In Section 14.3, you will learn how to find the area between two curves. Suppose you have two functions f(x) and g(x). Also assume that the higher curve is f(x). We are interested in finding the area from a point x = a to x = b between the two curves. We can do this by finding the area below f(x) and above the x-axis,
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,
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.