Integrals

01Antiderivatives: differentiation in reverse

An antiderivative of $f$ is any function $F$ with $F' = f$. Since $\frac{d}{dx}x^2 = 2x$, the function $x^2$ is an antiderivative of $2x$ — and so are $x^2 + 5$ and $x^2 - 17$, since constants die under differentiation. The MVT corollary from Module 08 guarantees that's the only ambiguity: any two antiderivatives differ by a constant. Hence the notation and its famous tail:

$$\int f(x)\,dx = F(x) + C$$

The symbol $\int f(x)\,dx$ is the indefinite integral — the whole family of antiderivatives — and omitting the $+C$ is the most reliably penalized error in Calc 1.

02The basic integration formulas

Run every derivative formula backwards. The reverse power rule: raise the exponent by one, divide by the new exponent:

$$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \quad (n \ne -1) \qquad\qquad \int \frac{1}{x}\,dx = \ln|x| + C$$

The $n = -1$ case is exactly the gap $\ln x$ fills — pleasingly, the one power the rule can't handle is covered by the one non-power derivative that produces a power. The rest of the card:

$$\int e^x dx = e^x + C \qquad \int \sin x\,dx = -\cos x + C \qquad \int \cos x\,dx = \sin x + C \qquad \int \sec^2 x\,dx = \tan x + C$$

Sign check for trig: differentiate your answer and see if you land back on the integrand — a five-second verification that catches the $\pm\cos$ confusion every time. Linearity works as before: integrate sums term by term and slide constants out front.

03Initial value problems

The $+C$ isn't decoration — it's a dial, and one data point sets it. If $f'(x) = 6x$ and $f(1) = 4$: integrate to get $f(x) = 3x^2 + C$, then impose the condition: $3 + C = 4$, so $C = 1$ and $f(x) = 3x^2 + 1$. Physics runs this twice: acceleration integrates to velocity (constant set by initial velocity), velocity integrates to position (constant set by initial position). A projectile with $a = -32$ ft/s², launched upward at $64$ ft/s from height $80$ ft, has $v(t) = -32t + 64$ and $s(t) = -16t^2 + 64t + 80$ — every constant earned from an initial condition.

04The area problem and Riemann sums

Now the second, apparently unrelated, story. To measure the area under $y = f(x)$ over $[a, b]$, slice the interval into $n$ strips of width $\Delta x = \frac{b-a}{n}$, erect a rectangle on each (height sampled at, say, each strip's left edge, right edge, or midpoint), and add them up:

$$\text{Area} \approx \sum_{i=1}^{n} f(x_i^*)\,\Delta x$$

That's a Riemann sum — the hero graph shows six right-endpoint rectangles under a curve. Coarse rectangles over- or under-shoot, but as $n \to \infty$ the error dies, and the limit is the definite integral:

$$\int_a^b f(x)\,dx = \lim_{n\to\infty}\sum_{i=1}^{n} f(x_i^*)\,\Delta x$$

Continuous functions (Module 03, one last payoff) are guaranteed to have this limit exist.

05What the definite integral means

$\int_a^b f(x)\,dx$ is a number, not a family of functions — note the contrast with the indefinite integral. It computes signed area: regions below the $x$-axis count negative, so $\int_0^{2\pi}\sin x\,dx = 0$ by symmetry. More broadly it accumulates anything: if $v(t)$ is velocity, $\int_a^b v\,dt$ is displacement (net change in position); if $r(t)$ is a flow rate in liters/min, the integral is total liters. Useful properties fall straight out of the picture: $\int_a^b = \int_a^c + \int_c^b$ (split the region), $\int_a^a = 0$, and swapping the bounds flips the sign.

06The Fundamental Theorem of Calculus

Here is the punchline of the entire course: the area problem and the antiderivative problem are the same problem. FTC Part 2 (the evaluation theorem): if $F$ is any antiderivative of a continuous $f$, then

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

No rectangles, no limits — find an antiderivative and subtract. Example: $\int_0^2 x^2\,dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{8}{3}$. Exact area under a parabola, three symbols of work. FTC Part 1 runs the bridge the other way: the accumulation function $g(x) = \int_a^x f(t)\,dt$ is an antiderivative of $f$, i.e. $g'(x) = f(x)$ — differentiating an integral hands back the integrand. Together the parts say differentiation and integration are inverse operations. Note the $+C$ vanishes in definite integrals: it cancels in the subtraction, so leave it out.

07u-substitution: the chain rule in reverse

Most integrands aren't on the formula card — they're chain-rule outputs, and substitution un-chains them. To compute $\int 2x(x^2+1)^5\,dx$: set $u = x^2 + 1$, so $du = 2x\,dx$, and the integral collapses to $\int u^5\,du = \frac{u^6}{6} + C = \frac{(x^2+1)^6}{6} + C$. The skill is spotting a function and its derivative sharing the integrand — the $2x$ standing next to $(x^2+1)^5$ is the fingerprint. If the constant is off by a factor, adjust: $\int x\,e^{x^2}dx$ has $du = 2x\,dx$, so write it as $\frac12\int e^u du = \frac12 e^{x^2} + C$. For definite integrals, convert the bounds to $u$-values and never return to $x$ at all.

08Area between the FTC and the future

Two closing applications. Exact areas: the area under $\sin x$ from $0$ to $\pi$ is $\int_0^\pi \sin x\,dx = [-\cos x]_0^\pi = 1 - (-1) = 2$ — a transcendental curve enclosing a perfectly clean area, courtesy of FTC. Net change: FTC rewritten as $F(b) = F(a) + \int_a^b F'(t)\,dt$ says any quantity's final value is its initial value plus the accumulated rate — the sentence that runs physics, economics, and biology. Everything in Calc 2 (integration techniques, volumes, series) grows from this module; if the FTC feels almost too convenient, that's the correct response to one of mathematics' genuinely great theorems.