1.2 The Definite Integral

Big idea

Rectangle sums gave a way to estimate an accumulated total, and letting the number of rectangles grow gave a single number the estimates approach. This section gives that number a name, a symbol, and a working set of rules. The name is the definite integral of $f$ from $a$ to $b$, and the symbol keeps the shape of the sum it came from: the stretched $S$ remembers the addition, $f(x)$ remembers the height, and $dx$ remembers the width.

Two things change once the limit is taken. First, the restriction that $f$ be nonnegative disappears. A rectangle whose height is negative contributes a negative amount, so the integral of a function that dips below the axis counts area below the axis as negative. The integral measures signed area, and a function that spends equal time above and below can integrate to zero even though its graph encloses plenty of region.

Second, the integral acquires an algebra. Because a Riemann sum is a sum, everything that is true about sums transfers: constants come out, sums split apart, and a partition of the interval splits the integral into pieces. These rules are the ones you will actually use, because they let you take an integral apart into recognizable chunks long before you know any method for evaluating one.

The section ends with a question the integral answers cleanly and nothing else does. What is the average value of a function across an interval? Averaging finitely many numbers means adding and dividing by how many. A function has infinitely many values, so the sum becomes an integral and the count becomes the length of the interval. The formula that results is the first genuinely new quantity the integral produces.

Decoder

The definite integral of $f$ from $a$ to $b$ is the limit of its Riemann sums as the subinterval width goes to zero, and it exists as one well-defined number whenever $f$ is continuous on $[a,b]$.

The phrase carrying the weight is “one well-defined number”. Every ingredient in a Riemann sum was a choice: how many pieces, where the sample points sit. The claim is that in the limit none of those choices survive. If two people build sums with different sample points and both let the width shrink, they reach the same number, and that is what allows a single symbol to stand for it.

The word “continuous” is a sufficient condition, not a necessary one. Functions with finitely many jumps are integrable too; badly behaved functions need not be. For everything in this course, continuous on a closed interval is the check to run, and it is almost always satisfied.

The classic mistake is reading the integral as area rather than signed area. If a problem asks for the total area between a graph and the axis, and the graph crosses the axis inside the interval, the integral over the whole interval is not the answer. Split at the crossing and change the sign on the pieces where the function is negative.

Definitions and results

The definite integral. For $f$ defined on $[a,b]$, with $\Delta x = (b-a)/n$ and sample points $x_i^{*}$,

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

when the limit exists. Here $a$ and $b$ are the limits of integration and $x$ is a dummy variable: $\int_a^b f(t)\,dt$ is the same number.

Integrability. If $f$ is continuous on $[a,b]$, or bounded with only finitely many discontinuities there, then $f$ is integrable on $[a,b]$.

Signed area. The integral counts region above the axis positively and region below it negatively. To get total unsigned area, integrate $|f|$, which in practice means splitting at the zeros of $f$.

Degenerate and reversed limits. Two conventions make the rest of the algebra work without special cases:

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

Linearity. Constants factor out and sums separate:

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

Splitting the interval. For any $c$, provided the pieces exist,

$$ \int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx $$

This holds whether or not $c$ lies between $a$ and $b$, thanks to the reversal convention.

Comparison. If $f(x) \le g(x)$ for all $x$ in $[a,b]$ with $a \le b$, then $\int_a^b f \le \int_a^b g$. In particular, if $m \le f(x) \le M$ on $[a,b]$ then

$$ m(b-a) \le \int_a^b f(x)\,dx \le M(b-a) $$

which gives a crude bound on any integral in one line.

Average value. The average value of $f$ on $[a,b]$ is

$$ f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx $$

Read it as a rectangle of width $b-a$ whose area matches the area under the curve: $f_{\text{avg}}$ is the height that rectangle needs.

Mean value theorem for integrals. If $f$ is continuous on $[a,b]$, then $f(c) = f_{\text{avg}}$ for at least one $c$ in $[a,b]$. A continuous function actually attains its own average somewhere.

Worked examples

Signed area from geometry

Evaluate $\int_0^4 (x-1)\,dx$ by looking at the region.

The graph is a line crossing the axis at $x = 1$. On $[0,1]$ it lies below the axis, forming a triangle with base $1$ and height $1$, so that piece contributes $-\tfrac{1}{2}$. On $[1,4]$ it lies above, forming a triangle with base $3$ and height $3$, contributing $+\tfrac{9}{2}$. The integral is

$$ \int_0^4 (x-1)\,dx = -\frac{1}{2} + \frac{9}{2} = 4 $$

The total unsigned area between the line and the axis is a different number, $\tfrac{1}{2} + \tfrac{9}{2} = 5$. Asking which one a problem wants is worth one extra second of reading.

A quarter of a familiar constant

Evaluate $\int_{-3}^{3} \sqrt{9-x^2}\,dx$.

Setting $y = \sqrt{9-x^2}$ and squaring gives $x^2 + y^2 = 9$ with $y \ge 0$, so the graph is the upper half of the circle of radius $3$. The region under it on $[-3,3]$ is a half disc, and

$$ \int_{-3}^{3}\sqrt{9-x^2}\,dx = \frac{1}{2}\pi(3)^2 = \frac{9\pi}{2} $$

No limit of sums is needed when the region is a shape you can already measure. Recognizing a semicircle, a triangle or a trapezoid under the integral sign is a skill worth keeping.

Using the rules instead of evaluating

Suppose $\int_0^5 f(x)\,dx = 12$ and $\int_0^2 f(x)\,dx = 5$. Find $\int_2^5 f(x)\,dx$ and $\int_5^2 3f(x)\,dx$.

Splitting at $2$ gives $\int_0^5 f = \int_0^2 f + \int_2^5 f$, so $\int_2^5 f = 12 - 5 = 7$. For the second, pull the constant out and reverse the limits:

$$ \int_5^2 3f(x)\,dx = 3\int_5^2 f(x)\,dx = -3\int_2^5 f(x)\,dx = -21 $$

Nothing about $f$ itself was used. The algebra of integrals runs on the numbers alone.

The average value of a linear function

Find the average value of $h(x) = 2x+1$ on $[1,4]$, and the point where it is attained.

The region under the line is a trapezoid with parallel sides $h(1) = 3$ and $h(4) = 9$ and width $3$, so $\int_1^4 (2x+1)\,dx = \tfrac{1}{2}(3+9)(3) = 18$. The average value is

$$ h_{\text{avg}} = \frac{1}{4-1}\int_1^4 (2x+1)\,dx = \frac{18}{3} = 6 $$

Solving $2c + 1 = 6$ gives $c = 2.5$, which does lie in $[1,4]$, as the mean value theorem for integrals promises.

Two checks. The average of a linear function should be its value at the midpoint of the interval, and $h(2.5) = 6$. And $h$ runs from $3$ to $9$ on this interval, so the comparison bound says the average must lie between $3$ and $9$; it does.

Practice

Start with the integral itself: reading it as signed area, evaluating simple ones from geometry, and applying the splitting, linearity and reversal rules to combine given values.

Then the average value. These ask for $f_{\text{avg}}$ on an interval and, in some cases, for the point where a continuous function attains it.

Quiz

Five items on the definition, signed area, the algebra rules, and average value.