Definite Integrals - Theory & Concepts

Definite Integrals

$Learning Objectives$

Define and interpret definite integrals as signed accumulation.
Explain Riemann sums, Darboux sums, and Riemann integrability accurately.
State and apply the Fundamental Theorem of Calculus (Part 1 and Part 2).
Apply essential properties of definite integrals, including symmetry.
Compute numerical approximations and distinguish them from exact values.
Apply average-value and mean-value results correctly.
Evaluate improper integrals and determine convergence or divergence.

Unlike an indefinite integral, which denotes a family of antiderivatives, a definite integral is a number. For a real-valued integrable function $f$ on $[a,b]$ , $\int_a^b f(x)\,dx$ measures signed accumulation: contributions above the $x$ -axis are positive and contributions below are negative. If $f\ge0$ on the interval, the signed integral agrees with ordinary geometric area; otherwise, geometric area is obtained by integrating $|f|$ or splitting the interval at sign changes.

$Definite Integral$

If $f$ is Riemann integrable on $[a,b]$ , then $\int_a^b f(x)\,dx$ is the common limit approached by all Riemann sums as the mesh of the partition tends to zero. Continuity is a common sufficient condition for integrability, but it is not part of the definition.

$Signed integral versus geometric area$

The identity

\int_a^b f(x)\,dx

gives a net or signed accumulation. If the goal is total geometric area between the graph and the $x$ -axis, use

\int_a^b |f(x)|\,dx,

provided $|f|$ is integrable. A signed integral can be zero even when substantial geometric area is present, as with an odd function over a symmetric interval.

For f of x equals x on minus two to two, the negative and positive triangular contributions cancel in the signed integral, while the total area is four. — Signed accumulation can cancel while area remains positive
On [−2, 0], the signed contribution is −2; on [0, 2], it is +2. Their net integral is 0, but integrating |x| gives total geometric area 4. Hatching and sign labels distinguish the regions.

Evaluation and Interpretation Workflow

Use this route to decide whether the integral is proper, which evaluation path fits the available information, whether every convergence condition is met, and whether the requested quantity is signed accumulation or total area.

Definite Integral Evaluation and Validation

Establish integrability, choose an evaluation method, handle every required improper limit, and report the selected integral quantity

Start with an integral question → Specify f, interval, quantity and data; Specify f, interval, quantity and data → Choose f or |f| for the requested quantity; Choose f or |f| for the requested quantity → Is the interval or integrand unbounded?; Is the interval or integrand unbounded? — Yes → Split at all infinite endpoints and singularities; Split at all infinite endpoints and singularities → Test each one-sided limit for g; Test each one-sided limit for g → Do all required limits converge?; Do all required limits converge? — Yes → Evaluate and combine limits; report the quantity; Do all required limits converge? — No → Report divergence of the selected integral; Is the interval or integrand unbounded? — No → Is g bounded and integrable here?; Is g bounded and integrable here? — Not established → Hold: establish integrability; Is g bounded and integrable here? — Yes → Use FTC or error-aware quadrature; Use FTC or error-aware quadrature → Is the result exact or sufficiently accurate?; Is the result exact or sufficiently accurate? — Yes → Report the result and requested quantity; Is the result exact or sufficiently accurate? — No → Refine the partition or order; Refine the partition or order → Is the result exact or sufficiently accurate?

Start with an integral question: terminator
Specify f, interval, quantity and data: input output
Choose f or |f| for the requested quantity: process. Set g = f for signed accumulation and g = |f| for total area before checking convergence.
Is the interval or integrand unbounded?: decision
Split at all infinite endpoints and singularities: process. For an interior singularity, form separate one-sided improper integrals. Apply the selected integrand g, including |f| when total area is requested.
Test each one-sided limit for g: process. Evaluate each required limit directly when possible; otherwise use an appropriate comparison test.
Do all required limits converge?: decision
Evaluate and combine limits; report the quantity: document. Evaluate and combine the finite one-sided limits for g; report the signed integral or total area requested. If using a numerical estimate, include discretization and tail-error evidence.
Report divergence of the selected integral: document. A divergent required one-sided limit means the requested improper integral does not converge. For total area, the selected integrand is |f|.
Is g bounded and integrable here?: decision
Hold: establish integrability: document. Establish boundedness and a Riemann-integrability criterion before evaluation.
Use FTC or error-aware quadrature: process. Use the Fundamental Theorem of Calculus when an appropriate antiderivative is available. Otherwise use quadrature with error evidence; composite Simpson's rule requires an even number of subintervals.
Is the result exact or sufficiently accurate?: decision. Accept an exact evaluation. For quadrature, require an error bound or refinement evidence within the stated tolerance.
Report the result and requested quantity: document. State an exact value when available; otherwise report the estimate with its error or refinement evidence.
Refine the partition or order: process. Recompute the estimate and its error evidence before checking the requested precision again.

$Interactive Simulation$

Explore how left-, right-, and midpoint Riemann sums approximate selected definite integrals. Exact reference values are displayed separately from decimal approximations.

Simulation: Riemann Sums & Definite Integrals

Rectangles (n)4

Coarse (2)Fine (100)

Integral Analysis

Riemann approximation:≈ 1.7500

Exact integral:8/3 ≈ 2.6667

Absolute error:0.9167

As the partition is refined, these Riemann sums approach the definite integral. The displayed decimal is a numerical approximation; the fraction or integer shown beside it is the exact value for the selected example.

Riemann Sums, Darboux Sums, and Integrability

Let $P: a=x_0<x_1<\cdots<x_n=b$ be a partition of $[a,b]$ . Write $\Delta x_i=x_i-x_{i-1}$ and choose a sample point $x_i^*\in[x_{i-1},x_i]$ . The corresponding Riemann sum is

\sum_{i=1}^{n} f(x_i^*)\Delta x_i.

Equal-width partitions are useful computationally, but the definition of Riemann integrability allows general partitions. The relevant refinement measure is the mesh $\|P\|=\max_i\Delta x_i$ .

Definite Integral via Riemann Sums

Riemann-sum definition using arbitrary tagged partitions.

\int_a^b f(x)\,dx =\lim_{\|P\|\to0}\sum_{i=1}^{n}f(x_i^*)\Delta x_i,

whenever this limit exists and is independent of the sample-point choices.

Variables

Symbol	Description	Unit
$P$	Partition of [a,b]	-
$x_i^*$	Sample point in the i-th subinterval	-
$Δx_i$	Width of the i-th subinterval	-
$\|\|P\|\|$	Mesh: largest subinterval width	-

$Common Tagged Riemann Sums$

Left Endpoint: $x_i^*=x_{i-1}$ .
Right Endpoint: $x_i^*=x_i$ .
Midpoint: $x_i^*=\frac{x_{i-1}+x_i}{2}$ .
For equal-width partitions, $\Delta x=(b-a)/n$ .

Darboux Sums and a Rigorous Integrability Criterion

Suppose $f$ is bounded on $[a,b]$ . On each subinterval $I_i=[x_{i-1},x_i]$ , define

M_i=\sup_{x\in I_i}f(x),\qquad m_i=\inf_{x\in I_i}f(x).

These are the supremum and infimum. They need not be attained as an actual maximum or minimum unless additional hypotheses guarantee attainment.

$Upper and Lower Darboux Sums$

Upper Darboux Sum: $U(f,P)=\sum_{i=1}^{n} M_i\Delta x_i$ .
Lower Darboux Sum: $L(f,P)=\sum_{i=1}^{n} m_i\Delta x_i$ .

A bounded function $f$ is Riemann integrable on $[a,b]$ if and only if for every $\varepsilon>0$ there exists a partition $P$ such that
$U(f,P)-L(f,P)<\varepsilon.$
Equivalently, the upper and lower Darboux integrals are equal.

$Continuity is sufficient, not necessary$

Every continuous function on a closed interval is Riemann integrable, and every bounded function with only finitely many discontinuities is Riemann integrable. These are sufficient conditions, not a complete classification. More generally, the Lebesgue criterion states that a bounded function on $[a,b]$ is Riemann integrable exactly when its set of discontinuities has measure zero. Thus, Riemann-integrable functions may have infinitely many discontinuities.

Numerical Integration Techniques

When an elementary antiderivative is unavailable, or when only tabulated data are given, numerical quadrature approximates the definite integral. An approximation should be reported with an approximation symbol, an appropriate number of digits, and—when available—an error estimate or convergence study.

$The Trapezoidal Rule$

For equally spaced points $x_i=a+i\Delta x$ ,

Composite Trapezoidal Rule

Numerical integration using trapezoids.

T_n=\frac{\Delta x}{2}\left[f(x_0)+2\sum_{i=1}^{n-1}f(x_i)+f(x_n)\right] \approx\int_a^b f(x)\,dx.

Variables

Symbol	Description	Unit
$Δx$	Equal subinterval width (b-a)/n	-
$T_n$	Trapezoidal approximation, not generally an exact value	-

$Composite Simpson's Rule$

Simpson's Rule uses quadratic interpolation on pairs of subintervals and requires an even number $n$ of equal subintervals.

Composite Simpson's Rule

Numerical integration using quadratic interpolation.

S_n=\frac{\Delta x}{3}\left[f(x_0)+4f(x_1)+2f(x_2)+\cdots+2f(x_{n-2})+4f(x_{n-1})+f(x_n)\right] \approx\int_a^b f(x)\,dx.

Variables

Symbol	Description	Unit
$n$	Even number of equal subintervals	-
$S_n$	Simpson approximation; exact for polynomials through degree 3 under the standard assumptions	-

$About approximation and exactness$

A decimal produced by a numerical rule is an approximation unless exactness follows from the rule and the integrand. For example, writing $2.6666$ does not make $8/3$ exact; $8/3$ is the exact value and $2.6666$ is only a truncated decimal representation.

The Fundamental Theorem of Calculus

Part 1: Differentiating an Accumulation Function

Let

G(x)=\int_a^x f(t)\,dt.

If $f$ is continuous on $[a,b]$ , then $G$ is continuous on $[a,b]$ , differentiable on $(a,b)$ , and

G'(x)=f(x).

Here $G$ is a signed accumulation function. When $f(x)<0$ , the accumulation decreases.

Fundamental Theorem of Calculus Part 1

Derivative of an accumulation function.

\frac{d}{dx}\left[\int_a^x f(t)\,dt\right]=f(x).

Variables

Symbol	Description	Unit
$t$	Dummy variable of integration	-
$x$	Variable upper limit	-

The graph f of t equals t changes from negative to positive at zero, while its accumulation function decreases to a minimum at zero and then rises; the tangent slope at x equals one is one. — The integrand is the slope of its accumulation function
For G(x) = ∫ from −2 to x of t dt = x²/2 − 2, G′(x) = f(x). The accumulation falls where f is negative, has a minimum where f is zero, and rises where f is positive.

Part 2: Evaluation by an Antiderivative

If $f$ is continuous on $[a,b]$ and $F'(x)=f(x)$ on that interval, then

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

The same evaluation identity extends to broader classes of Riemann-integrable functions when an appropriate antiderivative exists, but continuity is the standard elementary hypothesis.

Fundamental Theorem of Calculus Part 2

Evaluating a definite integral using an antiderivative.

\int_a^b f(x)\,dx=\left[F(x)\right]_a^b=F(b)-F(a).

Variables

Symbol	Description	Unit
$F$	Any antiderivative of f on the interval	-
$a,b$	Lower and upper integration limits	-

The Net Change Theorem

If $Q'(t)$ is the rate of change of a quantity $Q(t)$ , then

\int_a^b Q'(t)\,dt=Q(b)-Q(a).

The result is a net change, so positive and negative rates can cancel.

$Applications of Net Change and Accumulation$

If $v(t)=s'(t)$ , then $\int_a^b v(t)\,dt=s(b)-s(a)$ is displacement. Total distance is $\int_a^b|v(t)|\,dt$ .
If $C'(x)$ is marginal cost, then $\int_a^b C'(x)\,dx=C(b)-C(a)$ is the change in total cost.
If $\rho(x)$ is linear density, then $\int_a^b\rho(x)\,dx$ is the mass of the rod segment. If one defines cumulative mass $m(x)=\int_{x_0}^{x}\rho(s)\,ds$ , then $m'(x)=\rho(x)$ and the same statement can be written as $m(b)-m(a)$ .

Essential Properties of Definite Integrals

$Essential Properties$

Zero interval: $\int_a^a f(x)\,dx=0$ .
Reversing limits: $\int_b^a f(x)\,dx=-\int_a^b f(x)\,dx$ .
Linearity: $\int_a^b(\alpha f+\beta g)=\alpha\int_a^b f+\beta\int_a^b g$ .
Interval additivity: For any $c$ , $\int_a^c f+\int_c^b f=\int_a^b f$ , using the usual oriented-integral convention.
Positivity: If $f\ge0$ on $[a,b]$ , then $\int_a^b f\ge0$ .
Comparison: If $f\le g$ on $[a,b]$ , then $\int_a^b f\le\int_a^b g$ .

Symmetry: Even and Odd Functions

$Symmetry Properties$

If $f$ is even, $f(-x)=f(x)$ , then $\int_{-a}^{a}f(x)\,dx=2\int_0^a f(x)\,dx$ .
If $f$ is odd, $f(-x)=-f(x)$ , then $\int_{-a}^{a}f(x)\,dx=0$ .

The second identity expresses cancellation of signed accumulation; it does not imply zero geometric area.

Average Value and the Mean Value Theorem for Integrals

For an integrable function on $[a,b]$ with $a<b$ , its average value is

f_{\mathrm{avg}}=\frac{1}{b-a}\int_a^b f(x)\,dx.

If $f$ is continuous, then there exists at least one $c\in[a,b]$ such that $f(c)=f_{\mathrm{avg}}$ .

$Geometric interpretation$

The identity

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

says that a rectangle of signed height $f(c)$ and width $b-a$ has the same signed area as the integral. If $f\ge0$ , this is also an ordinary area statement. If $f$ takes negative values, describe it as signed accumulation rather than ordinary geometric area.

Improper Integrals

An improper integral occurs when the interval is unbounded or the integrand becomes unbounded at an endpoint or interior point. The symbol is defined through one or more limits; infinity is never substituted as though it were a number.

$Type 1: Infinite Intervals$

$\int_a^{\infty}f(x)\,dx=\lim_{t\to\infty}\int_a^t f(x)\,dx$ .
$\int_{-\infty}^{b}f(x)\,dx=\lim_{t\to-\infty}\int_t^b f(x)\,dx$ .
$\int_{-\infty}^{\infty}f(x)\,dx$ is defined by splitting at a finite $c$ ; both resulting improper integrals must converge. A symmetric Cauchy principal value, when it exists, is a different concept and does not by itself establish convergence of the improper integral.

$Type 2: Unbounded Integrands$

If $f$ is unbounded as $x\to b^-$ , use $\int_a^b f(x)\,dx=\lim_{t\to b^-}\int_a^t f(x)\,dx$ .
If $f$ is unbounded as $x\to a^+$ , use $\int_a^b f(x)\,dx=\lim_{t\to a^+}\int_t^b f(x)\,dx$ .
If $f$ is unbounded at an interior point $c$ , split at $c$ and require both one-sided improper integrals to converge.

The p-Tests

$Power-integral p-tests$

Two standard power integrals have opposite convergence inequalities:

Infinite tail:
$\int_1^{\infty}\frac{1}{x^p}\,dx$
converges if and only if $p>1$ ; it diverges for $p\le1$ .
Endpoint singularity:
$\int_0^1\frac{1}{x^p}\,dx$
converges if and only if $p<1$ ; it diverges for $p\ge1$ .

The boundary case $p=1$ diverges in both settings. The distinction is important: the same exponent can behave differently at infinity and near a finite singular endpoint.

Direct Comparison Test

$Comparison for nonnegative integrands$

Suppose $0\le f(x)\le g(x)$ for all sufficiently large $x$ .

If $\int_a^{\infty}g(x)\,dx$ converges, then $\int_a^{\infty}f(x)\,dx$ converges.
If $\int_a^{\infty}f(x)\,dx$ diverges, then $\int_a^{\infty}g(x)\,dx$ diverges.

Comparison determines convergence behavior; it does not usually determine the exact value of the integral.

$Convergence versus divergence$

An improper integral converges only when every defining limit exists as a finite real number. A limit equal to $+\infty$ or $-\infty$ , a nonexistent oscillatory limit, or failure of either side of a required split means the improper integral diverges.

$Interactive Simulation$

Use the simulation below to examine $\int_1^t x^{-2}\,dx=1-1/t$ and the limiting process $t\to\infty$ .

Improper Integral Visualization: ∫₁ᵗ 1/x² dx

Finite upper bound t: 2.0

Loading chart...

Convergence calculation

∫₁ᵗ x⁻² dx = [-1/x]₁ᵗ = 1 - 1/t

Finite integral = 0.5000

Remaining tail to the limiting value 1 = 1/t = 0.5000

As t increases, 1/t tends to 0. Therefore limₜ→∞ (1 - 1/t) = 1, so ∫₁∞ 1/x² dx converges to 1.

Key Takeaways

A definite integral is signed accumulation; geometric area requires nonnegativity or absolute values.
Darboux sums use suprema and infima, whether or not those bounds are attained.
Continuity and finitely many discontinuities are sufficient conditions for Riemann integrability, not necessary ones.
Numerical rules produce approximations unless exactness is established.
FTC connects signed accumulation and antiderivatives.
Improper integrals are defined by limits, and every required one-sided or infinite-tail limit must converge.
For power integrals, the infinite-tail p-test requires $p>1$ , while the finite-endpoint p-test near zero requires $p<1$ .
The p-tests and comparison test are convergence tools, not shortcuts to exact numerical values.

$References$

Previous TopicAntiderivatives and Indefinite Integrals - Examples & Applications

Quiz Me

Next TopicDefinite Integrals - Examples & Applications

Prev Next

Quiz Me