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1

Area so far

An accumulation, read as a function of its endpoint

Our running example is the robot on a line. Let $f(t)$ be its speed in metres per second at time $t$, and let the robot start at position $0$ at time $0$. The distance travelled by time $x$ is the area under the speed curve from $0$ to $x$. That area is not a single number — it depends on where you stop, so it is a function:

$$ A(x) \;=\; \int_a^x f(t)\,dt. $$

Here we take $a=0$ and $f(t)=3t^2$, so the area has a known closed form in reserve. Pick an endpoint $x$ and watch the shaded region grow; the lower panel traces $A(x)$ out point by point, each value obtained only by adding up thin strips — never by consulting a formula. The increment slider changes the width of the strip at the right edge and shows how much new area it contributes.

Top: speed $f(t)=3t^2$ with the travelled area shaded, and the current strip highlighted. Bottom: the accumulated distance $A(x)$, built numerically from the curve above.

💡 The whole theorem lives here: the bottom curve has a slope at every $x$, and that slope turns out to be the height of the top curve at $x$.
2

The reveal

The thin strip is $f(x)\,dx$, so the slope of the area is the speed

Part 1 of the Fundamental Theorem. If $f$ is continuous on $[a,b]$ and $A(x)=\int_a^x f(t)\,dt$, then $A$ is differentiable on $(a,b)$ and

$$ A'(x) \;=\; f(x). $$

Here is the argument in one line. Nudge the endpoint from $x$ to $x+dx$. The area gains only the thin strip between the two vertical lines, a near-rectangle of height $f(x)$ and width $dx$:

$$ A(x+dx) - A(x) \;\approx\; \underbrace{f(x)\,dx}_{\text{the strip}} \qquad\Longrightarrow\qquad \frac{A(x+dx)-A(x)}{dx} \;\to\; f(x). $$

The top panel plots $f(x)$ and the numerically differentiated area curve from Step 1 on the same axes. They are the same curve because differentiating undoes accumulating. The lower panel is the numerical story: shrink the step $dx$ and the finite-difference estimate falls onto $f(x)$ — until roundoff and accumulation error are amplified and the error floor appears. There is a best step size, and it is not zero.

Top: $f(x)$ (solid) and $\dfrac{dA}{dx}$ read off the accumulated curve (dashed) — indistinguishable. Bottom: the error $|\,\hat A{}'(x)-f(x)\,|$ against step size $h$ on log axes.

3

The odometer

Speed is the derivative of odometer; odometer is the integral of speed

An odometer is the Fundamental Theorem bolted to a wheel. At each instant it adds the tiny distance $f(t)\,dt$, and the total it shows is the accumulated area. Play the cart and watch three numbers agree: the instantaneous speed $f(t)$, the area accumulated so far, and the closed-form distance $t^3$. The speed curve is the one from Step 1; the odometer is the curve from Step 2's lower panel, read as a position instead of a graph.

The cart's speed at time $t$ is $f(t)=3t^2$; its odometer shows $\int_0^t f$, accumulated one frame at a time.

⚠️ The odometer never needs to know "which constant" — it only accumulates differences. That freedom is exactly the $+C$ you will meet the moment you start writing antiderivatives down.
4

Both directions

Evaluate an antiderivative and subtract

Part 2 of the Fundamental Theorem. If $F$ is any antiderivative of $f$ on $[a,b]$ — that is, $F' = f$ — then

$$ \int_a^b f(x)\,dx \;=\; F(b) - F(a). $$

Part 1 said that $F(x)=\int_a^x f$ is an antiderivative. Part 2 says every antiderivative differs from it by a constant, so any of them can evaluate a definite integral by subtracting endpoints. Slide $a$ and $b$: the shaded area on top changes, and the vertical rise of the cubic on the bottom changes by exactly the same amount. The numeric Riemann sum and the endpoint subtraction are computed independently and agree.

Top: $\int_a^b 3t^2\,dt$ as shaded area. Bottom: the rise $F(b)-F(a)$ of $F(t)=t^3$ across the same interval.

5

Where this shows up

Why integration stops being a limiting process

Part 2 is the reason you rarely compute an integral as a limit of sums. Finding an antiderivative converts a process into a subtraction, and that is what every technique in Part 10 is for: substitution and integration by parts are just the chain and product rules run backwards to recover the antiderivative you need. In Volume II, Part 12, the same statement becomes the link between a differential equation and its solution — $y' = f$ is solved by $y=\int f$ — while when no closed form exists the numeric accumulators from Step 1 are what actually run. The theorem also underwrites probability, where a density $p$ integrates to a distribution function and probabilities are read off by subtraction in Volume II, Part 11.

6

Antiderivatives worth knowing

$f(x)$$\int f(x)\,dx$Where it comes up
xⁿ, n ≠ −1xⁿ⁺¹ / (n+1) + CPolynomials everywhere; the term-by-term rule
1 / xln |x| + CLogarithms, relative change, entropy terms
eˣeˣ + CGrowth and decay, exponential families
cos xsin x + COscillators, Fourier analysis
sin x−cos x + COscillators, phase shifts
sec²xtan x + CTrigonometric substitutions
1 / (1 + x²)arctan x + CAngles, the Cauchy distribution
k (constant)kx + CUniform density and constant rates
7

Further reading

8

Check your understanding

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