The Definite Integral

A limit of Riemann sums, read as signed area and as accumulated change. The fundamental theorem, which makes the integral computable by antidifferentiation instead of by summing, and the substitution rule that inverts the chain rule.

Definition

Partition $[a, b]$ into $n$ subintervals of width $Δ x = (b - a) / n$ and choose a sample point $x_{i}^{*}$ in each. The Riemann sum is $\sum_{i = 1}^{n} f (x_{i}^{*}) Δ x$ , and the definite integral is its limit:

\int_{a}^{b} f (x) d x = lim_{n \to \infty} \sum_{i = 1}^{n} f (x_{i}^{*}) Δ x,

when the limit exists independently of how the sample points are chosen, in which case $f$ is integrable on $[a, b]$ . Every continuous function is integrable, as is every bounded function with finitely many discontinuities.

Signed area. Each term $f (x_{i}^{*}) Δ x$ is a rectangle's area taken with the sign of $f$ . Where $f$ is negative the contribution is negative, so the integral measures area above the axis minus area below, not total area, which requires $\int | f |$ .

Accumulated change. If $f$ is a rate, $\int_{a}^{b} f$ is the total change it accumulates over $[a, b]$ : a velocity integrates to displacement, a marginal cost to a cost increase.

The fundamental theorem. Two statements, inverse to each other.

Part 1. If $f$ is continuous and $G (x) = \int_{a}^{x} f (t) d t$ , then $G^{'} (x) = f (x)$ . Accumulating a function and then differentiating returns it.

Part 2. If $F$ is any antiderivative of $f$ on $[a, b]$ , then

\int_{a}^{b} f (x) d x = F (b) - F (a) .

This is what makes integration practical: a limit of sums is replaced by two evaluations.

Notation. $\int f (x) d x$ without limits denotes the indefinite integral, the family of antiderivatives $F + C$ . The constant is not decoration, antiderivatives differ by constants and only by constants, since a function with zero derivative on an interval is constant there.

Properties. $\int_{a}^{b} (f + g) = \int_{a}^{b} f + \int_{a}^{b} g$ and $\int_{a}^{b} c f = c \int_{a}^{b} f$ ; $\int_{a}^{b} f = - \int_{b}^{a} f$ ; $\int_{a}^{a} f = 0$ ; and $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ for any $c$ .

Substitution. If $u = g (x)$ is differentiable and $f$ is continuous,

\int_{a}^{b} f (g (x)) g^{'} (x) d x = \int_{g (a)}^{g (b)} f (u) d u,

which is the chain rule read backwards. The limits change with the variable; a definite integral whose limits are not converted is a common and silent error.

Assumptions and scope

The integral is signed. $\int_{a}^{b} f$ gives area above the axis minus area below, so a function crossing the axis needs $\int | f |$ or a split at the crossing to give total area.
Part 2 requires an antiderivative on the whole interval. Applying $F (b) - F (a)$ across a discontinuity of $f$ (as in $\int_{- 1}^{1} x^{- 2} d x$ ) produces a finite number for a divergent integral.
The constant of integration is not optional in an indefinite integral. Antiderivatives form a family, and dropping $C$ asserts a particular member for no reason.
In a definite integral by substitution the limits must be converted with the variable, or the antiderivative expressed back in $x$ before evaluating. Mixing the two gives a wrong number with no visible error.
Integrability does not require continuity, and differentiability of $f$ is never needed. The fundamental theorem's Part 1 does require continuity of $f$ at the point where $G^{'}$ is claimed.
Not every elementary function has an elementary antiderivative. $e^{- x^{2}}$ is the standard case, which is why numerical integration exists as a discipline rather than a fallback.

Worked material

Example

Integrals worth recognising

A constant, $\int_{a}^{b} c d x = c (b - a)$ . Every rectangle has the same height, so the sum is $c$ times the total width, with no limit needed. This is the case where accumulation reduces to multiplication, and it is the sanity check every other formula must reproduce when the integrand is flat.

A power, $\int_{0}^{1} x^{n} d x = \frac{1}{n + 1}$ for $n \neq - 1$ . The antiderivative is $x^{n + 1} / (n + 1)$ . At $n = 2$ this gives $\frac{1}{3}$ , matching the Riemann-sum derivation; at $n = 1$ it gives $\frac{1}{2}$ , which is the area of the triangle under $y = x$ on $[0, 1]$ , checkable by elementary geometry.

The exclusion $n \neq - 1$ is not a technicality: at $n = - 1$ the formula divides by zero, and $\int \frac{d x}{x}$ is $\ln | x | + C$ instead. Every other exponent follows one rule; this one does not.

A symmetric integrand over a symmetric interval, $\int_{- a}^{a} x^{3} d x = 0$ . The antiderivative $x^{4} / 4$ takes the same value at $- a$ and $a$ , so the difference vanishes. Geometrically the area below the axis on $[- a, 0]$ exactly cancels the area above on $[0, a]$ . Any odd function integrates to zero over an interval symmetric about the origin, which is worth spotting before computing, and is a case where the integral is zero while the total area is not.

An accumulation function, $G (x) = \int_{0}^{x} t^{2} d t = \frac{x^{3}}{3}$ . This is a function of the upper limit, not a number. Part 1 says $G^{'} (x) = x^{2}$ , and the explicit form confirms it. Recognising when a symbol denotes an accumulation rather than a value is what makes Part 1 usable.

A substitution-ready form, $\int_{0}^{2} 2 x (x^{2} + 1)^{3} d x = 156$ . The integrand contains an inner function $x^{2} + 1$ together with its derivative $2 x$ , which is the signature of a chain rule run backwards. Recognising that pattern is the whole skill; the arithmetic that follows is routine.

An integral with no elementary antiderivative, $\int_{0}^{1} e^{- x^{2}} d x$ . It exists, the integrand is continuous on a closed bounded interval, and it is approximately $0.7468$ , but no combination of elementary functions differentiates to $e^{- x^{2}}$ . This is not a gap in anyone's technique; it is a theorem. Such integrals are evaluated numerically, and this particular one is the reason the normal distribution's cumulative function is tabulated rather than written in closed form.

Four of the six are one power rule read backwards. The two exceptions mark the boundaries of the method: $1 / x$ , where the power rule's formula breaks down, and $e^{- x^{2}}$ , where antidifferentiation is not available at all and the limit of sums is the only definition left standing.

Non-example

Where the machinery misleads

The integral is not total area. $\int_{1}^{3} (x^{2} - 4 x) d x = - \frac{22}{3} \approx - 7.33$ , a negative number. The integrand is below the axis throughout, $f (1) = - 3$ , $f (2) = - 4$ , $f (3) = - 3$ , and each Riemann rectangle contributes a negative term. Reporting $+ 7.33$ as "the area" answers a different question: total area is $\int_{1}^{3} | f | = \frac{22}{3}$ , and the two coincide only when $f \geq 0$ throughout.

The distinction matters most when the integrand changes sign, where the integral can be zero while the total area is large. $\int_{- 1}^{1} x^{3} d x = 0$ , yet the region between the curve and the axis has area $\frac{1}{2}$ .

Part 2 across a discontinuity. Consider

\int_{- 1}^{1} \frac{d x}{x^{2}} .

Mechanically, $F (x) = - 1 / x$ gives $F (1) - F (- 1) = - 1 - 1 = - 2$ . This is nonsense twice over: the integrand is positive everywhere it is defined, so no negative answer is possible, and the integral in fact diverges. The area near 0 is unbounded. The failure is that $F$ is not an antiderivative on $[- 1, 1]$ , because neither $f$ nor $F$ is defined at 0. Part 2's hypothesis was not met, and the arithmetic proceeded anyway.

The warning sign is available before computing: the integrand blows up inside the interval. Checking that first is Step 1 of the procedure for this reason.

Substitution without converting the limits. For $\int_{0}^{2} 2 x (x^{2} + 1)^{3} d x$ with $u = x^{2} + 1$ , the correct evaluation is ${[\frac{u^{4}}{4}]}_{u = 1}^{u = 5} = \frac{624}{4} = 156$ . Evaluating the same antiderivative at the original limits gives $\frac{2^{4} - 0^{4}}{4} = 4$ . Nothing in the written work looks wrong; the answer is simply not the integral.

The constant of integration is not decoration. $\int 2 x d x = x^{2} + C$ . Writing $x^{2}$ alone names one antiderivative among infinitely many, and in solving a differential equation that constant is what an initial condition fixes. Omitting it there discards the only freedom the problem had.

A vanishing derivative does not make the integrand zero. If $\int_{a}^{b} f = 0$ it does not follow that $f = 0$ ; it follows only that the positive and negative contributions balanced. $\int_{0}^{2 π} \sin x d x = 0$ while $\sin$ is zero at only three points of the interval.

Not every integral has an elementary antiderivative. $\int e^{- x^{2}} d x$ cannot be written in closed form using elementary functions. A theorem, not a limitation of technique. Searching for a clever substitution is wasted effort, and the integral is evaluated numerically. Recognising the cases where antidifferentiation is unavailable is part of knowing the method's scope.

Common errors

Common misconception

A definite integral gives the total area between the curve and the axis, so its value is always positive.

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The Definite Integral

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Calculus

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Calculus: Early Transcendentals (2020)

The Definite Integral

Assumptions and scope

Worked material

Integrals worth recognising

Where the machinery misleads

Common errors

Related units

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Connected

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