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).
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.
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.
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]:
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:
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 x_{1} to x_{2} is the integral of F = kx with respect to x, from x_{1} to x_{2}:
Solution: This is just a simple integral:
There are many other examples of forces in physics, and the same process can be used to calculate net work with all of them.
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.
It might look like the graph on the right →
If n is large enough the marginal cost function can be approximated by the derivative of cost with respect to number, n:
The cost of producing items n_{1} to n_{2} is just the integral of the marginal cost function between those limits.
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:
... or we can use the marginal cost function, which is the derivative of the cost function to estimate (a good estimate) this cost:
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:
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.
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 T_{1} to T_{2} is:
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.
Solution: We'll solve the two problems side by side:
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.
Solution: We simply integrate the rates between t=0 and t=5 (i.e. 1945-1950) and between t=5 and t=10 (1950-1955):
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