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 ff on [a,b][a,b], ∫abf(x) dx\int_a^b f(x)\,dx measures signed accumulation: contributions above the xx-axis are positive and contributions below are negative. If f≥0f\ge0 on the interval, the signed integral agrees with ordinary geometric area; otherwise, geometric area is obtained by integrating ∣f∣|f| or splitting the interval at sign changes.

Definite Integral

If ff is Riemann integrable on [a,b][a,b], then ∫abf(x) dx\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

∫abf(x) dx\int_a^b f(x)\,dx

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

∫ab∣f(x)∣ dx,\int_a^b |f(x)|\,dx,

provided ∣f∣|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.

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

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 sum approximationRectangles approximate the definite integral of the selected nonnegative function over the displayed interval.0.000.251.002.25a=0b=2

Riemann Sums, Darboux Sums, and Integrability

Let P:a=x0<x1<⋯<xn=bP: a=x_0<x_1<\cdots<x_n=b be a partition of [a,b][a,b]. Write Δxi=xi−xi−1\Delta x_i=x_i-x_{i-1} and choose a sample point xi∗∈[xi−1,xi]x_i^*\in[x_{i-1},x_i]. The corresponding Riemann sum is

∑i=1nf(xi∗)Δxi.\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⁡iΔxi\|P\|=\max_i\Delta x_i.

Definite Integral via Riemann Sums

Riemann-sum definition using arbitrary tagged partitions.

∫abf(x) dx=lim⁡∥P∥→0∑i=1nf(xi∗)Δxi,\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

SymbolDescriptionUnit
PPPartition of [a,b]-
xi∗x_i^*Sample point in the i-th subinterval-
ΔxiΔx_iWidth of the i-th subinterval-
∣∣P∣∣||P||Mesh: largest subinterval width-

Common Tagged Riemann Sums

Darboux Sums and a Rigorous Integrability Criterion

Suppose ff is bounded on [a,b][a,b]. On each subinterval Ii=[xi−1,xi]I_i=[x_{i-1},x_i], define

Mi=sup⁡x∈Iif(x),mi=inf⁡x∈Iif(x).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)=∑i=1nMiΔxiU(f,P)=\sum_{i=1}^{n} M_i\Delta x_i.

  • Lower Darboux Sum: L(f,P)=∑i=1nmiΔxiL(f,P)=\sum_{i=1}^{n} m_i\Delta x_i.

    A bounded function ff is Riemann integrable on [a,b][a,b] if and only if for every ε>0\varepsilon>0 there exists a partition PP such that

    U(f,P)−L(f,P)<ε.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][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 xi=a+iΔxx_i=a+i\Delta x,

Composite Trapezoidal Rule

Numerical integration using trapezoids.

Tn=Δx2[f(x0)+2∑i=1n−1f(xi)+f(xn)]≈∫abf(x) dx.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

SymbolDescriptionUnit
ΔxΔxEqual subinterval width (b-a)/n-
TnT_nTrapezoidal 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 nn of equal subintervals.

Composite Simpson's Rule

Numerical integration using quadratic interpolation.

Sn=Δx3[f(x0)+4f(x1)+2f(x2)+⋯+2f(xn−2)+4f(xn−1)+f(xn)]≈∫abf(x) dx.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

SymbolDescriptionUnit
nnEven number of equal subintervals-
SnS_nSimpson 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 2.66662.6666 does not make 8/38/3 exact; 8/38/3 is the exact value and 2.66662.6666 is only a truncated decimal representation.

The Fundamental Theorem of Calculus

Part 1: Differentiating an Accumulation Function

Let

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

If ff is continuous on [a,b][a,b], then GG is continuous on [a,b][a,b], differentiable on (a,b)(a,b), and

G′(x)=f(x).G'(x)=f(x).

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

Fundamental Theorem of Calculus Part 1

Derivative of an accumulation function.

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

Variables

SymbolDescriptionUnit
ttDummy variable of integration-
xxVariable upper limit-

Part 2: Evaluation by an Antiderivative

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

∫abf(x) dx=F(b)−F(a).\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.

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

Variables

SymbolDescriptionUnit
FFAny antiderivative of f on the interval-
a,ba,bLower and upper integration limits-

The Net Change Theorem

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

∫abQ′(t) dt=Q(b)−Q(a).\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)v(t)=s'(t), then ∫abv(t) dt=s(b)−s(a)\int_a^b v(t)\,dt=s(b)-s(a) is displacement. Total distance is ∫ab∣v(t)∣ dt\int_a^b|v(t)|\,dt.
  • If C′(x)C'(x) is marginal cost, then ∫abC′(x) dx=C(b)−C(a)\int_a^b C'(x)\,dx=C(b)-C(a) is the change in total cost.
  • If ρ(x)\rho(x) is linear density, then ∫abρ(x) dx\int_a^b\rho(x)\,dx is the mass of the rod segment. If one defines cumulative mass m(x)=∫x0xρ(s) dsm(x)=\int_{x_0}^{x}\rho(s)\,ds, then m′(x)=ρ(x)m'(x)=\rho(x) and the same statement can be written as m(b)−m(a)m(b)-m(a).

Essential Properties of Definite Integrals

Essential Properties

Symmetry: Even and Odd Functions

Symmetry Properties

  • If ff is even, f(−x)=f(x)f(-x)=f(x), then ∫−aaf(x) dx=2∫0af(x) dx\int_{-a}^{a}f(x)\,dx=2\int_0^a f(x)\,dx.

  • If ff is odd, f(−x)=−f(x)f(-x)=-f(x), then ∫−aaf(x) dx=0\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][a,b] with a<ba<b, its average value is

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

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

Geometric interpretation

The identity

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

says that a rectangle of signed height f(c)f(c) and width b−ab-a has the same signed area as the integral. If f≥0f\ge0, this is also an ordinary area statement. If ff 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

  • ∫a∞f(x) dx=lim⁡t→∞∫atf(x) dx\int_a^{\infty}f(x)\,dx=\lim_{t\to\infty}\int_a^t f(x)\,dx.
  • ∫−∞bf(x) dx=lim⁡t→−∞∫tbf(x) dx\int_{-\infty}^{b}f(x)\,dx=\lim_{t\to-\infty}\int_t^b f(x)\,dx.
  • ∫−∞∞f(x) dx\int_{-\infty}^{\infty}f(x)\,dx is defined by splitting at a finite cc; 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 ff is unbounded as x→b−x\to b^-, use ∫abf(x) dx=lim⁡t→b−∫atf(x) dx\int_a^b f(x)\,dx=\lim_{t\to b^-}\int_a^t f(x)\,dx.
  • If ff is unbounded as x→a+x\to a^+, use ∫abf(x) dx=lim⁡t→a+∫tbf(x) dx\int_a^b f(x)\,dx=\lim_{t\to a^+}\int_t^b f(x)\,dx.
  • If ff is unbounded at an interior point cc, split at cc 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:

    ∫1∞1xp dx\int_1^{\infty}\frac{1}{x^p}\,dx

    converges if and only if p>1p>1; it diverges for p≤1p\le1.

  • Endpoint singularity:

    ∫011xp dx\int_0^1\frac{1}{x^p}\,dx

    converges if and only if p<1p<1; it diverges for p≥1p\ge1.

    The boundary case p=1p=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≤f(x)≤g(x)0\le f(x)\le g(x) for all sufficiently large xx.

  • If ∫a∞g(x) dx\int_a^{\infty}g(x)\,dx converges, then ∫a∞f(x) dx\int_a^{\infty}f(x)\,dx converges.

  • If ∫a∞f(x) dx\int_a^{\infty}f(x)\,dx diverges, then ∫a∞g(x) dx\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 ∫1tx−2 dx=1−1/t\int_1^t x^{-2}\,dx=1-1/t and the limiting process t→∞t\to\infty.

Improper Integral Visualization: ∫₁ᵗ 1/x² dx

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>1p>1, while the finite-endpoint p-test near zero requires p<1p<1.
  • The p-tests and comparison test are convergence tools, not shortcuts to exact numerical values.