Physical Meaning of Integrals

The whole of calculus can be compressed into a single sentence: integration is the mathematics of accumulation. In physics very few quantities are constant. Velocity changes with time, force changes with position, current fluctuates. Whenever we want to know how much of something has piled up over a stretch of time or a stretch of space, we need an integral.

The pattern is always the same. If you know a rate of change — velocity, power, current intensity — then integrating that rate with respect to the corresponding variable returns the accumulated total: displacement, work, charge. This is the physical content of the definite integral,

$$\text{accumulated total} = \int_{a}^{b} \text{rate of change}\, dx$$

This page works through the standard interpretations one at a time: accumulating a rate, measuring area under a curve, and computing the volume of a solid. By the end you should be able to read a physical story, say which quantity plays the role of the integrand, say which bounds fix the interval, and write down the integral without hesitating.

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1. Accumulated quantities

1.1 Displacement and velocity

Suppose a body moves along a straight line with velocity $v(t)$. During a very short time $dt$ the body moves by

$$ds = v(t)\, dt$$

so over the time interval $[t_1, t_2]$ the total displacement is

$$s = \int_{t_1}^{t_2} v(t)\, dt$$

Careful: total distance travelled is not the same thing as displacement. Distance ignores direction, so you must integrate the absolute value of the velocity: $\displaystyle \text{distance} = \int_{t_1}^{t_2} \lvert v(t) \rvert\, dt$.

Example 1: displacement for non-uniform motion.

A body moves along a straight line with velocity $v(t) = 3t^2 - 6t$ in metres per second. Find (a) the displacement from $t = 0$ to $t = 3$, and (b) the distance travelled over the same interval.

Solution (a), displacement:

$$s = \int_{0}^{3} (3t^2 - 6t)\, dt = \left[ t^3 - 3t^2 \right]_0^3 = (27 - 27) - (0 - 0) = 0 \text{ m}$$

A displacement of zero means the body ends up exactly where it started.

Solution (b), distance. First find when the velocity changes sign: $3t^2 - 6t = 0$ gives $3t(t - 2) = 0$, so $t = 0$ and $t = 2$.

• On $[0, 2]$ we have $v(t) \leq 0$, so the body moves backwards.
• On $[2, 3]$ we have $v(t) \geq 0$, so the body moves forwards.

$$\text{distance} = \int_{0}^{2} \lvert 3t^2 - 6t \rvert\, dt + \int_{2}^{3} \lvert 3t^2 - 6t \rvert\, dt$$ $$= -\int_{0}^{2} (3t^2 - 6t)\, dt + \int_{2}^{3} (3t^2 - 6t)\, dt$$ $$= -\left[t^3 - 3t^2\right]_0^2 + \left[t^3 - 3t^2\right]_2^3$$ $$= -(8 - 12) + (0 - (-4)) = 4 + 4 = 8 \text{ m}$$

The body covers 8 metres but ends up back at its starting point. Displacement and distance are genuinely different quantities, and the sign of $v(t)$ is what separates them.

1.2 Force and work

When a force $F(x)$ acts on a body along the direction of the $x$-axis, the work done over an infinitesimal displacement $dx$ is $dW = F(x)\, dx$. The total work from position $a$ to position $b$ is therefore

$$W = \int_{a}^{b} F(x)\, dx$$

Example 2: work done against a spring.

By Hooke’s law the restoring force of a spring is $F(x) = -kx$, where $k$ is the spring constant. How much work does it take to stretch the spring a distance $d$ from its natural length?

Solution: the applied force must overcome the restoring force, so the external force is $F_{\text{ext}} = kx$.

$$W = \int_{0}^{d} kx\, dx = k \left[ \frac{x^2}{2} \right]_0^d = \frac{1}{2}kd^2$$

Instance: with $k = 200$ N/m and $d = 0.1$ m, the work is $W = \frac{1}{2} \times 200 \times 0.01 = 1$ J.

Example 3: work done by a variable force.

A body is acted on by a force $F(x) = 6x^2 + 2x$ newtons, directed along the motion. Find the work done as the body moves from $x = 0$ to $x = 2$ metres.

Solution:

$$W = \int_{0}^{2} (6x^2 + 2x)\, dx = \left[ 2x^3 + x^2 \right]_0^2 = (16 + 4) - 0 = 20 \text{ J}$$

1.3 Current and charge

The current $I(t)$ is the rate at which charge flows, that is $I(t) = \frac{dQ}{dt}$. Hence the total charge passing through a cross-section of a conductor during the interval $[t_1, t_2]$ is

$$Q = \int_{t_1}^{t_2} I(t)\, dt$$

Example 4: charge carried by an alternating current.

The current through a conductor is $I(t) = 5\sin(2\pi t)$ amperes. Find the total charge passing through the cross-section from $t = 0$ to $t = 0.5$ seconds.

Solution:

$$Q = \int_{0}^{0.5} 5\sin(2\pi t)\, dt$$

Substitute $u = 2\pi t$, so $du = 2\pi\, dt$:

$$Q = \frac{5}{2\pi} \int_{0}^{\pi} \sin u\, du = \frac{5}{2\pi} \left[-\cos u\right]_0^{\pi}$$ $$= \frac{5}{2\pi} (-\cos\pi + \cos 0) = \frac{5}{2\pi} (1 + 1) = \frac{5}{\pi} \approx 1.59 \text{ C}$$

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2. Area

The most basic geometric meaning of the definite integral: the definite integral $\displaystyle\int_a^b f(x)\,dx$ is the signed area between the curve $y = f(x)$ and the $x$-axis, from $x = a$ to $x = b$.

• When $f(x) \geq 0$ the integral is positive and equals the area under the curve.
• When $f(x) \leq 0$ the integral is negative; the magnitude of the area under the curve is the area below $\lvert f(x)\rvert$.
• The area between two curves $y = f(x)$ and $y = g(x)$ is $\displaystyle\int_a^b \lvert f(x) - g(x) \rvert\, dx$.

Example 5: area enclosed with the axis.

Find the area enclosed between the parabola $y = x^2 - 4x + 3$ and the $x$-axis.

Solution: first find the intersections: $x^2 - 4x + 3 = 0$ gives $(x-1)(x-3) = 0$, so $x = 1$ and $x = 3$.

On $[1, 3]$ the parabola opens upward and lies at or below the axis between its two roots, so $y = x^2 - 4x + 3 \leq 0$ there. The area is

$$A = -\int_{1}^{3} (x^2 - 4x + 3)\, dx = -\left[ \frac{x^3}{3} - 2x^2 + 3x \right]_1^3$$ $$= -\left[ \left(\frac{27}{3} - 18 + 9\right) - \left(\frac{1}{3} - 2 + 3\right) \right]$$ $$= -\left[ (9 - 18 + 9) - \left(\frac{1}{3} + 1\right) \right] = -\left[ 0 - \frac{4}{3} \right] = \frac{4}{3}$$

The curve and the axis therefore enclose an area of $\dfrac{4}{3}$ square units.

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3. Volume

Integrals compute the volume of solids of every shape. The standard method is the cross-section method (also called the slicing method): if the area of the cross-section of a solid at position $x$ is $A(x)$, then the volume between $x = a$ and $x = b$ is

$$V = \int_{a}^{b} A(x)\, dx$$

3.1 Volumes of revolution, disk method

Revolving the curve $y = f(x)$ (with $f(x) \geq 0$) through a full turn about the $x$-axis produces a solid whose volume is

$$V = \pi \int_{a}^{b} \bigl[f(x)\bigr]^2\, dx$$

The reason is that the cross-section at position $x$ is a disk of radius $f(x)$, hence of area $A(x) = \pi \bigl[f(x)\bigr]^2$.

Example 6: volume of a solid of revolution.

Revolve the curve $y = \sqrt{x}$, for $0 \leq x \leq 4$, about the $x$-axis. Find the volume of the resulting solid.

Solution: the cross-section is a disk of radius $r = \sqrt{x}$, so its area is $A(x) = \pi(\sqrt{x})^2 = \pi x$.

$$V = \pi \int_{0}^{4} x\, dx = \pi \left[ \frac{x^2}{2} \right]_0^4 = \pi \cdot \frac{16}{2} = 8\pi$$

The volume is $8\pi$ cubic units, about $25.13$.

3.2 Volumes of revolution, shell method

Revolving the region bounded by the curve $y = x^2$ and the line $y = 0$, for $0 \leq x \leq 2$, about the $y$-axis, is more naturally attacked with cylindrical shells.

Example 7: shell method about the $y$-axis.

Take a vertical strip of width $dx$ at position $x$ and revolve it about the $y$-axis. It sweeps out a cylindrical shell with

• radius $r = x$
• height $h = x^2$
• volume element $dV = 2\pi x \cdot x^2\, dx = 2\pi x^3\, dx$

Therefore

$$V = \int_{0}^{2} 2\pi x^3\, dx = 2\pi \left[ \frac{x^4}{4} \right]_0^2 = 2\pi \cdot \frac{16}{4} = 8\pi$$

Again the volume is $8\pi$ cubic units. That the disk method and the shell method agree is a useful consistency check: the same solid can be described in two ways, and both descriptions must give the same number.

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At a glance

Three ideas tie the table together:

1. Unity. Whether the total is a physical quantity, an area, or a volume, the mechanism is identical: a limit of sums of infinitesimal pieces.
2. From rate to total. Integration inverts differentiation. Given a rate, the integral recovers the net change of the quantity it describes.
3. Modelling skill. The real work in any application is identifying the integrand and the limits. Once those two are right, the integral is routine.

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Exercises

Exercise 1

A particle moves along a straight line with velocity $v(t) = 2t - 6$ metres per second for $0 \leq t \leq 4$. Find (a) the displacement and (b) the total distance travelled over this interval.

Exercise 2

A force $F(x) = 3x^2$ newtons acts on a particle along the direction of the $x$-axis as it moves from $x = 1$ to $x = 4$ metres. Find the work done.

Exercise 3

Revolve the region under the curve $y = 2x$ for $0 \leq x \leq 3$ about the $x$-axis. Find the volume of the resulting solid.

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Summary

Integrals are accumulation made precise. The same formal object — the definite integral of a rate — shows up as displacement, work, charge, area, and volume, and the modelling task in every case is the same: identify the integrand, identify the interval, and integrate.

Symbols Used in This Article