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Axioma

Mathematics · Intermediate · 40 min · Chapter 7 of 28

Integral calculus

The integral accumulates. The fundamental theorem says differentiation and integration are inverse operations.

Read first: Differential calculus

The idea

Cut the interval [a, b] into thin pieces of width Δx. On each piece pick a sample height f(xᵢ) and add f(xᵢ) Δx. That sum is a Riemann sum. The definite integral ∫ from a to b of f(x) dx is the limit of these sums as the mesh goes to zero, when the limit exists. Continuous functions on a closed interval always qualify. Area below the axis contributes negatively, so the integral is signed area, not paint.

An antiderivative of f is any function F with F′ = f. Antiderivatives differ by a constant, because the derivative of a constant is zero. The fundamental theorem has two faces. If you define G(x) as the integral of f from a to x, then G′ = f. And if F is any antiderivative, the definite integral from a to b equals F(b) − F(a). Differentiation and definite integration undo each other.

Technique is pattern matching against the derivative rules. The power rule backwards says the integral of xⁿ is xⁿ⁺¹/(n + 1) for n ≠ −1, and the integral of 1/x is ln|x|. Substitution handles the chain rule: if you see g′(x) multiplied by something made of g(x), set u = g(x). Integration by parts undoes the product rule and is the tool for products the substitution will not eat. None of this is required to understand what the integral means.

Hold onto this

  1. A definite integral is the signed area under a curve, defined as a limit of Riemann sums.
  2. The fundamental theorem recovers that area from any antiderivative.
  3. Substitution is the chain rule run backwards.
  4. Signed area counts region below the axis as negative.

Formulas

Fundamental theorem

Evaluate any antiderivative at the ends and subtract. This is the working rule for definite integrals.

F
Any antiderivative of f
a, b
Endpoints

Power rule for integrals

Raise the exponent by one and divide by the new exponent. The +C records the family of antiderivatives.

C
An arbitrary constant

Integral of the reciprocal

The missing case n = −1. The absolute value keeps the formula valid for negative x.

|x|
Absolute value

Substitution

Set u = g(x), so du = g′(x) dx. The integral becomes one in the variable u.

u
The inner function

Area between curves

When f stays above g, the integrand is the vertical gap and the integral is the enclosed area.

f − g
Upper minus lower

Worked examples

1. A definite integral by antiderivative

Compute the integral of x² from 0 to 2.

  1. An antiderivative of x² is x³/3.
  2. Evaluate: (2)³/3 − (0)³/3 = 8/3.
  3. The region lies above the axis, so 8/3 is also the geometric area.

8/3.

2. Substitution

Compute the integral of 2x(x² + 1)³ dx.

  1. Let u = x² + 1. Then du = 2x dx, which is exactly the remaining factor.
  2. The integral becomes ∫ u³ du = u⁴/4 + C.
  3. Substitute back: (x² + 1)⁴ / 4 + C.

(x² + 1)⁴ / 4 + C.

Try these

Find the area under y = sin x from 0 to π.

Why does the integral of an odd function over [−a, a] often vanish?

Common mistakes

  • Forgetting to convert the limits when a substitution is used on a definite integral. Either switch the limits to u-values or substitute back before evaluating.
  • The integral from a to b of a function that dips below the axis is not the geometric area. Take the absolute value, or split the interval, if you want paint.
  • An antiderivative is a family. Definite integrals kill the +C because it subtracts away; indefinite integrals must keep it.