1. Integrals And Integration
The fundamental theorem of calculus, as presented in Section
9.4 Theorem 2.1, provides us a powerful method to
calculate definite integrals, by the formula:
is any antiderivative of f.
This method places the burden of calculation squarely on finding an
antiderivative F of
If an antiderivative of a function is known or found then definite
integrals of that function can readily be computed. So the
task is often simply to find antiderivatives, not definite integrals. An antidifferentiable function f has infinitely many
antiderivatives (any 2 of them differ from each other by a constant). Now the general antiderivative of f represents all the
antiderivatives of f. Thus the task is actually simply to find general antiderivatives, when no definite integral is needed at
9.4 Definition 3.1 we saw that the general antiderivative is the indefinite
integral. Consequently in terms of
integrals the task is simply to find indefinite integrals, when no definite integral is needed at the moment. If the indefinite
integral of a function is known or found, definite integrals of that function can readily be calculated. Many known indefinite
integrals are recorded in integral tables for all to use.
The word “integral” refers to the definite integral or the
indefinite integral. Which one it refers to will be clear from the
context. Integration as defined in Section 9.4 Definition 4.1 is the process of finding a definite integral or an indefinite
integral. Which process it is will be clear from the context.
Note that we don't need to insert a constant into the antiderivative (x3/3) + ex; see Section 9.4 Evaluation.
a. Find this derivative:
If the derivative of x2ex is
(2x + x2)ex, then an
antiderivative of (2x
+ x2)ex is x2ex, so the general
indefinite integral of (2x + x2)ex is x2ex + C, where C is an arbitrary constant. Remark that it's not necessary to include
the constant C of the indefinite integral in the computation of the definite integral; see Section 9.4 Evaluation.
2. Integration By Inspection
We know this differentiation formula: (d/dx)
x2 = 2x, that is, the derivative of x2 is 2x. By inspection we see that 2x is
the derivative of x2, so that an antiderivative of 2x is x2. Thus the general antiderivative or indefinite integral of 2x is x2 +
C, where C is an arbitrary constant. This is an example of integration by inspection.
We use a trigonometric identity to render the integrand into
a form whose antiderivative is obvious. This is another
example of integration by inspection.
Integration by inspection refers to the situation
where we by inspecting the integrand see right away what its
antiderivative is, as in Example 2.1, or see that it can be rendered into a form whose antiderivative is obvious, as in
Example 2.2. Integration by inspection clearly requires that we know differentiation formulas and rules. For example we
know these differentiation formulas: (d/dx) x2 = 2x, (d/dx) (1/2)x = 1/2, and (d/dx) sin x = cos x and the chain rule
(for (d/dx) sin 2x = (cos 2x)(2) = 2 cos 2x). Part 4 below gives a table of basic integrals corresponding to some
differentiation formulas and rules.
3. Integrals Requiring Integration Techniques
Now consider this integral:
This is hard to be evaluated by inspection. There's a
technique to find it. There are integrals that require techniques. The
next several sections present various techniques to find integrals that are hard to evaluate or can't be evaluated by
inspection. In this section we handle only integrals that can be evaluated by inspection.
The following integrals are some of the basic ones that derive from differentiation formulas and rules and that should be memorized. Each one can be verified by simply differentiating the right-hand side to obtain the integrand on the left-hand side. Here are 2 examples:
For formula 1:
Some Basic Integrals
5. Integrals Of Some Operations On Functions
Here are integrals of some operations on functions, namely of sum, difference, and constant multiple of functions:
where a and b are constants.
These formulas can easily be proved by deriving them from differentiation formulas. For example for formula 1:
Problems & Solutions
1. Calculate the following indefinite integrals.
2. Compute this indefinite integral:
3. Evaluate this indefinite integral:
We insert the constant C
only after the last integral has been evaluated. There's no point to insert it
before then, because
if we did, then we would have to add up those constants to get just a single constant.