Integration by Substitution
There are a couple of tricky substitutions that are not intuitive. Here are some examples:
The normal course of action is to make the expression inside of the radical equal to our new variable. This is the correct way to start but students may halt when they see the result:
Which they will see as being useless to substitute back into the original integral. The trick here is to solve for x before taking the derivative:
Another problem where we can apply the same “trick” is the, at first, innocent looking problem:
Integration by Parts
Typically the average person's experience has income arriving in discreet groupings, for example bi-monthly or monthly paychecks from employers. It is not the same for larger businesses, which owing to their size and the amount of their transactions think of income coming more as a stream. Businesses will often model the income with a function to help in making future projections. Since the income is often deposited into interest earning accounts, the value of a company can't be strictly computed just by how much money they are taking in currently. Economists will look at Present and Future values to determine the value of investments considering the “Time Preference” of money being worth more in-hand today than the same amount in the future. The Present and Future Values functions for businesses with income streams are:
This problem is presented as an application that requires parts to solve. Many times the income stream is expressed as a constant or linear function, which may not require parts, but the multiplication of the exponential is frequently going to. I'll compute the Present Value here:
The integral can be split, and the first term will require parts with
Parts again with:
Distributing the numbers removes the parenthesis and allows us to “wrap around” the integral:
Very number intensive. The key here is record keeping, but the math is the same as simpler parts problems. The future value function works very much the same way.
Integration by Partial Fractions
What if there is an irreducible quadratic term in the denominator after factoring? For example:
This is still a partial fractions problem. If there is an irreducible factor that is a quadratic in the denominator, then the numerator needs to be a linear term. In this case, the separation by partial fractions looks like:
Once the problem is set up correctly, it is solved in the same manner as all other partial fractions problems.
After finding common denominators set the numerators equal
Gather and factor terms with variables with the same power
Then set the coefficients of each variable equal on both sides of the equation
There are a couple of ways to solve a particular integral which will illustrate good practices with trig identities and integration.
It is possible that students can brainstorm the different ways, but it is also a good activity to assign different groups the methods of solution.
Method 1: Substitution
This is the most straightforward method.
Method 2: Integration by Parts
Since these are two functions that are multiplied, it makes sense to use parts:
Method 3: Trig identities
The last example illustrates the importance of always including the constant added term, and remembering that any constant can be rolled into it, since it is not determined.
The best problems in mathematics are often the ones that can be solved using different methods. There is something that captures my imagination about the truth and totality of the major theorems, like those presented by Euclid, that can be proven by straightedge and compass, and then thousands of years later with Galois groups. This is not nearly on the level of such classical problems, but it is valuable and entertaining for students to have the opportunity to verify facts using different methods. Especially those methods that may seem like they were dreamed up for the entertainment of torturing math students.
This does not fit the substitution for sin exactly, but the subtraction indicates that the sine substitution is the one we need.
Using the rules of logs, then substituting back in x using trig identities, we can find the same answer as above:
Coulomb's Law is an equation that gives the electrostatic force between two charged particles. The scalar form of Coulomb's Law is:
I placed all of the quantities in, but it will probably be easier to integrate using constant variables rather than using all of the very large, or very small numbers involved. The only thing to be careful of is to remember what is a constant, and what is the variable. Also, notice that this is an improper integral, so we will need to express it as a limit:
Now we can see that as n approaches infinity, that term goes to zero, so the integral does converge. Substituting in the quantities left out:
We do expect the integral to converge. As the distance between the particles advances to infinity, the force becomes minimal, and with the squared term in the denominator, this is a classic converging integral.
Ordinary Differential Equations
A common application of differential equation is fluid mixing problems. Given information of about the rate of increase or decrease of both the concentration and the fluid being mixed in sets up as a fairly common separable equation. Example:
We need to find the rate at which the chromium leaves the lake. Since the amount that leaves at any single time will depend on the current concentration, the rate that the contaminant leaves will be equal to the rate of water leaving times the concentration of the contaminant. Or put in variables:
Which is a separable differential equation:
Solving then for the time: