$\int$   means sum.


To integrate, in the sense of calculus, means to sum. The integral symbol itself is roughly an S-shape ( ), and it was originally intended to stand for "sum." Integrals are used in a great many continuous summing situations. This section contains a few examples of the use of the definite integral as a sum.

We'll begin with a simple example, the relationship between speed (the derivative of position) and distance traveled (the integral of speed between definite limits).


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.

constant velocity graph


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.

The velocity is the first derivative of position, so if we want position, we must integrate the velocity function, and we'll do it between the limits [2, 5]:

$$ \begin{align} \int_2^5 &(11t - 5)\, dt = \frac{11t^2}{2} - 5t \bigg|_2^5 \\ \\ &= \frac{11\cdot 5^2}{2} - 5\cdot 5 - \left( \frac{11\cdot 2^2}{2} - 5\cdot 2 \right) \\ \\ &= \frac{275}{2} - \frac{50}{2} - \left( \frac{44}{2} - \frac{20}{2} \right) \\ \\ &= \frac{225}{2} - \frac{24}{2} = \frac{201}{2} = 100.5 \;m \end{align}$$


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.

In physics, work (w) is force multiplied by distance, so the total work done in stretching a spring from a length of x1 to x2 is the integral of F = kx with respect to x, from x1 to x2:

$$w = \int_{x_1}^{x_2} kx \, dx$$



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.


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

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 nth 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.

marginal cost graph


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 301st 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.


Solution: We can calculate the cost of making the 301st stove in two ways. The first is just to calculate the cost of making 301 stoves and subtract the cost of making 300:

$$ \begin{align} C(x) &= 5000 + 190 n - 0.15n^2 \\ C(301) - C300) &= 190(301) - 0.15(301)^2 \\ &- [190(300) - 0.15(300)^2)] \\ &= \$ \, 99.85 \end{align}$$

... or we can use the marginal cost function, which is the derivative of the cost function to estimate (a good estimate) this cost:

$$ \begin{align} C'(x) &= 190 - 0.3 n \\ C'(301) &= 190 - 0.3(301) = \$ \, 99.70 \end{align}$$

The small difference in these calculations of the same quantity arise because the derivative as marginal cost is an approximation — but a good one.

To compare the cost of the first 100 and last 100 items, we integrate the marginal cost function between 0 and 100, and between 400 and 500:

$$ \begin{align} Cost_{0-100} &= \int_0^{100} C'(n) \, dn \\ \\ &= \int_0^{100} [190 - 0.3n] \, dn \\ \\ &= 190n - 0.075 n^2 \bigg|_0^{100} \\ \\ &= 19,000 - 750 = \$ \, 18,250 \end{align}$$

and

$$ \begin{align} Cost_{400-500} &= \int_{400}^{500} [190 - 0.3n] \, dn \\ \\ &= 190n - 0.075 n^2 \bigg|_{400}^{500} \\ \\ &= 57,500 - 52,000 = \$ \, 5,500 \end{align}$$

The cost of making the last 100 stoves is significantly lower than making the first. In many types of businesses, one seeks to sell enough items to be able to keep the marginal cost of manufacturing optimally low.


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.

For a heat capacity, C, that does not vary with temperature, the heat q absorbed by mass m of a substance in warming from temperature T1 to T2 is:

$$q = \int_{T_1}^{T_2} mC \, dT = mC(T_2 - T_1)$$

Now if the heat capacity is variable as a function of temperature, our integral just gets a little more complicated, but we can still do it.


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


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

Variable heat capacity


$$ \begin{align} q &= \int_{10}^{90} m(1.4 _ 0.2\sqrt{T}) \, dT \\ \\ &= m(1.4T + 0.3T^{3/2}) \bigg|_{10}{90} \\ \\ &= 10[1.4(90) + 0.13(90^{3/2}) \\ \\ &\; \; - (1.4(10) + 0.13(10^{3/2}))] \\ \\ &= \bf 2189 \; cal \end{align}$$

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 (ire. 1945-1950) and between t=5 and t=10 (1950-1955):

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

The births between 5 and 10 years (1950 and 1955) are found with the same integral, just evaluated at the new limits:

$$ \begin{align} &= 4.5t + 0.95t^2 \bigg|_5^{10} \\ \\ &= 4.5(10) + 0.95(10^2) \\ \\ &= \bf 140 \; births \end{align}$$


Practice problems

1.

Suppose that a cylindrical tank has a radius of 2 m and a height of 5 m and is full to the top with water. Calculate the amount of work that would be required to remove all of the water from the top of the tank. The mass of water is 100 Kg·m-3, and 1 Joule (J) of work is 1 Kg m2 s-2

Solution

2.

Two cars move along the same road with the velocity (mi./h) vs. time (hours) functions shown in this graph. If the cars leave from the same place at the same time, how much farther does the faster car travel between hours 1 and 4?

Solution

Solution begins here

The distance traveled is the integral of the velocity function over time. We simply do the two integrals over the interval [1, 4]:

$$d_1 = \int_1^4 ln|2x - 1| \, dx$$

let u = 2x - 1, then dew = 2dx

$$ \begin{align} \frac{1}{2} \int_1^7 ln(u) \, dew &= \frac{1}{2} \left[ u ln(u) + u \right] \bigg|_1^7 \\ &\uparrow \text{ changed to u limits} \\ &= \frac{1}{2}[7 ln(7) + 7 - ln(1) - 1] \\ &= \frac{1}{2}[7 ln(7) + 6] = 19.62 \; mi. \end{align}$$

The distance traveled by the second car is

$$ \begin{align} d_2 = \int_1^4 ln(x) \, dx &= \left[ x ln(x) + x \right] \bigg|_1^4 \\ &= 4 ln(4) + 3 = 8.545 ;\ mi. \end{align}$$

The difference (the light-shaded area between the curves) is 11.08 mi.

3.

The rate at which rainwater flows into a drain pipe is modeled by $R(t) = 20 sin\left( \frac{t^2}{35} \right)$, in feet3/hour. How many cubic feet of rainwater flow into the pipe during the 8-hour time interval t = 0 to t = 8?

This problem is adapted from the free-response section of the 2015 AP Calculus exam. You can access past free-response questions at AP Central .

Solution

The amount that flows is the integral of the rate function, R(t) between the limits t = 0 and t = 8:

$$V = \int_0^8 20 sin \left( \frac{t^2}{35} \right) \, dt$$

This integral is difficult, if not impossible to do analytically (with paper and pencil), but we can do it numerically on a calculator. Here's a screen shot of the result on a TI-84 calculator.

so the volume is 76.6 ft.3


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