^3 (3x^2 - 4x + 5)\,dx\). Try it: Evaluate this integral.

Example 2. Find \(\displaystyle\int \Bigl(2\sqrt{x} - \frac{3}{x} + e^{2x}\Bigr)\,dx\).

First rewrite \(\sqrt{x} = x^{1/2}\). Then

Check: \(\frac{d}{dx}\bigl(\frac43 x^{3/2}\bigr) = \frac43\cdot\frac32 x^{1/2} = 2\sqrt{x}\); the other two terms differentiate back correctly too.

Example 3: using an initial condition. A function satisfies \(f'(x) = 6x - 2\) and \(f(1) = 5\). Find \(f\).

If you did UK Mechanics you memorised the constant-acceleration equations. They are just two integrations. Let acceleration be the constant \(a\), with initial velocity \(u\) and initial position 0 (the old notation used \(s\) for displacement and \(v\) for velocity).

Velocity is the antiderivative of acceleration:

Displacement is the antiderivative of velocity:

A ball thrown straight up at \(u = 20\text{ m/s}\) with \(a = -9.8\text{ m/s}^2\) has \(v(2) = 20 - 19.6 = 0.4\) m/s and \(s(2) = 40 - 19.6 = 20.4\) m. We will reproduce those numbers numerically in Section 19.13. The area under the velocity–time graph is the displacement, exactly as the Mechanics syllabus said, and now you know why.

What is \(\int 2x\cos(x^2)\,dx\)? It is not in the table, and it is a product, so linearity does not help. But stare at it: \(x^2\) sits inside the cosine, and \(2x\), the derivative of \(x^2\), sits outside. That pattern is the fingerprint of the chain rule.

The chain rule (Chapter 17) says

Read it backwards: if an integrand has the form \(f(g(x))\,g'(x)\) and \(F\) is an antiderivative of \(f\), then

Leibniz notation makes this almost automatic. Name the inner function \(u = g(x)\). Then \(\frac{du}{dx} = g'(x)\), which we write as the differential statement \(du = g'(x)\,dx\). The integrand’s \(g'(x)\,dx\) becomes \(du\) and the inner function becomes \(u\):

Example 1. \(\int 2x\cos(x^2)\,dx\). Try it: See u = x² stretch strips; Integrate 2x cos(x²).

Check: \(\frac{d}{dx}\sin(x^2) = \cos(x^2)\cdot 2x\). Correct.

Example 2: a missing constant. \(\int \frac{x}{x^2 + 1}\,dx\). Try it: Integrate x/(x² + 1).

Example 3: a definite integral with changed limits. \(\int_0^1 x(x^2 + 1)^3\,dx\). Try it: Substitution strips for Example 3; Evaluate this integral.

Alternative approach for Example 3. You could also find the indefinite integral \(\tfrac18(x^2+1)^4 + C\), back-substitute, and evaluate with the original limits: \(\tfrac18(2^4) - \tfrac18(1^4) = \tfrac{15}{8}\). Same answer. Converting the limits is usually less work and avoids one source of error; back-substituting is handy when you want the antiderivative anyway. A third alternative, expanding \((x^2+1)^3\) into a polynomial, works here but becomes hopeless for \((x^2+1)^{20}\) and impossible for \((x^2+1)^{1/2}\).

Substitution handles "function of a function times the inner derivative". It cannot do \(\int x e^x\,dx\): there is no inner function whose derivative is lying around. This is a product of two unrelated kinds of function, a polynomial and an exponential, and its reverse partner is the product rule.

The product rule says \((uv)' = u'v + uv'\). Integrate both sides:

Rearrange to isolate one of the integrals:

In differential shorthand, with \(du = u'\,dx\) and \(dv = v'\,dx\):

The formula trades one integral for another. It helps when the new integral \(\int v\,du\) is easier than the original. The trick is to choose \(u\) as the factor that gets simpler when differentiated (\(x \to 1\), \(\ln x \to 1/x\)) and \(dv\) as the factor you can easily integrate without it getting worse (\(e^x\), \(\sin x\)).

Example 1. \(\int x e^x\,dx\). Algebraic beats exponential in LIATE, so \(u = x\). Try it: Integrate x eˣ.

Check: \(\frac{d}{dx}\bigl(e^x(x-1)\bigr) = e^x(x - 1) + e^x\cdot 1 = x e^x\). Correct.

What if we had chosen the other way, \(u = e^x\), \(dv = x\,dx\)? Then \(v = x^2/2\) and the new integral is \(\int \frac{x^2}{2}e^x\,dx\), which is worse. That is the signal to swap.

Example 2: the "invisible" second factor. \(\int \ln x\,dx\). There seems to be only one factor, but write it as \(\ln x \cdot 1\). Try it: Integrate ln x.

Example 3: a definite integral. \(\int_0^{\pi} x\sin x\,dx\).

Example 4: going round in a circle on purpose. \(I = \int e^x\sin x\,dx\). Parts twice, with \(u\) the trig factor both times:

The original integral \(I\) has reappeared. Treat it as an unknown and solve: \(2I = e^x(\sin x - \cos x)\), so

Where you’d use this. Integration by parts is how averages of "\(x\) times a density" are computed. If the waiting time for the next request at a server has the exponential density \(\lambda e^{-\lambda x}\) (see Section 19.10), the mean wait is \(\int_0^\infty x\,\lambda e^{-\lambda x}\,dx\). Parts with \(u = x\), \(dv = \lambda e^{-\lambda x}dx\) gives \(\bigl[-xe^{-\lambda x}\bigr]_0^\infty + \int_0^\infty e^{-\lambda x}\,dx = 0 + \frac{1}{\lambda}\): at \(\lambda = 0.5\) requests per second, the average wait is 2 seconds. The same calculation gives the mean lifetime of a radioactive nucleus and the expected time to failure of a component in reliability engineering. Substitution, meanwhile, is how the normal distribution is standardised: \(z = (x - \mu)/\sigma\) turns every bell-curve integral into one with \(\mu = 0\), \(\sigma = 1\), which is why a single \(z\)-table (or one library function) serves them all.

Try it: Partial fractions of 1/(x² − 1).

In practice, professionals reach for a computer algebra system (SymPy, Mathematica) for hard symbolic integrals, and for numerical methods whenever only a number is needed. The point of learning substitution and parts is not to compete with software; it is to read the formulas in papers and textbooks (many derivations are a substitution or parts step in disguise) and to be able to check what software tells you.

Problem. Find the area enclosed between \(y = x\) and \(y = x^2\). Try it: Area between x and x².

Slice and approximate. The curves meet where \(x = x^2\), that is \(x(x - 1) = 0\), so at \(x = 0\) and \(x = 1\). Between them, \(x > x^2\) (try \(x = 0.5\): \(0.5 > 0.25\)), so a thin vertical slice at position \(x\) has height \(\text{top} - \text{bottom} = x - x^2\) and area \((x - x^2)\,\Delta x\).

Integrate.

In general, \(\text{Area} = \int_a^b \bigl(\text{top}(x) - \text{bottom}(x)\bigr)\,dx\). Because we subtract heights, it does not matter whether the curves are above or below the axis. If the curves cross inside the interval, split at the crossings so that "top" really is on top in each piece.

The average of \(n\) numbers is their sum divided by \(n\). For a function, sample it at \(n\) evenly spaced points and average:

The right side is a Riemann sum divided by \(b - a\). Letting \(n \to \infty\):

Example. The average value of \(\sin x\) over one arch \({[0, \pi]}\) is \(\frac{1}{\pi}\cdot 2 = \frac{2}{\pi} \approx 0.6366\). Electrical engineers use exactly this (and the related root-mean-square average) to relate peak and average AC voltages. The car’s average speed was \(\tfrac13\cdot 120 = 40\) mph.

Spin the curve \(y = f(x)\), \(a \le x \le b\), around the \(x\)-axis. The solid it sweeps out can be sliced like a salami into thin discs. The slice at \(x\) is a cylinder of radius \(f(x)\) and thickness \(\Delta x\), with volume \(\pi f(x)^2\,\Delta x\). Sum and integrate (the disc method): Try it: Spin a semicircle into a sphere.

Example: a sphere. The upper half of a circle of radius \(r\) is \(y = \sqrt{r^2 - x^2}\) for \(-r \le x \le r\). Spinning it gives a sphere:

The school formula for the volume of a sphere, which you probably memorised without proof, falls straight out.

In physics, the work done by a constant force \(F\) moving an object a distance \(d\) in the direction of the force is \(W = Fd\) (in joules when \(F\) is in newtons and \(d\) in metres). When the force varies with position, slice:

Example: stretching a spring. Hooke’s law says a spring stretched by \(x\) metres pulls back with force \(F = kx\). For \(k = 200\) N/m, the work to stretch it 0.1 m from rest is

Example: winding up a hanging chain. A 10 m chain of mass 2 kg/m hangs from a winch. The slice of chain at depth \(y\) (thickness \(\Delta y\)) weighs \(2 \times 9.8\,\Delta y\) newtons and must be lifted \(y\) metres, so it needs work \(19.6\,y\,\Delta y\). Total:

Here the force is constant but the distance varies by slice, a different variation from the spring. The slice method does not care which.

For a continuous random quantity, such as the waiting time until the next request hits a server, it makes no sense to ask for the probability of exactly 1.000000 seconds. Instead a probability density function \(p(x)\) describes how probability is spread out, and probabilities are areas:

Example. Waiting times are often modelled by the exponential density \(p(x) = \lambda e^{-\lambda x}\) for \(x \ge 0\). With \(\lambda = 0.5\) per second, the probability of waiting at most 1 second is

The integral of the density over \({[0, \infty)}\) is \(1\), an improper integral (next section). The whole of Chapter 30 is built on this idea, including the normal (bell) curve, whose integral is the famous non-elementary \(\int e^{-x^2}\,dx\).

The car example generalises: the integral of a rate of change gives the total change. If \(Q'(t)\) is a rate,

This is FTC Part 2 read physically, and it is sometimes called the net change theorem. Water flowing into a tank at \(r(t) = 10 + 2t\) litres per minute adds \(\int_0^5 (10 + 2t)\,dt = \big[10t + t^2\big]_0^5 = 50 + 25 = 75\) litres in the first 5 minutes. The same pattern turns power (watts) into energy (joules or kWh), a bandwidth curve (bytes per second) into total data transferred, and a births-minus-deaths rate into a population change.

Can a region that stretches out to infinity have finite area? Our definition used a closed interval \({[a,b]}\), so we need to extend it. And we need to, because probability densities live on unbounded intervals.

Integrate up to a finite cut-off \(b\), then let \(b \to \infty\):

If the limit exists and is finite, the integral converges; otherwise it diverges.

Example 1: \(\int_1^\infty \frac{1}{x^2}\,dx\). Try it: ∫ 1/x² to infinity; ∫ 1/x to infinity.

It converges: the infinitely long region has area exactly 1.

Example 2: \(\int_1^\infty \frac{1}{x}\,dx\).

It diverges. Both curves fall to zero, and \(1/x\) is only a little larger than \(1/x^2\) far out, but \(1/x\) doesn’t fall fast enough: the area keeps growing (slowly, like a logarithm) forever.

The general rule (the p-test): \(\int_1^\infty x^{-p}\,dx\) converges exactly when \(p > 1\), with value \(\frac{1}{p - 1}\). The borderline case \(p = 1\) diverges.

Example 3: an integrand that blows up. \(\int_0^1 \frac{1}{\sqrt{x}}\,dx\). The integrand is infinite at \(x = 0\), so cut off at \(c > 0\): Try it: A spike with finite area; Evaluate this improper integral.

A spike to infinity with finite area 2. (For \(\int_0^1 \frac1x\,dx\) the same calculation gives \(-\ln c \to \infty\): divergent.)

Try it: Gabriel’s horn’s volume.

Antiderivatives fail us in three common situations:

In all three we want a number, and the Riemann-sum idea gives it directly. The question is how to get high accuracy from few function evaluations. This is called numerical integration or quadrature.

Intuition. Instead of flat-topped rectangles, join neighbouring points on the curve with straight lines. Each strip becomes a trapezoid, which hugs a smooth curve much better than a rectangle.

Why it works. A trapezoid with parallel sides \(f(x_{i-1})\) and \(f(x_i)\) and width \(h\) has area \(\frac{h}{2}\bigl(f(x_{i-1}) + f(x_i)\bigr)\). Add the \(n\) strips. Every interior point is shared by two neighbouring trapezoids, so it is counted twice; the two endpoints are counted once:

It is also exactly the average of the left and right Riemann sums.

Intuition. Straight lines cannot follow curvature. Parabolas can. Take the points in groups of three, fit the unique parabola through each triple, and integrate the parabola exactly.

Why it works. Shift a triple so the middle node is at 0: nodes \(-h, 0, h\) with heights \(y_0, y_1, y_2\). For a parabola \(q(x) = Ax^2 + Bx + C\) through them,

The \(Bx\) term integrates to zero (odd function). From the three points, \(q(0) = C = y_1\) and \(y_0 + y_2 = q(-h) + q(h) = 2Ah^2 + 2C\), so \(2Ah^2 = y_0 + y_2 - 2y_1\). Substitute:

The middle point gets weight 4, the ends weight 1. Chaining pairs of strips across \({[a,b]}\) (so \(n\) must be even) gives weights \(1, 4, 2, 4, 2, \dots, 4, 1\):

We know \(\int_1^2 \frac{1}{x}\,dx = \ln 2 \approx 0.693147\). Pretend we don’t, and use \(n = 4\), \(h = 0.25\). Try it: Trapezoid and Simpson for ln 2.

Nodes and values:

Trapezoid:

Error: \(0.697024 - 0.693147 = 0.003877\).

Simpson:

Error: \(0.693254 - 0.693147 = 0.000107\). The same five function values give a result 36 times more accurate, just by weighting them more cleverly.

Try it: Watch the errors shrink.

Read the columns as ratios. Each doubling of \(n\) divides the trapezoid error by about 4 (\(0.1039 / 0.02577 \approx 4.03\)) and the Simpson error by about 16 (\(2.692\times10^{-4} / 1.659\times10^{-5} \approx 16.2\)). That is what "error \(\propto h^2\)" and "error \(\propto h^4\)" mean in practice. On a log–log plot those become straight lines of slope \(-2\) and \(-4\); the left Riemann sum has slope \(-1\).

Try it: The log–log error plot.

Where you’d use this: drug exposure. In clinical trials, blood is sampled at a few times after a dose and the area under the curve (AUC) of concentration measures the body’s total exposure to the drug. With samples at 0, 1, 2, 4 and 8 hours of 0, 8, 10, 6 and 2 mg/L (uneven spacing, so trapezoids), the AUC is \(\tfrac{0 + 8}{2}(1) + \tfrac{8 + 10}{2}(1) + \tfrac{10 + 6}{2}(2) + \tfrac{6 + 2}{2}(4) = 4 + 9 + 16 + 16 = 45\) mg·h/L. Regulators compare generic and brand-name drugs by exactly this number. The same trapezoid rule turns a GPS speed log into the distance on a running app, and a sampled power trace into the energy a server consumed.

So far the rate of change was a known function of time: \(v(t)\) was given, and we integrated to get position. Often, though, the rate depends on the current amount. A population grows faster when there are more organisms. A hot cup of coffee cools faster when it is much hotter than the room. A falling object’s acceleration depends on its velocity through air resistance. Equations relating a quantity to its own rate of change are called differential equations, and they are how science writes down the laws of change.

Intuition. The rate of growth is proportional to the current size: double the population, double the births per year. You already know a function with that property: \(e^t\) is its own derivative (Chapter 17).

Solving by separation of variables. Move all the \(y\) to one side and all the \(t\) to the other (treating \(dy\) and \(dt\) as Leibniz did, which can be justified by the chain rule), then integrate both sides:

Check: \(\frac{d}{dt}\bigl(y_0 e^{kt}\bigr) = k\,y_0 e^{kt} = ky\). Correct.

Example. Bacteria double every 20 minutes. Then \(e^{20k} = 2\), so \(k = \frac{\ln 2}{20} \approx 0.03466\) per minute. After 2 hours (6 doubling periods) the population is \(e^{120k} = e^{6\ln 2} = 2^6 = 64\) times its initial size. With \(k < 0\) the same equation describes radioactive decay, drug elimination and (via Newton’s law of cooling, \(\frac{dT}{dt} = -k(T - T_{\text{room}})\)) that cooling coffee. See Chapter 12 for more on exponential models.

\(F = ma\) says acceleration (the second derivative of position) is force divided by mass:

When \(F\) is constant we integrated twice in Section 19.5 to get the suvat equations. When \(F\) depends on position or velocity (springs, drag, gravity between planets), exact solutions are rare and we must step forward numerically.

Intuition. If you know where you are and how fast you are changing right now, assume the rate stays constant for a short time \(\Delta t\) and step forward. Then recompute the rate and repeat. It is a left Riemann sum built on the fly, where each rectangle’s height depends on the result so far.

Worked example by hand. \(\frac{dy}{dt} = 0.5y\), \(y(0) = 1\), step \(\Delta t = 1\): Try it: Step through this example.

The exact solution is \(y(4) = e^{2} \approx 7.389\). Each step multiplies by 1.5 instead of \(e^{0.5} \approx 1.6487\), so Euler lags behind because it uses the rate at the start of each step, while the true rate keeps increasing during the step. Smaller steps fix this, as the figure shows.

For motion we track two quantities, position and velocity, each with its own rate: \(\frac{dy}{dt} = v\) and \(\frac{dv}{dt} = a = F/m\). Try it: Compare the three integrators.

The velocity is exact (its rate is constant, so a left sum is perfect), but the position is off by nearly a metre: each step uses the velocity from the start of the step, which is always a little too high on the way up.

Try it: Euler vs symplectic vs Verlet.

Try it: Watch Euler explode.

How many times does the inner statement run in this loop? Try it: The loop as a staircase.

The exact answer is \(\sum_{i=1}^{n} i = \frac{n(n+1)}{2}\), which is 36 for \(n = 8\). For algorithm analysis we usually care about growth: \(\frac{n(n+1)}{2} \approx \frac{n^2}{2}\), so the loop is \(O(n^2)\). Notice that \(\int_0^n x\,dx = \frac{n^2}{2}\) gives the leading term immediately.

A sum \(\sum_{k=1}^{n} f(k)\) is a Riemann sum with width 1. If \(f\) is increasing, comparing each unit rectangle with the area under the curve on either side gives the integral bounds

(For decreasing \(f\) the inequalities flip.) The sum is trapped between two integrals that are easy to compute, so it grows at the same rate as the integral.

For the harmonic sum, \(1/x\) is decreasing, so the bounds read \(\ln(n+1) \le H_n \le 1 + \ln n\). For \(n = 1000\) that is \(6.909 \le H_{1000} \le 7.908\); the true value is about \(7.485\). Try it: Bound the harmonic sum.

The last row is worth a moment. \(\int_1^n \ln x\,dx\) uses the integration-by-parts result from Section 19.8, and gives \(\log_2(n!) \approx n\log_2 n\). Since a comparison sort must distinguish all \(n!\) orderings and each comparison gives one bit, no comparison sort can beat roughly \(n\log_2 n\) comparisons. An integration-by-parts exercise proves a fundamental limit of computer science. Try it: ln n! against an integral.