Suppose that the profit for a company is increasing at a rate of
where the company has been in operation for t years. What is the total change in profit over the first three years?
In this problem, we are given the rate at which profit is changing over time. This is confirmed by the fact that the function is defined as P′(t), the derivative of profit. However, the question is about the corresponding profit function P(t). So we need to find this profit function by taking the antiderivative of P′(t),
Remember, the antiderivative undoes the derivative so the antiderivative of P′(t) is P(t). To do this antiderivative, we need to use the Substitution Method.
This means that
You might think that the total change in profit over the first three years is P(3), but this is the profit at the end of the third year. To find the total change in profit we need to calculate P(3) – P(0),
Sometimes a seemingly easy problem can get fairly complicated with the addition of a few extra requirements. For instance, suppose we want to find the area between the functions y = x2 – 4 and y = 3x. If we graph the two functions, we see that they appear to cross at x = -1 and x = 4.
We can verify these two points by setting the functions equal to each other and solving for x.
The area between these curves lies above the parabola and below the line.
To find the area of the shaded region, take the definite integral from x = -1 to x = 4 of the higher function minus the lower function,
We can evaluate this integrand using the Fundamental Theorem of Calculus,
Let’s now modify this problem by finding the area enclosed by the functions y = x2 – 4 and y = 3x as well as the lines x = -5 and x = 1. Graph each of these equations.
The region enclosed by these graphs is more complicated since the functions cross at x = -1.
To find the area of the enclosed region, we need to break it into two parts. The first part runs from x = -5 to the point of intersection at x = -1. The area of this part is
The second part extends from x = -1 to x = 1 and has area
The area enclosed by the functions y = x2 – 4 and y = 3x as well as the lines x = -5 and x = 1 is the sum of these parts or 206/3. Adding the vertical lines on either side of the point of intersection requires the use of two definite integrals since the parabola is higher on the left side of x = -1 and the line is higher on the right side of x = -1.
Thus the area between the curves is 184/3 + 22/3 or 206/3.
In Section 14.3, I carry out several examples where the producers’ or consumers’ surplus is calculated. I want to give you a few more examples including some of the examples worked out by students in class.
Let’s take a look at producers’ surplus. To get a good idea of this concept, let’s visualize what area on a supply or demand graph represents. In the graph below, we have a supply function $latex displaystyle S(Q)=0.9Q$. The supply and demand are in equilibrium when 100 units are produced at 90 dollars per unit.
On this graph heights are in dollars per unit and widths are in units. This means the units on any area will be
and that the equilibrium quantity is Q = 16. Find the producers’ surplus.
First find the equilibrium price (black). Then find the area under the supply curve (red) and the area under the equilibrium price (green). The difference between these amounts (blue) is the producers’ surplus.
Notice that the producers’ surplus is the area between the equilibrium price and the the supply curve. We can compute the surplus by computing this area.
This gives rise to the formula often quoted for the producers’ surplus,
Only one problem on the Section 14.1 Homework was missed by very many people. In that problem you were given the rate of change of profit, P‘(t), and asked to calculate how much the profit changed. Since this is a question about P(t), you need to undo the derivative with an antiderivative in the form of the Fundamental Theorem of Calculus. With this function, we would write it as