What an integral measures, starting with a rectangle, integrals from easy to advanced lesson

A height of $2$ held constant from $x = 0$ to $x = 3$ encloses a rectangle, and the integral

\int_0^3 2\,dx

asks for the total that this height accumulates across the interval. This page reads each part of the notation, finds the value $6$ from the rectangle, and explains when an integral counts as positive or negative and which unit its answer carries.

Key terms

Definite integral

Definition Let $f$ be a function on a closed interval $[a, b]$ with $a < b$ . Cut $[a, b]$ into pieces, choose a point in each piece, and add up the products of the height of $f$ at the chosen point and the width of the piece. If these sums approach one number as the widest piece shrinks to width $0$ , whatever cuts and points are used, then $f$ is integrable on $[a, b]$ and that number is the definite integral

\int_a^b f(x)\,dx.

Every continuous function on $[a, b]$ is integrable. A constant function $f(x) = c$ is the simplest case, because every such sum equals $c$ times the total width of the pieces, so

\int_a^b c\,dx = c\,(b - a).

The page on the Riemann-sum limit makes this limit precise.

In plain words A definite integral adds up height times width across an interval, using thinner and thinner strips. The result is one number, fixed by the function and the two endpoints. The integral sign is a stretched letter S, a reminder that it comes from a sum.

Example For the integral $\int_0^3 2\,dx$ , the integrand is the constant $2$ on $[0, 3]$ . Cut the interval into strips of any widths $\Delta x_1, \Delta x_2, \dots, \Delta x_n$ . Every strip has height $2$ , so

\begin{aligned} &2\,\Delta x_1 + \dots + 2\,\Delta x_n \\[0.4em] &\quad = 2\,(\Delta x_1 + \dots + \Delta x_n) \\[0.4em] &\quad = 2 \cdot 3 = 6, \end{aligned}

because the widths fill the interval from $0$ to $3$ . Three broad strips and a thousand thin ones give the same total, so every sum is exactly $6$ , the limit is $6$ , and

\int_0^3 2\,dx = 2\,(3 - 0) = 6.

This is the area of the rectangle with width $3$ and height $2$ . The value is positive because the height $2$ is positive and the limits increase from $0$ to $3$ .

Integrand and limits of integration

Definition In the definite integral $\int_a^b f(x)\,dx$ , the function $f(x)$ is the integrand, the number $a$ is the lower limit of integration and the number $b$ is the upper limit. The symbol $dx$ marks $x$ as the variable of integration, the variable that runs from $a$ to $b$ . That variable is a placeholder, so $\int_a^b f(t)\,dt$ is the same number. When $a < b$ , the limits are the endpoints of the interval $[a, b]$ , whose width is $b - a$ .

In plain words The integrand gives the height at each position, and the limits say where the accumulation starts and where it stops. The $dx$ belongs to the width of each strip, not to its height, so it never multiplies the integrand by an extra factor of $x$ .

Example In $\int_1^4 5\,dx$ the integrand is $5$ , the lower limit is $1$ and the upper limit is $4$ . The width is $4 - 1 = 3$ , so

\int_1^4 5\,dx = 5 \cdot 3 = 15.

Using the upper limit as the width would give $5 \cdot 4 = 20$ , which is the integral from $0$ to $4$ , a different interval. Renaming the variable changes nothing, since $\int_1^4 5\,dt = 15$ as well.

Signed area

Definition Let $f$ be integrable on $[a, b]$ with $a < b$ . The integral counts the region between the graph of $f$ and the $x$ -axis with signs. Area above the axis counts as positive and area below the axis counts as negative,

\begin{aligned} \int_a^b f(x)\,dx &= A_{\text{above}} \\[0.4em] &\quad - A_{\text{below}}, \end{aligned}

where $A_{\text{above}}$ is the area of the part of the region above the axis and $A_{\text{below}}$ is the area of the part below it. The direction also carries a sign. By definition

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

and $\int_a^a f(x)\,dx = 0$ . The geometric area between the graph and the axis, with no signs, is $\int_a^b \lvert f(x) \rvert\,dx$ .

In plain words An integral is a net total. Parts of the graph below the axis subtract, and running the interval from right to left flips the sign of the whole result. The sign comes from the height and the direction of travel, not from whether the positions on the axis are negative.

Example The integral of $-2$ over $[0, 3]$ describes a rectangle of the same size as before, but below the axis, so

\int_0^3 (-2)\,dx = -6,

although the shaded rectangle has geometric area $6$ . Reversing the limits of the original integral also gives $\int_3^0 2\,dx = -6$ . For a step function equal to $2$ on $[0, 1]$ and to $-1$ on $(1, 3]$ , the signed total is $2 \cdot 1 + (-1) \cdot 2 = 0$ , while the geometric area is $2 + 2 = 4$ . In $\int_0^3 2\,dx$ nothing cancels, since every strip has positive height and positive width.

Accumulation

Definition Let $r(t)$ be the rate at which a quantity $Q$ changes, measured per unit of $t$ , so that $Q'(t) = r(t)$ , and suppose $r$ is continuous on $[a, b]$ . The accumulation of the rate over $[a, b]$ is the definite integral

\int_a^b r(t)\,dt,

and it equals the net change $Q(b) - Q(a)$ . For a constant rate $r$ every sum of contributions $r\,\Delta t$ equals $r\,(b - a)$ , so the net change is $r\,(b - a)$ . The general case is the fundamental theorem of calculus.

In plain words An integral collects rate times a small change of input, piece by piece, and adds the pieces. The height of the graph is a rate, the width of a strip is an amount of input, and their product is an amount of the quantity. A negative rate means the quantity decreases, so its contribution is negative.

Example A pump fills a tank at a steady 2 liters per second from $t = 0$ to $t = 3$ seconds. In any short stretch of $\Delta t$ seconds the tank gains $2\,\Delta t$ liters, and these gains add up to

\int_0^3 2\,dt = 2\,(3 - 0) = 6

liters. This is the rectangle of $\int_0^3 2\,dx$ with new labels on the axes. If the pump instead drains the tank at 2 liters per second, the rate is $-2$ and the volume changes by $\int_0^3 (-2)\,dt = -6$ liters.

Units of an integral

Definition If the variable of integration $x$ is measured in one unit and the integrand $f(x)$ in another, then $\int_a^b f(x)\,dx$ is measured in the product of the two units. Writing $[q]$ for the unit of a quantity $q$ ,

[\text{integral}] = [f(x)]\,[x].

When neither the integrand nor the variable carries a unit, the integral is a pure number.

In plain words Each strip contributes height times width, so its unit is the unit of the height times the unit of the width. Read the labels on both axes before naming the unit of an answer. The answer is an area in square units of length only when both axes measure length.

Example If both axes of the rectangle under the height $2$ from $0$ to $3$ measure centimeters, the integral is $6$ square centimeters. With time in seconds and a rate in liters per second it is $6$ liters, because the seconds cancel. A heater that draws a constant 1.5 kilowatts for 4 hours uses

\int_0^4 1.5\,dt = 6

kilowatt-hours of energy. The integral $\int_0^3 2\,dx$ itself has no labels, so its value $6$ is a pure number.

Check yourself

Name the integrand, the limits of integration and the variable of integration in $\int_{-2}^{3} 4\,dt$ , and evaluate the integral.

Show the answer

The integrand is $4$ , the lower limit is $-2$ , the upper limit is $3$ , and the variable of integration is $t$ . The width of the interval is $3 - (-2) = 5$ , so

\int_{-2}^{3} 4\,dt = 4 \cdot 5 = 20.

Taking the upper limit $3$ as the width would give $12$ , which is wrong. The value is positive even though part of the interval lies at negative positions, because the height $4$ is positive and the limits increase.

Find $\int_1^4 (-2)\,dx$ and $\int_4^1 (-2)\,dx$ . What is the geometric area of the region between the line $y = -2$ and the $x$ -axis for $1 \le x \le 4$ ?

Show the answer

The region is a rectangle of width $3$ and height $2$ lying below the axis, so its signed area is

\int_1^4 (-2)\,dx = -2 \cdot 3 = -6.

Reversing the limits flips the sign, so $\int_4^1 (-2)\,dx = 6$ . The geometric area is $6$ in both cases. It is the size of the region, which is never negative, and it equals $\int_1^4 \lvert -2 \rvert\,dx = 6$ .

A tank leaks water at a steady 0.5 liters per minute for 2 hours. Write the change in the volume of water in the tank as an integral and evaluate it with its unit.

Show the answer

The rate is measured per minute, so the time must be measured in minutes too, and 2 hours is 120 minutes. The volume decreases, so the rate is $-0.5$ liters per minute, and

\begin{aligned} \int_0^{120} (-0.5)\,dt &= -0.5 \cdot 120 \\[0.4em] &= -60. \end{aligned}

Liters per minute times minutes gives liters, so the tank loses 60 liters. Multiplying $0.5$ by $2$ would pair a rate per minute with a time in hours, and the result $1$ is not a number of liters. Working in hours instead gives the rate $-30$ liters per hour and the same change, $-30 \cdot 2 = -60$ liters.

What an integral measures, starting with a rectangle

Key terms

Definite integral

Integrand and limits of integration

Signed area

Accumulation

Units of an integral

Check yourself

How this lesson connects

Used later in