Area and the Definite Integral

Differential calculus began with the slope of a tangent and answered it with a limit of secant slopes. Integral calculus begins with a different question whose answer is also a limit. Given a function $f$ on $[a, b]$ , what is the area of the region under its graph? For a rectangle or a triangle the answer is elementary. For a region with a curved upper boundary there is no formula, so area itself has to be defined, and the definition is a limit.

Defining area precisely isolates one type of limit: a sum of many small contributions, taken as each contribution shrinks. The same limit recurs whenever a quantity accumulates from a varying rate. Distance from velocity, mass from density, work from force all reduce to it. That limit is the definite integral.

Approximating an area by rectangles

Let $f$ be continuous and nonnegative on $[a, b]$ , and let $S$ be the region bounded above by $y = f (x)$ , below by the $x$ -axis, and on the sides by the lines $x = a$ and $x = b$ . To estimate the area of $S$ , cut $[a, b]$ into $n$ subintervals of equal width

Δ x = \frac{b - a}{n},

and over each subinterval erect a rectangle whose height is a value of $f$ on that subinterval. Choosing the height at the right endpoint $x_{i} = a + i Δ x$ gives the right-endpoint sum

R_{n} = f (x_{1}) Δ x + f (x_{2}) Δ x + \dots + f (x_{n}) Δ x .

Choosing the left endpoint gives $L_{n}$ ; choosing the midpoint gives $M_{n}$ . Each is an approximation to the area, exact only in the limit.

Right-endpoint rectangles under

y = x^{2}

[0, 1]

for

n = 4

and

n = 8

; refining the partition drives the total rectangle area toward the true area

1/3

For $f (x) = x^{2}$ on $[0, 1]$ the two estimates bracket the area from opposite sides, because $x^{2}$ is increasing: the left-endpoint rectangles sit under the curve and the right-endpoint rectangles poke above it. A short computation with $n = 4$ gives

L_{4} = 0.21875, R_{4} = 0.46875,

so $0.21875 < A < 0.46875$ . Doubling to $n = 8$ tightens this to $0.2734 < A < 0.3984$ , and $n = 1000$ traps $A$ between $0.33283$ and $0.33383$ . Both sequences close on the same value, computed exactly below to be $\frac{1}{3}$ .

Evaluating the limit for a power

The definition is only useful if the limit can sometimes be computed. For $f (x) = x^{2}$ on $[0, 1]$ the right endpoints are $x_{i} = i / n$ and $Δ x = 1/ n$ , so

R_{n} = i = 1 \sum n (\frac{i}{n})^{2} \frac{1}{n} = \frac{1}{n ^{3}} i = 1 \sum n i^{2} .

The sum of the first $n$ squares has a closed form,¹

i = 1 \sum n i^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6},

and substituting collapses the whole expression to a ratio of polynomials in $n$ :

R_{n} = \frac{1}{n ^{3}} \cdot \frac{n ( n + 1 ) ( 2 n + 1 )}{6} = \frac{( n + 1 ) ( 2 n + 1 )}{6 n ^{2}} = \frac{1}{6} (1 + \frac{1}{n}) (2 + \frac{1}{n}) .

As $n \to \infty$ the two correction terms vanish:

A = n \to \infty lim R_{n} = \frac{1}{6} (1) (2) = \frac{1}{3} .

The lower sums $L_{n}$ approach the same $\frac{1}{3}$ from below, confirming the estimate. Computing an area this way is exact but laborious; it requires a summation formula tailored to the integrand. The Fundamental Theorem removes that labor entirely, but the limit is what the answer means.

A cubic integrand pushes the same method one step further and uses two power-sum identities at once, $\sum i^{3} = [n (n + 1) /2]^{2}$ and $\sum i = n (n + 1) /2$ .

Sample points and sigma notation

Nothing in the construction requires the height to come from an endpoint. Let $x_{i}^{*}$ (also written $c_{i}$ ) be any point in the $i$ -th subinterval $[x_{i - 1}, x_{i}]$ , called a sample point, and the sum $\sum_{i} f (x_{i}^{*}) Δ x$ still approximates the area. Endpoints and midpoints are the convenient choices, but the limit is independent of the choice.

A partition of

[a, b]

into

n

subintervals of width

Δ x

, with an arbitrary sample point

c_{i} = x_{i}^{*}

in each; the shaded rectangle has height

f (c_{i})

and width

Δ x

Sigma notation compresses these sums. Writing the area three equivalent ways,

A = n \to \infty lim i = 1 \sum n f (x_{i}) Δ x = n \to \infty lim i = 1 \sum n f (x_{i - 1}) Δ x = n \to \infty lim i = 1 \sum n f (x_{i}^{*}) Δ x,

with right endpoints, left endpoints, and arbitrary samples respectively.

The distance problem gives the same limit

An object moves along a line with velocity $v = f (t) \geq 0$ on $a \leq t \leq b$ . If the velocity were constant the distance would be velocity times time. It is not constant, so partition the time interval, treat the velocity as approximately constant on each short piece, and add up the piecewise distances. With readings at $t_{0}, t_{1}, \dots, t_{n}$ spaced $Δ t$ apart, the distance over $[t_{i - 1}, t_{i}]$ is about $f (t_{i - 1}) Δ t$ , and the total is

d \approx i = 1 \sum n f (t_{i - 1}) Δ t .

The estimate sharpens as the readings grow more frequent, and in the limit

d = n \to \infty lim i = 1 \sum n f (t_{i}) Δ t .

This is the area limit with $f$ a velocity and $Δ x$ a $Δ t$ : distance traveled equals the area under the velocity graph. Area, distance, mass, and work share one structure — partition, sample, sum, take the limit — which the definite integral captures once and for all.

A step-function estimate of distance: each rectangle's area,

f (t_{i - 1}) Δ t

, is the distance covered in one time interval at the velocity read at its left endpoint.

The definite integral

The limit of Riemann sums is given its own name and symbol.

The pieces of the notation are named:

Integral sign $\int$ : an elongated S, chosen by Leibniz because the integral is a limit of sums.
Integrand $f (x)$ : the function being integrated.
Limits of integration $a$ (lower) and $b$ (upper).
Differential $d x$ : for now it carries no value of its own; it marks $x$ as the variable of integration and corresponds to the $Δ x$ in the sum.

The sum $\sum_{i = 1}^{n} f (x_{i}^{*}) Δ x$ is a Riemann sum, after Bernhard Riemann. A definite integral is the number a Riemann sum approaches; equivalently, a Riemann sum approximates the integral to any accuracy once $n$ is large enough.

Two facts make the definition workable in practice. First, the integral is a number, not a function of $x$ — the variable is a placeholder, so $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (t) d t$ . Second, continuity is enough to guarantee the limit exists.

For an integrable function the sample points may as well be right endpoints, which simplifies calculation:

a \int b f (x) d x = n \to \infty lim i = 1 \sum n f (x_{i}) Δ x, Δ x = \frac{b - a}{n}, x_{i} = a + i Δ x .

A limit of sums can be read backwards as an integral. The limit $lim_{n \to \infty} \sum_{i = 1}^{n} (x_{i}^{3} + x_{i} sin x_{i}) Δ x$ over $[0, π]$ is, by matching against the definition, exactly $\int_{0}^{π} (x^{3} + x sin x) d x$ ; the integrand is read straight off the summand.

The integral as net area

When $f \geq 0$ the integral is the area under the graph. When $f$ dips below the axis the Riemann sum subtracts: a rectangle below the axis has negative height $f (x_{i}^{*}) < 0$ , so it contributes a negative amount. The integral therefore measures net area — area above the axis counted positive, area below counted negative.

The integral of a sign-changing function is the area above the axis minus the area below it; here the region

A_{1}

adds and

A_{2}

subtracts.

For example, $\int_{0}^{3} (x^{3} - 6 x) d x = - 6.75$ : the function is negative over most of $[0, 3]$ , so the region below the axis outweighs the sliver above it, and the integral comes out negative even though it is built from areas.

Properties of the definite integral

Before doing any evaluation, two conventions extend the definition to non-increasing limits. Reversing the endpoints flips $Δ x$ and hence the sign; equal endpoints give a degenerate partition of width zero:

b \int a f (x) d x = - a \int b f (x) d x, a \int a f (x) d x = 0.

The following properties all descend from the definition and the fact that a limit of sums respects sums and constant factors. Take $f$ and $g$ integrable.

#	Property	Reading
1	$a \int b c d x = c (b - a)$	integral of a constant is $c$ times the width
2	$a \int b [f + g] d x = a \int b f d x + a \int b g d x$	integral of a sum is the sum of integrals
3	$a \int b c f d x = c a \int b f d x$	a constant factor comes outside
4	$a \int b [f - g] d x = a \int b f d x - a \int b g d x$	integral of a difference splits
5	$a \int c f d x + c \int b f d x = a \int b f d x$	adjacent intervals join

Property 5 is the additive law over subintervals: the area from $a$ to $c$ plus the area from $c$ to $b$ is the area from $a$ to $b$ . It holds for any ordering of $a, b, c$ once the sign conventions above are in force. As a worked instance, if $\int_{0}^{10} f = 17$ and $\int_{0}^{8} f = 12$ , then

8 \int 10 f (x) d x = 0 \int 10 f (x) d x - 0 \int 8 f (x) d x = 17 - 12 = 5.

Comparison properties

Three more properties compare sizes and require $a \leq b$ .

Nonnegativity. If $f (x) \geq 0$ on $[a, b]$ , then $\int_{a}^{b} f d x \geq 0$ .
Monotonicity. If $f (x) \geq g (x)$ on $[a, b]$ , then $\int_{a}^{b} f d x \geq \int_{a}^{b} g d x$ — a bigger integrand has a bigger integral.
Bounding. If $m \leq f (x) \leq M$ on $[a, b]$ , then $m (b - a) \leq a \int b f (x) d x \leq M (b - a) .$

Monotonicity orders two integrals without evaluating either.

The bounding property is the quickest way to estimate an integral without any summation. The integral is trapped between the areas of two rectangles spanning $[a, b]$ : one of height $m$ , the minimum of $f$ , and one of height $M$ , the maximum.

The bounding property traps

\int_{a}^{b} f d x

between the rectangle of height

m

(the minimum of

f

) and the rectangle of height

M

(the maximum), both of width

b - a

The Midpoint Rule

When a numerical estimate is the goal rather than an exact limit, midpoints beat endpoints. Take the sample point $\overset{x}{ˉ}_{i} = \frac{1}{2} (x_{i - 1} + x_{i})$ at the center of each subinterval.

Midpoint rectangles: each height is

f (\overset{x}{ˉ}_{i})

sampled at the center of its subinterval, so the overshoot on one half offsets the undershoot on the other.

A midpoint rectangle overshoots on one half of its subinterval and undershoots on the other, so the errors partly cancel and the estimate is typically much better than $L_{n}$ or $R_{n}$ at the same $n$ . For $\int_{1}^{2} (1/ x) d x$ with $n = 5$ , the midpoints are $1.1, 1.3, 1.5, 1.7, 1.9$ and

1 \int 2 \frac{1}{x} d x \approx \frac{1}{5} (\frac{1}{1.1} + \frac{1}{1.3} + \frac{1}{1.5} + \frac{1}{1.7} + \frac{1}{1.9}) \approx 0.6919.

Estimating the same integral $\int_{0}^{3} (x^{3} - 6 x) d x$ discussed above with $n = 40$ midpoints gives $M_{40} \approx - 6.7563$ , already closer to the exact $- 6.75$ than the right-endpoint sum $R_{40} \approx - 6.3998$ . The approximate-integration lesson develops the error bounds that make this precise.

Stewart, §4.1 and Appendix E (Sigma Notation), where the identity $\sum_{i = 1}^{n} i^{2} = n (n + 1) (2 n + 1) /6$ is proved by induction. ↩