Definite Integral as Area Under a Curve

⚡ Key Takeaways

The definite integral $\int_a^b f(x)\,dx$ measures the signed area between the curve $y=f(x)$ and the $x$ -axis from $x=a$ to $x=b$ . Key result: it is the limit of summing the areas of infinitely many thin rectangles under the curve. Applies to computing areas, accumulated quantities (distance, work, volume) wherever a rate is integrated. Origin: rigorously formalized by Bernhard Riemann (1854), building on Newton, Leibniz, Cauchy.

💡 The Core Insight

Slice the region under a curve into many thin vertical rectangles. Each has height $f(x_i)$ and width $\Delta x$ , so area $f(x_i)\,\Delta x$ . Add them all up to estimate the total area. Then let the rectangles become infinitely thin ( $\Delta x \to 0$ , number $\to \infty$ ). The limit of these sums — if it exists — is the definite integral. The genius is that "infinitely many infinitely thin pieces" converges to a finite, exact answer.

DefinitionRiemann Sum

Partition $[a,b]$ into $n$ subintervals using points

1

\quad a = x_0 < x_1 < x_2 < \cdots < x_n = b

Derivation hint: any increasing sequence of points starting at $a$ and ending at $b$ defines a valid partition. Let $\Delta x_i = x_i - x_{i-1}$ be the width of subinterval $i$ , and choose a sample point $x_i^* \in [x_{i-1}, x_i]$ . The Riemann sum is

2

\quad S_n = \sum_{i=1}^{n} f(x_i^*)\,\Delta x_i

Derivation hint: each term $f(x_i^*)\Delta x_i$ is the area of one rectangle with height $f(x_i^*)$ and width $\Delta x_i$ . For a regular partition (equal widths), $\Delta x = \dfrac{b-a}{n}$ and $x_i = a + i\,\Delta x$ , giving

3

\quad S_n = \sum_{i=1}^{n} f\!\left(a + i\,\Delta x\right)\,\Delta x, \qquad \Delta x = \frac{b-a}{n}

Derivation hint: substitute equal width $\Delta x=(b-a)/n$ into [2] and use $x_i = a + i\Delta x$ .

DefinitionThe Definite Integral

The definite integral of $f$ from $a$ to $b$ is the limit of Riemann sums as the largest subinterval width (the norm $\|P\|=\max_i \Delta x_i$ ) shrinks to zero:

4

\quad \int_a^b f(x)\,dx = \lim_{\|P\|\to 0} \sum_{i=1}^{n} f(x_i^*)\,\Delta x_i

Derivation hint: the integral exists precisely when this limit gives the same value regardless of how partitions are refined or sample points chosen. Here $a$ is the lower limit, $b$ the upper limit, $f(x)$ the integrand, $x$ the variable of integration, and $dx$ the symbol indicating the infinitesimal width. The function $f$ is then called integrable on $[a,b]$ .

TheoremIntegrability of Continuous Functions

If $f$ is continuous on the closed interval $[a,b]$ , then $f$ is integrable on $[a,b]$ ; the limit in [4] exists and is a unique finite number. (More generally, a bounded function with only finitely many discontinuities is integrable.)

TheoremBasic Properties of the Definite Integral

For integrable functions $f,g$ and constants $c,k$ :

5

\quad \int_a^b \big[\,c\,f(x) + k\,g(x)\,\big]\,dx = c\int_a^b f(x)\,dx + k\int_a^b g(x)\,dx

Derivation hint: linearity of the limit and of finite sums in [2] passes through to the integral.

6

\quad \int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx, \qquad a \le c \le b

Derivation hint: split each partition at the point $c$ ; the two pieces' Riemann sums add to the whole.

7

\quad \int_a^a f(x)\,dx = 0, \qquad \int_b^a f(x)\,dx = -\int_a^b f(x)\,dx

Derivation hint: zero-width interval gives zero area; reversing limits flips the sign of every $\Delta x$ .

ProofArea of a Region Equals the Limit of Rectangle Sums

We show that summing thin rectangles converges to the true area for a continuous, non-negative $f$ on $[a,b]$ .

Step 1 — Bound the area. On each subinterval $[x_{i-1},x_i]$ , since $f$ is continuous it attains a minimum $m_i$ and a maximum $M_i$ . The true area $A_i$ of the strip satisfies

8

\quad m_i\,\Delta x \le A_i \le M_i\,\Delta x

Derivation hint: the strip contains a rectangle of height $m_i$ and is contained in one of height $M_i$ .

Step 2 — Form lower and upper sums. Summing over all strips:

9

\quad L_n = \sum_{i=1}^{n} m_i\,\Delta x \le A \le \sum_{i=1}^{n} M_i\,\Delta x = U_n

Derivation hint: add inequality [8] over $i=1,\dots,n$ ; the middle sum telescopes to total area $A$ .

Step 3 — Squeeze the gap. The difference is

10

\quad U_n - L_n = \sum_{i=1}^{n} (M_i - m_i)\,\Delta x

Derivation hint: subtract $L_n$ from $U_n$ term by term. Because $f$ is continuous on a closed interval it is uniformly continuous, so for fine enough partitions each $M_i - m_i$ becomes arbitrarily small (less than any chosen $\varepsilon$ ). Then $U_n - L_n \le \varepsilon \sum \Delta x = \varepsilon(b-a) \to 0$ .

Step 4 — Conclude. Since $L_n \le A \le U_n$ and $U_n - L_n \to 0$ , both sums converge to the same limit, and any Riemann sum [2] (whose value lies between $L_n$ and $U_n$ ) is squeezed to that limit. That common value is, by definition [4], $\int_a^b f(x)\,dx = A$ . $\blacksquare$

∎

BR

Bernhard Riemann

1826–1866

The German mathematician Bernhard Riemann gave the integral its modern rigorous footing in his 1854 Habilitationsschrift "On the representation of a function by a trigonometric series." Earlier, Newton and Leibniz had used integration intuitively (1660s–80s), and Cauchy had defined the integral for continuous functions (1823); Riemann's contribution was a precise definition — the limit of sums in [4] — that worked even for discontinuous functions, with an exact criterion for when the limit exists. The construction is now universally called the Riemann integral. Riemann's broader genius spanned geometry (Riemannian manifolds, later the language of general relativity) and analysis (the Riemann zeta function), all before his death from tuberculosis at age 39.

🧠 Intuition

Think of painting a region with vertical brushstrokes. Each stroke is a tall thin rectangle: its height is how far the curve sits above the axis at that spot, its width is how thick the brush is. A thick brush gives a rough, blocky estimate of the painted area; a thin brush traces the curve more faithfully. The integral is what you get in the imaginary limit of an infinitely thin brush — perfect fidelity. A second picture: imagine water flowing into a tank at a rate $f(t)$ litres per second. Over a tiny time $dt$ , the volume added is $f(t)\,dt$ . The total water after time $a$ to $b$ is the sum of all those tiny contributions — exactly $\int_a^b f(t)\,dt$ . This is why integrals appear everywhere a rate accumulates into a total: the "area under the rate curve" is the accumulated amount.

ExampleArea under a parabola from the definition

Given: Compute $\displaystyle\int_0^2 x^2\,dx$ directly from the Riemann sum (regular partition, right endpoints).

Step 1 — Set up the partition. $\Delta x = \dfrac{2-0}{n} = \dfrac{2}{n}$ , and $x_i = 0 + i\Delta x = \dfrac{2i}{n}$ .

Step 2 — Write the sum.

S_n = \sum_{i=1}^{n} \left(\frac{2i}{n}\right)^2 \cdot \frac{2}{n} = \frac{8}{n^3}\sum_{i=1}^{n} i^2

Step 3 — Use the sum-of-squares formula $\sum_{i=1}^n i^2 = \dfrac{n(n+1)(2n+1)}{6}$ :

S_n = \frac{8}{n^3}\cdot \frac{n(n+1)(2n+1)}{6} = \frac{8(n+1)(2n+1)}{6n^2} = \frac{4}{3}\cdot\frac{(n+1)(2n+1)}{n^2}

Step 4 — Take the limit. As $n\to\infty$ , $\dfrac{(n+1)(2n+1)}{n^2}\to 2$ :

\int_0^2 x^2\,dx = \frac{4}{3}\cdot 2 = \frac{8}{3}

Boxed answer: $\displaystyle\int_0^2 x^2\,dx = \boxed{\dfrac{8}{3} \approx 2.667}$ (square units). Check via antiderivative: $[x^3/3]_0^2 = 8/3$ . ✓

ExampleSigned area and a real accumulation

Given: A particle moves with velocity $v(t) = t - 3$ metres per second for $t\in[0,5]$ seconds. Find (a) the net displacement and (b) the total distance travelled.

Step 1 — Net displacement is the signed integral.

\int_0^5 (t-3)\,dt = \left[\frac{t^2}{2} - 3t\right]_0^5 = \left(\frac{25}{2} - 15\right) - 0 = -2.5

The negative sign means the particle ends up $2.5\text{ m}$ behind its start.

Step 2 — Identify where $v$ changes sign. $v=0$ at $t=3$ . On $[0,3]$ , $v<0$ (moving backward); on $[3,5]$ , $v>0$ (forward).

Step 3 — Total distance uses absolute values of each piece.

\int_0^3 (t-3)\,dt = \left[\tfrac{t^2}{2}-3t\right]_0^3 = (4.5-9) = -4.5 \;\Rightarrow\; \text{distance } 4.5\text{ m}

\int_3^5 (t-3)\,dt = \left[\tfrac{t^2}{2}-3t\right]_3^5 = (12.5-15)-(4.5-9) = -2.5 + 4.5 = 2.0 \;\Rightarrow\; 2.0\text{ m}

Step 4 — Add the unsigned pieces. Total distance $= 4.5 + 2.0 = 6.5\text{ m}$ .

Boxed answers: Net displacement $= \boxed{-2.5\text{ m}}$ , total distance $= \boxed{6.5\text{ m}}$ .

⚠️ Common mistake

"The integral always gives a positive area." False. The definite integral gives signed area: where $f(x)<0$ , the rectangles have negative height and contribute negatively. To get the geometric (always-positive) area between the curve and the axis, you must integrate $|f(x)|$ — splitting the interval at every point where $f$ crosses the axis. Confusing $\int_a^b f\,dx$ with "geometric area" is the single most common error.

" $dx$ is just decoration." No — $dx$ records which variable you integrate and represents the infinitesimal width. It is essential for substitution, units, and reading the integral as a sum of $f(x)\cdot dx$ products. The integral $\int_a^b f(x)\,dx$ carries units of [height]×[width].

Counterexample

Consider $f(x) = \sin x$ over $[0, 2\pi]$ . One might guess the "area under one full wave" is large and positive. But

\int_0^{2\pi} \sin x\,dx = [-\cos x]_0^{2\pi} = (-1) - (-1) = 0.

The positive hump on $[0,\pi]$ exactly cancels the negative trough on $[\pi,2\pi]$ . The signed area is zero, while the geometric area is $\int_0^{2\pi}|\sin x|\,dx = 4$ . This vividly shows the integral is not "the area you see."

connection

The definite integral connects to a wide web of ideas. It is glued to differentiation by the Fundamental Theorem of Calculus, which says integration and differentiation are inverse operations — turning the hard limit of [4] into an easy antiderivative subtraction. The technique of finding those antiderivatives is the subject of Indefinite Integral and Antiderivatives. The Riemann sum [2] is a special case of the broader theory in Riemann Sums and Numerical Integration, where rules like the trapezoidal and Simpson's rule estimate integrals that have no closed form. Geometrically, the integral generalizes to Area Between Two Curves (integrate the difference of functions) and to Volumes of Revolution (integrate cross-sectional area). The limit concept underlying [4] is the same one studied in Limits and Continuity.

🔬 Where this appears

⏳ History

- c. 250 BCE — Archimedes: Computes the area of a parabolic segment by the method of exhaustion, summing inscribed triangles — a Riemann sum in spirit two millennia early.

- 1635 — Cavalieri: "Method of indivisibles" treats areas as sums of infinitely many line segments, a bold (if shaky) precursor.

- 1660s–1680s — Newton & Leibniz: Independently develop calculus; Leibniz introduces the elongated-S sign

\int

(1675) for summa (sum) and the

dx

notation.

- 1820s — Cauchy: Gives the first rigorous definition of the integral for continuous functions as a limit of sums, anchoring it to limits rather than infinitesimals.

- 1854 — Riemann: Defines the integral via arbitrary partitions and sample points [4], with a precise integrability criterion; works for many discontinuous functions.

- 1875 — Darboux: Introduces upper and lower sums (our [9]), giving a clean equivalent formulation now standard in textbooks.

- 1902 — Lebesgue: Creates a more powerful integral by partitioning the range instead of the domain, integrating wildly discontinuous functions the Riemann integral cannot handle.

- 20th century: Numerical integration (trapezoidal, Simpson, Gaussian quadrature) and computer methods make definite integrals routinely computable in science and engineering.

open-problem

The Riemann integral, for all its power, has boundaries. The function that is $1$ on rationals and $0$ on irrationals (the Dirichlet function) is bounded yet not Riemann-integrable: its upper sums are always $1$ and lower sums always $0$ , so the squeeze in [10] never closes. The Lebesgue integral repairs this — yet even Lebesgue integration cannot give finite, well-defined answers to many integrals arising in modern physics, such as the path integrals of quantum field theory, where the relevant "integral over all paths" still lacks a fully rigorous mathematical definition in the general case. Likewise, improper integrals over infinite intervals or with infinite integrands require careful limit definitions and may diverge (e.g. $\int_1^\infty \frac{1}{x}\,dx = \infty$ ), forcing distinctions between conditional and absolute convergence that remain technically delicate. Defining "area" itself for sufficiently pathological sets connects to deep open questions in measure theory.

Definite Integral as Area Under a Curve

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also