Linear Function Example in Integral Calculus

Every course on integration opens with the same question: what does the symbol $\int$ actually mean? The most honest answer is that it is a shorthand for a limit of sums, and the most convincing way to see that is to carry out the limit once, slowly, on a function that is simple enough to finish the arithmetic. That is the purpose of this page: $y = x$ is the easiest integrand there is, and yet computing $\int_a^b x\,dx$ from the definition takes real work and produces a result we can verify geometrically in two ways.

Once the machinery has been run once by hand, later methods look like variations on the same theme. The value of doing it this way is that you end up with a formula, $\int_a^b x\,dx = \frac{b^2 - a^2}{2}$, that you can check at a glance — and that you understand exactly what the limit in the definition is doing.

Definition

For a function $f$ continuous on $[a,b]$, the definite integral is the limit of the Riemann sums

$$\int_a^b f(x)\,dx = \lim_{n\to\infty}\sum_{i=1}^{n} f(x_i)\,\Delta x,$$

where the interval is cut into $n$ pieces of equal width $\Delta x = \frac{b-a}{n}$, and $x_i$ is a point of the $i$-th piece. The limit exists because, for a continuous $f$, the value of the sum does not depend on where inside each piece you pick $x_i$ — the pieces are thin enough for that to be true in the limit.

For this page the integrand is $f(x) = x$, and we take the right endpoint of each piece, so $x_i = a + i\Delta x$.

The Problem

The shaded region below is the area under $y = x$ between $x = a$ and $x = b$. Find it.

Two routes lead to the answer, and this page takes both. The geometric route uses the trapezoid area formula. The integral route builds the area from infinitely many thin strips. They must agree, and the agreement is the check that the definition of the integral is doing its job.

The Geometric Route

For $y = x$ on $[a,b]$ with $a < b$, the region bounded by the curve, the axis and the two vertical lines at $a$ and $b$ is a trapezoid: the two non-parallel sides are parallel to each other, the lower base has length $a$ (the height of the curve at the left end), the upper base has length $b$ (its height at the right end), and the height of the trapezoid is the horizontal width $b - a$.

Substituting the three lengths gives

$$S = \frac{(a + b)(b - a)}{2} = \frac{b^2 - a^2}{2}.$$

The Integral Route

The idea behind the integral is the same picture seen at closer range: cut the region into many narrow vertical strips, approximate each strip by a rectangle, and add up the areas.

1. Divide $[a,b]$ into $n$ narrow strips of width $\Delta x$.
2. Approximate the $i$-th strip by a rectangle of height $f(x_i)$ and width $\Delta x$, so its area is $f(x_i)\cdot\Delta x$.
3. Add the areas of all $n$ strips, and then let $n$ grow without bound.

For $f(x) = x$ this gives the height of the $i$-th rectangle as $x_i$ and its area as $x_i \cdot \Delta x$, so the total area is

$$\sum_{i=1}^{n} x_i \cdot \Delta x.$$

When the strips are made thin enough, the sum of the rectangle areas approaches the true area of the region, and the precise area is

$$\lim_{n \to \infty} \sum_{i=1}^{n} x_i \cdot \Delta x.$$

The Summation Formula We Need

Before the limit can be evaluated, one summation formula has to be in hand.

Carrying Out the Limit

Now everything is in place. For $y = x$ we have $x_i = a + i\Delta x$ and $\Delta x = \frac{b-a}{n}$, so

$$\sum_{i=1}^{n} x_i \cdot \Delta x = \sum_{i=1}^{n} \left(a + i \cdot \frac{b-a}{n}\right) \cdot \frac{b-a}{n} = \sum_{i=1}^{n} \left(a \cdot \frac{b-a}{n} + i \cdot \frac{(b-a)^2}{n^2}\right)$$ $$= n \cdot a \cdot \frac{b-a}{n} + \frac{(b-a)^2}{n^2} \cdot \sum_{i=1}^{n} i = a(b-a) + \frac{(b-a)^2}{n^2} \cdot \frac{n(n+1)}{2}.$$

Taking the limit as $n \to \infty$, the factor $\frac{n+1}{n}$ tends to $1$:

$$\lim_{n \to \infty} \left[a(b-a) + \frac{(b-a)^2}{2}\cdot\frac{n+1}{n}\right] = a(b-a) + \frac{(b-a)^2}{2} = \frac{(b-a)(a+b)}{2} = \frac{b^2 - a^2}{2}.$$ $$\int_a^b x \, dx = \frac{b^2 - a^2}{2}$$

The symbol $\int$ is the integral sign, read as “integral”; the notation $\int_a^b$ denotes the definite integral from $a$ to $b$, that is, the limit of the sums of thin strips described above.

Verification

The integral route produced $\frac{b^2 - a^2}{2}$ and the geometric route produced $\frac{(a+b)(b-a)}{2}$. These are the same number, since $b^2 - a^2 = (b-a)(b+a)$, so the two independent arguments agree.

Two quick numerical checks are worth doing. With $a = 0$ and $b = 1$ the region is a right triangle with base $1$ and height $1$, so the area must be $\frac{1}{2}$, and the formula gives $\frac{1^2 - 0^2}{2} = \frac{1}{2}$. With $a = 5$ and $b = 8$ the region is a trapezoid of area $\frac{(5+8)(8-5)}{2} = \frac{39}{2} = 19.5$, and the formula gives $\frac{64 - 25}{2} = \frac{39}{2}$ as well.

For functions less tame than a straight line, no such shortcut exists, and the strip-by-strip computation becomes considerably heavier. That is what the remaining chapters are for.

Exercises

Exercise 1

Compute the definite integral of $y = x$ over the interval $[0,2]$.

Exercise 2

Evaluate the definite integral of $y = x$ over $[0,1]$ and explain the result geometrically.

Exercise 3

Evaluate $\int_{-1}^{2} x\,dx$, and then find the total area of the region between $y = x$ and the $x$-axis on $[-1,2]$. Why are the two answers different?

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Summary

The definite integral is a limit of Riemann sums, and running that limit on $y = x$ makes the machinery concrete: cut the interval into strips, approximate each by a rectangle, sum the areas, and let the strips become infinitely thin. The two summation formulas involved — $\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$ and the geometric series behind it — are what make the limit evaluable, and the result $\int_a^b x\,dx = \frac{b^2 - a^2}{2}$ reproduces the trapezoid area formula exactly.

Two lessons are worth carrying forward. First, always check an integral against an independent argument when one is available; the agreement between the strip computation and the trapezoid formula is what makes the definition trustworthy. Second, remember what the integral computes: signed area, not total area. When the curve dips below the axis, the two differ, and confusing them is one of the most consequential errors in the whole subject.

Symbols Used in This Article