Integrals

Divide the interval $[a,b]$ into $n$ thin strips of width $\Delta x = (b-a)/n$. In each strip, build a rectangle with height $f(x_i)$.

The total area of all rectangles is approximately $\sum_{i=1}^{n} f(x_i)\Delta x$. This is a Riemann sum.

As $n \to \infty$ (strips get infinitely thin), the Riemann sum converges to the definite integral: $\int_a^b f(x)\,dx = \lim_{n\to\infty} \sum_{i=1}^{n} f(x_i)\Delta x$.

This is why integrals represent accumulation: you are adding up infinitely many infinitesimally small contributions.

FTC Part 1: if $F(x) = \int_a^x f(t)\,dt$, then $F'(x) = f(x)$. Differentiation and integration are inverse operations.

FTC Part 2: $\int_a^b f(x)\,dx = F(b) - F(a)$ where $F$ is any antiderivative of $f$ (meaning $F'=f$).

Part 2 is the computational powerhouse: instead of computing limits of Riemann sums, just find an antiderivative and evaluate at the bounds.

The connection to derivatives: if position is $s(t)$ and velocity is $v(t) = s'(t)$, then total displacement is $\int_a^b v(t)\,dt = s(b) - s(a)$. Integration recovers position from velocity.

The definite integral $\int_a^b f(x)\,dx$ computes net (signed) area: regions where $f(x) > 0$ (above the $x$-axis) contribute positively, and regions where $f(x) < 0$ (below the $x$-axis) contribute negatively. Positive and negative areas can cancel.

If you want total area (always positive, no cancellation), integrate the absolute value: $\int_a^b |f(x)|\,dx$.

In practice, find where $f(x) = 0$ (the zeros), split the integral at those zeros, and negate the integral on sub-intervals where $f < 0$. This turns every piece positive before adding.

Example: $\int_0^{2\pi} \sin x\,dx = 0$ because the positive area on $[0,\pi]$ exactly cancels the negative area on $[\pi,2\pi]$. But the total area is $\int_0^\pi \sin x\,dx + \int_\pi^{2\pi} |\sin x|\,dx = 2 + 2 = 4$.

Physical analogy: if velocity is positive (forward) for a while and then negative (backward), the integral of velocity gives displacement (net distance from start), which could be small or zero. The integral of $|v|$ gives total distance traveled.

An antiderivative of $f(x)$ is any function $F(x)$ such that $F'(x) = f(x)$. The indefinite integral $\int f(x)\,dx$ represents the entire family of antiderivatives.

Since the derivative of a constant is $0$, any two antiderivatives of the same function differ by a constant: if $F'(x) = f(x)$ and $G'(x) = f(x)$, then $F(x) - G(x) = C$ for some constant. That's why every indefinite integral includes $+C$.

The $+C$ is not a formality — it's essential. In applications, the value of $C$ is determined by an initial condition. For instance, if an object has velocity $v(t) = 3t^2$ and position $s(0) = 5$, then $s(t) = t^3 + C$ and $s(0) = C = 5$, so $s(t) = t^3 + 5$. Without $+C$, you'd get $s(t) = t^3$ and miss that the object started at position 5.

On definite integrals, the $+C$ cancels: $F(b) + C - (F(a) + C) = F(b) - F(a)$. So you only need $+C$ for indefinite integrals.

These are the integral formulas you must know by heart. Every other technique (substitution, by-parts, etc.) ultimately reduces an integral to one of these:

Power rule: $\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$. Works for any real $n$ — including negative and fractional exponents like $\int x^{-3}\,dx$ or $\int \sqrt{x}\,dx = \int x^{1/2}\,dx$.

The missing case: $\int \frac{1}{x}\,dx = \ln|x| + C$. The power rule gives $\frac{x^0}{0}$, which is undefined, so $\ln|x|$ fills this gap. The absolute value is needed because $\ln$ requires a positive input.

Exponentials: $\int e^x\,dx = e^x + C$ and $\int a^x\,dx = \frac{a^x}{\ln a} + C$ for $a > 0, a \neq 1$.

Trigonometric: $\int \sin x\,dx = -\cos x + C$, $\int \cos x\,dx = \sin x + C$, $\int \sec^2 x\,dx = \tan x + C$, $\int \csc^2 x\,dx = -\cot x + C$, $\int \sec x \tan x\,dx = \sec x + C$, $\int \csc x \cot x\,dx = -\csc x + C$.

Inverse trig: $\int \frac{1}{\sqrt{1-x^2}}\,dx = \arcsin x + C$ and $\int \frac{1}{1+x^2}\,dx = \arctan x + C$. These arise constantly after trig substitutions and partial fractions.

Verification: you can always check an antiderivative by differentiating it. If $\frac{d}{dx} F(x) = f(x)$, then $\int f(x)\,dx = F(x) + C$.

If the integrand looks like $f(g(x)) \cdot g'(x)$, substitute $u = g(x)$ so $du = g'(x)\,dx$.

The integral becomes $\int f(u)\,du$, which is usually simpler.

Example: $\int 2x \cos(x^2)\,dx$. Let $u = x^2$, $du = 2x\,dx$. Integral becomes $\int \cos u\,du = \sin u + C = \sin(x^2) + C$.

For definite integrals, either convert the bounds to $u$-values, or back-substitute and use the original bounds.

Tip: look for an "inner function" whose derivative (or a constant multiple of it) appears as a factor in the integrand.

Formula: $\int u\,dv = uv - \int v\,du$. This comes from the product rule run backwards.

Choose $u$ and $dv$ using the LIATE heuristic: Logarithms, Inverse trig, Algebraic ($x^n$), Trig, Exponentials. Pick $u$ from higher in this list.

Example: $\int x e^x\,dx$. Let $u = x$ (algebraic), $dv = e^x\,dx$. Then $du = dx$, $v = e^x$.

Result: $x e^x - \int e^x\,dx = x e^x - e^x + C = e^x(x-1) + C$.

Sometimes you need to apply by-parts twice (e.g., $\int x^2 e^x\,dx$) or use the tabular method for repeated applications.

A special case: $\int e^x \sin x\,dx$ requires by-parts twice, then solving for the integral algebraically.

Integrals of the form $\int \sin^m x \cos^n x\,dx$ come up constantly and have a systematic strategy based on whether $m$ and $n$ are odd or even.

Case 1 — one power is odd: peel off one factor of the odd-powered function, convert the remaining even power using $\sin^2 x + \cos^2 x = 1$, and substitute with $u$ = the other function. Example: $\int \cos^3 x\,dx$. Peel: $\int \cos^2 x \cdot \cos x\,dx = \int(1-\sin^2 x)\cos x\,dx$. Let $u = \sin x$: $\int(1-u^2)\,du = u - u^3/3 + C = \sin x - \frac{\sin^3 x}{3} + C$.

Case 2 — both powers are even: use half-angle identities to reduce. $\sin^2 x = \frac{1-\cos(2x)}{2}$ and $\cos^2 x = \frac{1+\cos(2x)}{2}$. Repeat as needed. These can be tedious but are completely mechanical.

Other trig integrals: $\int \tan^n x\,dx$ and $\int \sec^n x\,dx$ have their own patterns. For $\tan$, peel off $\tan^2 x = \sec^2 x - 1$ and reduce. For $\sec$, use integration by parts or reduction formulas.

Key results worth knowing: $\int \tan x\,dx = -\ln|\cos x| + C = \ln|\sec x| + C$ and $\int \sec x\,dx = \ln|\sec x + \tan x| + C$ (this last one is famously non-obvious).

When the integrand involves $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$, or $\sqrt{x^2 - a^2}$, no algebraic trick will eliminate the square root. Trig substitution works by exploiting Pythagorean identities to remove the radical.

$\sqrt{a^2 - x^2}$: let $x = a\sin\theta$, $dx = a\cos\theta\,d\theta$. Then $\sqrt{a^2-x^2} = a\cos\theta$. (Think: $1 - \sin^2\theta = \cos^2\theta$.)

$\sqrt{a^2 + x^2}$: let $x = a\tan\theta$, $dx = a\sec^2\theta\,d\theta$. Then $\sqrt{a^2+x^2} = a\sec\theta$. (Think: $1 + \tan^2\theta = \sec^2\theta$.)

$\sqrt{x^2 - a^2}$: let $x = a\sec\theta$, $dx = a\sec\theta\tan\theta\,d\theta$. Then $\sqrt{x^2-a^2} = a\tan\theta$. (Think: $\sec^2\theta - 1 = \tan^2\theta$.)

After substituting, the radical vanishes, leaving a trig integral. Evaluate it, then convert back to $x$ using a reference triangle: draw a right triangle where the side lengths are chosen so $\sin\theta$, $\cos\theta$, or $\tan\theta$ equals $x/a$ (or the appropriate ratio).

Recognition tip: the expression doesn't have to be a clean $\sqrt{a^2-x^2}$. Complete the square first if needed. For instance, $\sqrt{6x-x^2} = \sqrt{9-(x-3)^2}$ after completing the square, and now the $a^2 - u^2$ pattern is visible with $u = x-3$, $a = 3$.

To integrate a rational function $\frac{P(x)}{Q(x)}$ where $\deg P < \deg Q$, decompose it into simpler fractions.

Factor the denominator, then write: $\frac{1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$.

Solve for $A$ and $B$ by clearing denominators and comparing coefficients or plugging in strategic $x$ values.

Repeated factors: $\frac{1}{(x-1)^2}$ requires terms $\frac{A}{x-1} + \frac{B}{(x-1)^2}$.

Irreducible quadratics: for $(x^2+1)$ in the denominator, the numerator is $Ax+B$, not just $A$.

If $\deg P \geq \deg Q$, perform polynomial long division first, then decompose the remainder.

An improper integral has an infinite limit of integration or an integrand that blows up within the interval. In either case, you can't just plug in bounds — you need a limit.

Type 1 — infinite bound: replace $\infty$ with a variable and take a limit. $\int_1^{\infty} \frac{1}{x^2}\,dx = \lim_{b\to\infty} \int_1^b \frac{1}{x^2}\,dx = \lim_{b\to\infty} [-1/x]_1^b = \lim_{b\to\infty}(-1/b+1) = 1$. Converges.

Type 2 — infinite discontinuity: the integrand blows up at a point in $[a,b]$. Approach the bad point with a limit. $\int_0^1 \frac{1}{\sqrt{x}}\,dx = \lim_{a\to 0^+} \int_a^1 x^{-1/2}\,dx = \lim_{a\to 0^+} [2\sqrt{x}]_a^1 = 2 - 0 = 2$. Converges.

If the limit is finite, the improper integral converges. If the limit is $\pm\infty$ or does not exist, it diverges.

Key result: $\int_1^{\infty} \frac{1}{x^p}\,dx$ converges if $p > 1$ and diverges if $p \leq 1$. This is the continuous analog of the p-series test ($\sum 1/n^p$) and serves as a benchmark for comparison.

Comparison test for improper integrals: if $0 \leq f(x) \leq g(x)$ and $\int g$ converges, then $\int f$ converges. If $\int f$ diverges, then $\int g$ diverges. Same logic as for series.

When facing an unfamiliar integral, work through this decision tree in order:

1. Simplify first: can algebra, trig identities, or long division reduce it? Expanding a product or splitting a fraction may reveal a standard form. Never skip this step.

2. Check the table: is it already a standard form you know? $\int \sec^2 x\,dx$, $\int e^x\,dx$, $\int 1/(1+x^2)\,dx$, etc. If so, write down the answer immediately.

3. Substitution ($u$-sub): is there a function composition $f(g(x))$ with $g'(x)$ (or a constant multiple of it) present in the integrand? If yes, let $u = g(x)$.

4. Trig integrals: powers of $\sin$ and $\cos$ (or $\tan$/$\sec$)? Use the odd/even strategies and trig identities.

5. Trig substitution: see $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$, or $\sqrt{x^2 - a^2}$? Match the pattern to the right substitution.

6. Integration by parts: product of two different types (e.g., $x \cdot e^x$, $x^2 \cdot \sin x$, $\ln x \cdot x^n$)? Use LIATE to choose $u$.

7. Partial fractions: rational function $P(x)/Q(x)$ where $Q$ can be factored? Decompose and integrate each piece.

8. If none of these work: try a creative rewrite (multiply by $1$ in a clever form, add and subtract a term, complete the square) and re-enter the decision tree. Practice is the only way to develop speed and intuition.

Integration extends naturally from 'area under one curve' to 'area between two curves' and 'volume of a solid.' The strategy is always the same: slice, approximate each slice, and integrate.

Area between curves: the area between $y = f(x)$ (top) and $y = g(x)$ (bottom) from $x = a$ to $x = b$ is $\int_a^b [f(x) - g(x)]\,dx$. First determine which function is on top in each sub-interval. If the curves cross, split the integral at the crossing points.

Volume by disks/washers: revolve a region around the $x$-axis. Each cross-section is a disk (or washer if there's a hole). $V = \pi\int_a^b [f(x)]^2\,dx$ for a solid disk of radius $f(x)$. For a washer: $V = \pi\int_a^b ([R(x)]^2 - [r(x)]^2)\,dx$ where $R$ is the outer radius and $r$ the inner.

Volume by cylindrical shells: revolve around the $y$-axis. Instead of slicing perpendicular to the axis, wrap thin cylindrical shells around it. $V = 2\pi\int_a^b x \cdot f(x)\,dx$.

These topics are expanded in the Applications of Integration module. The key insight here: integration is not just about area — it's a universal tool for accumulating any quantity that can be sliced into thin pieces.