
The techniques of integration are basically those of differentiation looked at backwards.
The rule for differentiating a sum: that you get the sum of the derivatives of the summands, gives rise to the same fact for integrals: the integral of a sum as integrand is the sum of the integrals of the summands.
1. The Product Rule Backwards
The product rule, says that the derivative of a product is the sum gotten by differentiating each factor as if the other were constant and adding up the results.
We can read this backwards as a way to handle an integrand of the form fg', when we know how to handle the integrand f 'g. For, we can write the product rule as
fg' = (fg)'  f 'g
and integrating both sides tells us
This statement is called "integrating by parts" and is useful for integrands like x^{k}exp(x) or ln(x) or xln(x).
For example, to integrate ln(x), set f(x) = ln(x) and g'(x)= 1. Then and g(x) = x.
We can conclude that the integral of ln(x) from a to b is bln(b)  aln(a)  (b  a).
Exercise 18.7 Do the other integrals mentioned just above: with integrands x^{k}exp(x) for k = 1 and k = 2, and also xln(x).
2. The Chain Rule Backwards
The chain rule tells us how to differentiate f(g(x)) and the answer is .
This tells us that we if we can recognize an integrand as having the form , we can integrate it to get f(g(x)) evaluated at b less its evaluation at a.
What can we recognize this way?
Here are examples you should mull over: . Try guessing what to choose for g(x) and see if you can get it to work. If you fail, try again.
Using the chain rule backwards is sometimes called the method of substitution.
We will not dwell on this topic. For further details see the corresponding sections of course 18.013A.
We do note that, by an appropriate magical substitution, you can turn any rational function of sines and cosines into a rational function, which you can actually integrate, with enough effort.
Is there anything we cannot integrate?
Yes definitely. The integrands are examples, for which there is no solution that can be expressed in terms of standard spreadsheet available functions.
Nowadays, you can consult any of a number of available programs, such as Maple, Mathematica, and Matlab and they will give you formal solutions to any doable integrals, and solutions to arbitrary accuracy for those that cannot be integrated exactly in terms of the functions we have defined.
We now turn to the question: how feasible is it to integrate, that is, to determine areas under curves, numerically?
