The in-person class has spent some time at the board doing substitution problems. I took pictures so that you would have more examples handy. Several of these examples are similar to ones you all have had trouble with on the homework.

There is a lot of good work here…any mistakes? Which solution is easiest to read…and perhaps the best?

Example 1 $latex \displaystyle \int{\left( x+1 \right){{\left( {{x}^{2}}+2x \right)}^{2}}dx}$

Example 2 $latex \displaystyle \int{4x{{e}^{{{x}^{2}}+9}}dx}$

Example 3 $latex \displaystyle \int{6x{{e}^{2{{x}^{2}}+1}}dx}$

Example 4 $latex \displaystyle \int{5x{{e}^{10{{x}^{2}}+4}}dx}$

Example 5 $latex \displaystyle \int{\frac{\sqrt{2+\ln \left( x \right)}}{x}dx}$

Example 6 $latex \displaystyle \int{\left( {{x}^{3}}+2x \right){{\left( {{x}^{4}}+4{{x}^{2}}+7 \right)}^{8}}dx}$

Example 7 $latex \displaystyle \int{\left( 6-6z \right){{e}^{2z-{{z}^{2}}}}dz}$

Example 8 $latex \displaystyle \int{\left( x+1 \right){{\left( {{x}^{2}}+2x \right)}^{3}}dx}$

Example 9 $latex \displaystyle \int{\left( 3{{x}^{2}}+4 \right){{\left( 2{{x}^{3}}+8x \right)}^{19}}dx}$

Choose the expression for u. This is generally the inside part of a composition in the integrand. Use the derivative to find an other expression for du.

Match the integrand with u and du. All variables in the original integrand must change to u.

Change the integrand so that it is written in terms of u.

Work out the antiderivative in terms of u.

Put in the expression for u so that the antiderivative is written in terms of the original variable.

Now let’s look at the examples carried out in class.

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

This is the amount of money a supplier would be willing to receive if each of the units yielded revenue according to the prices on the supply curve from 0 to 100 units.

However, if the market is in equilibrium all 100 units with earn 90 dollars per unit yielding

Since the market is in equilibrium, the supplier actually receives a higher price per units giving an additional 4500 dollars in revenue. This extra amount is called the producers’ surplus.

The consumer would be willing to pay more than the equilibrium price. The amount they save by paying the equilibrium price is called the consumers’ surplus.

Problem 1 Suppose the supply curve for a particular product is

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,