$\int$   means sum.

To integrate, in the sense of calculus, means to sum. The integral symbol itself is roughly an S-shape ( ∫ ). It was originally intended to stand for "sum" or "summation." Integrals are used in a great many continuous summing situations, in which we add up an infinite number of infinitessimally small things, whatever they may be.

Distance traveled from a velocity function

As a first example of the integral as a sum, here's a plot of a trip taken at constant velocity over some arbitrary time.

Distance traveled is velocity x time, so if we sum up the areas of all of the rectangles shown here, we'll get the distance traveled.

The whole blue area, of course, is just one big rectangle, so we could just find the distance directly, but carving it up into smaller rectangles will help to visualize the next step.

Variable velocity

It's no great leap to extend the method to a velocity function that's curved. In this graph, a trip is made for a period of time, during which the velocity increases at increasing rates for a while, then slows. Recall that any change in velocity is known as acceleration.

Roll over the graph to compare the left and right Riemann sums. We know that such a sum converges to the integral as Δx → 0, giving us the exact area under the curve, and thus the exact distance traveled.

Example 1

Let's say that a particle is moving with a velocity of v(t) = 11t - 5 m/s in a straight line. How much distance does the particle cover between 2 s and 5 s?

Solution: Notice that the velocity depends linearly on time. Except for a small correction of -5 m/s to every velocity calculated from this function, the velocity is 11 times the time, so the particle is accelerating.

Example 2 – work

In physics, we do work when we exert a force over a distance, like the force applied over a distance x in stretching the spring below. Hooke's law, $F = kx,$ says that the force needed to compress or stretch the spring is proportional to its length, x:

• The more we compress the spring the more force we feel
• The more we stretch it, the more force we feel.

Of course, we could always overdo this and stretch a real spring to the point where it deforms, but we'll assume we won't do that.

Example problem: For a particular spring, the spring constant k is k = 250 N/m (Newtons per meter). Calculate the work done in stretching the spring from 0.1 m to 0.2 m.

$$ \begin{align} w &= \int_{x_1}^{x_2} kx \, dx = \int_{0.1}^{0.2} 250 x \, dx \\[7pt] &= \frac{250 x^2}{2} \bigg|_{0.1}^{0.2} = 125 (0.2^2 - 0.1^2) \\[7pt] &= 3.75 \; N\cdot m \end{align}$$

There are many other examples of forces in physics, and the same process can be used to calculate net work with all of them.

Example 3 — Marginal cost

In economics, marginal cost is the cost of producing one more unit of goods or services.

Think about opening your own bicycle factory. The first bike you produce will be very expensive, because to produce it you'll have to buy machinery, draw up plans, make prototypes, hire workers, &c. These costs will be shared between two bikes when you make a second, three for the third, and so on, so that the share of the cost of making the n^th bike will grow smaller, as the number increases.

If n is large enough the marginal cost function can be approximated by the derivative of cost with respect to number, n:

$$\text{marginal cost} = \frac{dC}{dn}$$

The cost of producing items n₁ to n₂ is just the integral of the marginal cost function between those limits.

Example problem

Suppose that the total cost incurred by an appliance manufacturer per week for manufacturing n stoves is C(x) = 5000 + 190n - 0.15n² for 0 ≤ n ≤ 500.

1. Find the actual cost of manufacturing the 301^st stove.


2. Find the rate of change of the total cost function when n = 250 units.


3. If 500 stoves are to be manufactured in a week, find the cost of manufacturing the first 100 stoves and compare it to the cost of manufacturing the last 100.

Example 4 — Heat Capacity

The heat capacity of a substance is a measure of its ability to absorb heat energy without an associated rise in temperature. Substances like water, in which the molecules have many varieties of internal motion, have high heat capacities. Substances like diamond, where the atoms are locked more-or-less rigidly in place, have low heat capacities. It takes much more heat input to raise the temperature of a gram of water by 1˚C than it does a 1 g diamond. The heat capacity is a constant for most substances over a small temperature range, but does change somewhat with temperature — a moving target.

Example problem

Let the heat capacity of a substance be $C = 1.4 + 0.2\sqrt{T}$. Calculate the heat absorbed (in calories) in heating 10 g of the substance from 10˚C to 90˚C using both C as given, and a constant heat capacity of 1.4 cal/g·˚C

Solution: We'll solve the two problems side-by side:

Constant heat capacity

$$ \begin{align} q &= \int_{10}^{90} mC \, dT \\[5pt] &= mC(90 - 10) \\[5pt] &= (10 \, g)\left( 1.4 \frac{cal}{g˚C} \right)(80˚C) \\[5pt] &= \bf 1120 \; cal \end{align}$$

The variable heat capacity produces a much higher heat input because the ability of the substance to absorb heat without an accompanying raise in temperature rises as the sample gets hotter.

Example 5 — Birth Rate

Suppose that the birth rate for the ten years after the end of WWII, the so-called "baby boom" was b(t) = 4.5 + 1.9t in millions of births per year (0 ≤ t ≤ 10) where t is time in years since the beginning of 1945. Find the number of births between 1945 and 1950, and between 1950 and 1955 (from Jan. 1 of the beginning year through Dec. 31 of the ending year for both).

Solution : We simply integrate the rates between t=0 and t=5 (1945-1950) and between t=5 and t=10 (1950-1955):

$$ \begin{align} B &= \int_0^5 b(t) \, dt \\[5pt] &= \int_0^5 (4.5 + 1.9t) \, dt \\[5pt] &= 4.5t + 0.95t^2 \bigg|_0^5 \\[5pt] &= 4.5(5) + 0.95(5^2) \\[5pt] &= \bf 46.25 \; births \end{align}$$

Practice problems