Grade 12

Definite Integral

The definite integral measures the net accumulation of a continuously varying quantity—from the signed area under a curve to the volume of a solid of revolution—and is linked to antiderivatives by the Fundamental Theorem of Calculus.

IntuitionSlice thin, add many: measuring curved regions

How do you find the area of a region with a curved boundary, or the total distance travelled when your speed changes every instant? Elementary geometry only gives formulas for flat-sided shapes like rectangles and triangles. The core idea of integration is slice thin, add many: cut the curved region under $y = f(x) \ge 0$ on $[a, b]$ into $n$ vertical strips of equal width $\Delta x = (b - a)/n$ , and replace each strip by a flat-topped rectangle. With a few wide rectangles, the staircase is a rough approximation; as $n$ grows and $\Delta x \to 0$ , the step error shrinks and the sum of rectangle areas settles onto a single exact number—the definite integral.

Left-endpoint rectangle sum for

y = x^2

[0, 2]

with

n

slices of width

\Delta x = 2/n

. Increase

n

to see the staircase area approach the exact integral

\int_0^2 x^2\,dx = 8/3 \approx 2.667

Historical note

Around 250 BCE, Archimedes used the method of exhaustion—packing a curved shape with ever-smaller polygons—to prove that a parabolic segment has area $4/3$ of its inscribed triangle. In the late 17th century, Isaac Newton and Gottfried Wilhelm Leibniz independently discovered that accumulating infinitesimal slices is the exact inverse of differentiation; Leibniz introduced the elongated S symbol $\int f(x)\,dx$ (from Latin summa, "sum" of thin strips of height $f(x)$ and width $dx$ ). In 1854, Bernhard Riemann separated integration from antiderivatives by defining $\int_a^b f(x)\,dx$ directly as the limit of finite sums over arbitrary partitions.

Archimedes of Syracuse Isaac Newton Gottfried Wilhelm Leibniz Bernhard Riemann

SchoolThe Newton–Leibniz formula and signed area

Definition: Definite integral (Newton–Leibniz formula)

Let $f$ be a continuous function on the closed interval $[a, b]$ , and let $F$ be any antiderivative of $f$ on $[a, b]$ (so $F'(x) = f(x)$ for all $x \in [a, b]$ ). The definite integral of $f$ from $a$ to $b$ is the difference $F(b) - F(a)$ . Because any other antiderivative has the form $G(x) = F(x) + C$ for a constant $C$ , the constant cancels in subtraction: $G(b) - G(a) = (F(b) + C) - (F(a) + C) = F(b) - F(a)$ , so the definite integral does not depend on which antiderivative is chosen.

\int_a^b f(x)\,dx = F(x)\Big|_a^b = F(b) - F(a), \qquad F'(x) = f(x)

Why does subtracting an antiderivative at two endpoints compute the limit of rectangle sums? Consider the area function $A(x)$ giving the area under $y = f(t)$ from $a$ to $x$ . Increasing $x$ by a tiny $h$ adds a thin strip of height $\approx f(x)$ and width $h$ , so $\Delta A = A(x+h) - A(x) \approx f(x)\,h$ . Dividing by $h$ and letting $h \to 0$ gives $A'(x) = f(x)$ : the area function is an antiderivative of $f$ ! Since $A(a) = 0$ , we must have $A(x) = F(x) - F(a)$ . For $f(x) = x^2$ on $[0, 2]$ , taking $F(x) = x^3/3$ immediately yields $\int_0^2 x^2\,dx = 2^3/3 - 0 = 8/3$ . When $f(x)$ dips below the $x$ -axis ( $f(x) < 0$ ), rectangle heights are negative, so the integral computes signed area (area above the axis minus area below): for instance, $\int_0^{2\pi} \sin x\,dx = (-\cos 2\pi) - (-\cos 0) = 0$ because the positive arch on $[0, \pi]$ and the negative arch on $[\pi, 2\pi]$ cancel each other.

Fundamental algebraic and interval properties of the definite integral
Property	Formula	Geometric meaning
Constant multiple	$\int_a^b k f(x)\,dx = k \int_a^b f(x)\,dx$	Scaling height by $k$ scales signed area by $k$
Sum and difference	$\int_a^b [f(x) \pm g(x)]\,dx = \int_a^b f(x)\,dx \pm \int_a^b g(x)\,dx$	Stacking heights adds or subtracts the slice areas
Interval additivity	$\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx$	Splitting the region at $x = c$ into two adjacent pieces
Zero width and reversed limits	$\int_a^a f(x)\,dx = 0, \quad \int_b^a f(x)\,dx = -\int_a^b f(x)\,dx$	Integrating backward reverses the sign of $\Delta x$

Example: Signed integral versus geometric area on $[0, 2]$

For $f(x) = x^2 - 1$ on $[0, 2]$ , compute (a) the definite integral $I = \int_0^2 (x^2 - 1)\,dx$ and (b) the total geometric area $S = \int_0^2 |x^2 - 1|\,dx$ between the graph and the $x$ -axis.

Solution

(a) Using the antiderivative $F(x) = \frac{x^3}{3} - x$ , the Newton–Leibniz formula gives $I = \left(\frac{2^3}{3} - 2\right) - 0 = \frac{2}{3}$ . (b) On $[0, 2]$ , $x^2 - 1 = 0$ at $x = 1$ , with $x^2 - 1 \le 0$ on $[0, 1]$ and $x^2 - 1 \ge 0$ on $[1, 2]$ . Splitting at $x = 1$ gives $S = \int_0^1 (1 - x^2)\,dx + \int_1^2 (x^2 - 1)\,dx = \frac{2}{3} + \frac{4}{3} = 2$ .

Example

A variable force $F(x) = 3x^2 + 2$ newtons acts on an object as it moves along the $x$ -axis from $x = 0$ m to $x = 3$ m. (a) Write the work done as a definite integral. (b) Compute the exact value. (c) Find the average force over this displacement.

Solution

Step 1 — Set up the integral. The work done by variable force $F(x) = 3x^2 + 2$ along a straight path is $W = \int_0^3 F(x)\,dx = \int_0^3 (3x^2 + 2)\,dx$ .

Step 2 — Compute using the Newton–Leibniz formula. An antiderivative is $G(x) = x^3 + 2x$ . Applying the formula: $W = G(3) - G(0) = (27 + 6) - 0 = 33$ J.

Step 3 — Find the average force. By the Mean Value Theorem for Integrals, the average value of the force over $[0,3]$ is $\bar{F} = \dfrac{W}{3-0} = \dfrac{33}{3} = 11$ N. This is the constant force that does the same total work over the same displacement — a direct application of the integral mean value theorem.

UndergraduateRiemann integrability and the Fundamental Theorem of Calculus

Definition: Partition, Riemann sum, and Riemann integral

A partition of $[a, b]$ is a finite sequence $P = (x_0, x_1, \dots, x_n)$ with $a = x_0 < x_1 < \dots < x_n = b$ , subinterval widths $\Delta x_i = x_i - x_{i-1}$ , and mesh $\|P\| = \max_{1 \le i \le n} \Delta x_i$ . Choosing a sample point $x_i^* \in [x_{i-1}, x_i]$ in each subinterval gives the Riemann sum $S(f, P, x^*) = \sum_{i=1}^n f(x_i^*)\,\Delta x_i$ . A bounded function $f : [a, b] \to \mathbb{R}$ is Riemann integrable with integral $I = \int_a^b f(x)\,dx$ if for every $\varepsilon > 0$ there exists $\delta > 0$ such that $|S(f, P, x^*) - I| < \varepsilon$ whenever $\|P\| < \delta$ , regardless of how the sample points $x_i^*$ are chosen.

Theorem: Fundamental Theorem of Calculus

Let $f : [a, b] \to \mathbb{R}$ be continuous. (1) The accumulation function $A(x) = \int_a^x f(t)\,dt$ is differentiable on $[a, b]$ and $A'(x) = f(x)$ . (2) If $F$ is any antiderivative of $f$ on $[a, b]$ , then $\int_a^b f(x)\,dx = F(b) - F(a)$ .

Why is it true?

The rate at which accumulated area $A(x)$ grows with $x$ is the current height $f(x)$ of the curve at the moving right boundary. Once we know $A'(x) = f(x)$ , any other antiderivative $F$ shares the same derivative, so $A(x)$ and $F(x)$ differ by a constant.

Proof

For $h \neq 0$ , interval additivity gives $A(x+h) - A(x) = \int_x^{x+h} f(t)\,dt$ . Since $\int_x^{x+h} 1\,dt = h$ , we can write $\frac{A(x+h) - A(x)}{h} - f(x) = \frac{1}{h} \int_x^{x+h} [f(t) - f(x)]\,dt$ . By continuity of $f$ at $x$ , for any $\varepsilon > 0$ there is $\delta > 0$ such that $|t - x| \le |h| < \delta$ implies $|f(t) - f(x)| < \varepsilon$ , hence $\left|\frac{A(x+h) - A(x)}{h} - f(x)\right| \le \varepsilon$ , proving $A'(x) = f(x)$ . For Part (2), $(A - F)' = f - f = 0$ on $[a, b]$ , so $A(x) - F(x) = C$ ; setting $x = a$ gives $0 - F(a) = C$ , and setting $x = b$ gives $A(b) = F(b) - F(a)$ .

\bar{f} = \frac{1}{b-a}\int_a^b f(x)\,dx, \qquad \frac{d}{dx}\int_a^x f(t)\,dt = f(x)

Theorem: Mean Value Theorem for Integrals

If $f : [a, b] \to \mathbb{R}$ is continuous, there exists at least one point $c \in (a, b)$ such that $\int_a^b f(x)\,dx = f(c)(b - a)$ . Equivalently, $f$ attains its average value $\bar{f} = \frac{1}{b - a} \int_a^b f(x)\,dx$ at some interior point $c \in (a, b)$ .

Why is it true?

Apply Lagrange's Mean Value Theorem to the antiderivative $A(x) = \int_a^x f(t)\,dt$ on $[a, b]$ : there exists $c \in (a, b)$ with $A(b) - A(a) = A'(c)(b - a) = f(c)(b - a)$ . Geometrically, a rectangle of width $b - a$ and height $f(c)$ has the exact same signed area as the curved region under $y = f(x)$ .

Proof

Step 1 — Define the area function and apply the FTC. Let $A(x) = \int_a^x f(t)\,dt$ . By the FTC, $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ with $A'(x) = f(x)$ .

**Step 2 — Apply Lagrange's Mean Value Theorem to $A(x) = \int_a^x f(t)\,dt$ .** Since $A(x) = \int_a^x f(t)\,dt$ is continuous on $[a,b]$ and differentiable on $(a,b)$ , there exists $c \in (a,b)$ with $A(b) - A(a) = A'(c)(b-a) = f(c)(b-a)$ . Since $A(a) = 0$ , this gives $\int_a^b f(x)\,dx = f(c)(b-a)$ .

**Step 3 — Conclude that $f$ attains its average value.** Dividing by $(b-a) > 0$ yields $f(c) = \frac{1}{b-a}\int_a^b f(x)\,dx = \bar{f}$ . By the Extreme Value Theorem, $f$ attains minimum $m$ and maximum $M$ on $[a,b]$ ; since $m \le \bar{f} \le M$ , the IVT guarantees $f$ equals $\bar{f}$ at some $c \in (a,b)$ .

AdvancedSlicing 3D volumes and stepping beyond Riemann

Theorem: Cavalieri's Principle and Volume of a Solid of Revolution

If a solid lies between planes perpendicular to the $x$ -axis at $x = a$ and $x = b$ , and the cross-section at each $x \in [a, b]$ has integrable area $S(x)$ , then the solid's volume is $V = \int_a^b S(x)\,dx$ . In particular, rotating the planar region under $y = f(x)$ on $[a, b]$ around the $x$ -axis produces circular cross-sections of radius $|f(x)|$ and area $S(x) = \pi [f(x)]^2$ , so $V = \pi \int_a^b [f(x)]^2\,dx$ .

Why is it true?

Slice the solid into $n$ thin slabs of thickness $\Delta x_i$ ; the $i$ -th slab is approximately a cylinder of base area $S(x_i^*)$ and volume $S(x_i^*)\,\Delta x_i$ . Summing over all slabs gives a Riemann sum $\sum_{i=1}^n S(x_i^*)\,\Delta x_i \to \int_a^b S(x)\,dx$ . Because the integral depends only on $S(x)$ , any two solids with equal cross-sectional areas at every height have the same volume (Cavalieri's principle, 1635).

Proof

Step 1 — Slice into thin slabs. Partition $[a,b]$ into $n$ equal subintervals of width $\Delta x = (b-a)/n$ . On the $i$ -th subinterval $[x_{i-1},x_i]$ , the solid is approximated by a slab of base area $S(x_i^*)$ and thickness $\Delta x$ , giving slab volume $S(x_i^*)\,\Delta x$ .

Step 2 — Form the Riemann sum and take the limit. Summing all slabs: $V_n = \sum_{i=1}^n S(x_i^*)\,\Delta x$ . Since $y = f(x)$ is integrable on $[a,b]$ , this converges as $n \to \infty$ regardless of sample points $x_i^*$ : $V = \lim_{n\to\infty} V_n = \int_a^b S(x)\,dx$ .

Step 3 — Apply to a solid of revolution and state Cavalieri's principle. Rotating the region under $y = f(x)$ about the $x$ -axis produces cross-sections of radius $|f(x)|$ and area $S(x) = \pi[f(x)]^2$ . Substituting: $V = \pi\int_a^b [f(x)]^2\,dx$ . Cavalieri's principle (1635) states: if two solids satisfy $S_A(x) = S_B(x)$ for all $x \in [a,b]$ , then $\int_a^b S_A(x)\,dx = \int_a^b S_B(x)\,dx$ , so they have equal volumes regardless of shape.

Solid of revolution generated by rotating

y = \sqrt{x}

[0, 4]

around the

x

-axis. Each slice at

x

is a disk of radius

\sqrt{x}

and area

\pi x

, giving volume

V = \pi \int_0^4 x\,dx = 8\pi

. Adjust the rotation angle or drag to inspect the 3D surface.

What happens when Riemann slices are no longer enough? If a function jumps on a dense set—such as Dirichlet's indicator function $\mathbf{1}_{\mathbb{Q}}(x)$ on $[0, 1]$ , which equals $1$ on rationals and $0$ on irrationals—every subinterval contains both rationals and irrationals, so upper Riemann sums equal $1$ , lower Riemann sums equal $0$ , and the Riemann integral does not exist. In 1902, Henri Lebesgue resolved this by slicing the range (values $y$ ) instead of the domain $x$ and measuring the set of $x$ 's where $f(x) \approx y$ (see [Measure theory and Lebesgue integration](/ly-thuyet-do-do)): since $\mathbb{Q}$ is countable and has Lebesgue measure $0$ , $\int_{[0,1]} \mathbf{1}_{\mathbb{Q}}\,d\mu = 0$ , and powerful convergence theorems allow passing limits inside the integral ( $\lim \int f_n = \int \lim f_n$ ). In higher dimensions ([Multiple integrals](/tich-phan-boi)), integration over curves, surfaces, and manifolds is unified by differential forms and the generalized Stokes theorem $\int_{\partial \Omega} \omega = \int_\Omega d\omega$ , of which the Newton–Leibniz formula $\int_a^b dF = F(b) - F(a)$ is the 1-dimensional case.

ResearchIntegration at the modern research frontier

Research frontier as of 2026

As of 2026, three active research directions push integration far beyond classical calculus: (1) High-dimensional numerical integration and Quasi-Monte Carlo (QMC). In statistical physics, financial mathematics, and Bayesian machine learning, one must evaluate integrals $\int_{[0,1]^d} f(x)\,dx$ in hundreds or thousands of dimensions ( $d \gg 100$ ). Classical grid quadrature needs $N^d$ evaluations (the curse of dimensionality), whereas randomized QMC methods and higher-order digital nets (Owen, Dick, Pillichshammer) achieve nearly $O(N^{-\alpha})$ error rates independent of dimension under weighted Sobolev norms. (2) Rough paths and singular stochastic integration. When integrating $\int_0^T Y_t\,dX_t$ driven by highly irregular signals $X_t$ (such as fractional Brownian motion with Hurst parameter $H \le 1/2$ ), Riemann–Stieltjes sums diverge; Terry Lyons' rough path theory (1998) and Martin Hairer's regularity structures (Fields Medal 2014, with ongoing advances through 2026 in singular stochastic PDEs and quantum field theory) restore well-posed integration by augmenting the path with its iterated integrals. (3) Formalized measure theory and symbolic integration. Following Robert Risch's decision algorithm (1969–1970) for elementary antiderivatives, interactive theorem provers—most notably the Lean 4 `mathlib` library (2020–2026)—have fully formalized the Bochner integral, Haar measure, and the Fundamental Theorem of Calculus, enabling machine-verified proofs in modern analysis and probability.

By the Newton–Leibniz formula, what is the value of $\int_0^2 x^2\,dx$ ?

Why does $\int_0^{2\pi} \sin x\,dx = 0$ even though the curve $y = \sin x$ encloses a non-zero geometric area with the $x$ -axis?

Let $A(x) = \int_1^x \sqrt{t^3 + 1}\,dt$ for $x > 0$ . By the Fundamental Theorem of Calculus, what is $A'(2)$ ?

Rotating the region under $y = \sqrt{x}$ on $[0, 4]$ around the $x$ -axis forms a solid of revolution. What is its volume $V$ ?

References

Michael Spivak (2008). Calculus
James Stewart (2015). Calculus: Early Transcendentals
Bernhard Riemann (1868). Ueber die Darstellbarkeit einer Function durch eine trigonometrische Reihe
Carl B. Boyer (1959). The History of the Calculus and Its Conceptual Development