## 7.2. Integration Techniques | IRA |

This section provides integration techniques, i.e. methods for finding
the actual value of an integral. We have already found the most basic
technique (the *Integral Evaluation Shortcut* or *First Fundamental
Theorem of Calculus*): to evaluate an integral over an interval
*[a, b]*, find an antiderivate *F* of the integrand
*f* and compute *F(b) - F(a)*.

Definition 7.2.1: AntiderivativeFor a given function fthe functionFwith the property thatF '(x) = f(x)is called theantiderivative of.f

Not all functions have a simple antiderivative, and any two antiderivatives
of the same function can differ only by a constant (**prove this**). But
different from finding a derivative, where a couple of rules can be used to
find virtually every derivative, finding an *anti*derivative is basically
a guessing game.

As is typical for finding the antiderivatives, the answers for the above examples can not be deduced. They must be guessed, or learned. However, there are two standard mechanisms that can be useful: substitution and integration by parts.

Example 7.2.2: Standard AntiderivativesFind the following antiderivatives:

(a) x^{r}dx(b) 1/x dx(c) e^{x}dx(d) sin(x) dx(e) cos(x) dx(f) tan(x) dx(g) (h) (i)

This theorem, in words, says that if you can identify a composition of functions as well as the derivative of one of the composed functions, there's a good chance you can find the antiderivative and evaluate the corresponding integral.

Theorem 7.2.3: Substitution RuleIf fis a continuous function defined on[a, b], andsa continuously differentiable function from[c, d]into[a, b]. Then

Think about the substitution rule as a "change of variable": if a particular
expression in *x* makes an integrand difficult, change it to
*u*. Then compute the derivative of *u*, i.e.
*du/dx = u'(x)* and "solve" it for *du*:
*du = u'(x) dx*. If you can use *u'(x) dx* to change
the integrand into an expression involving only *u* and no
*x*, then the substitution worked (and has hopefully simplified
the integrand). Otherwise try a different substitution or a
different technique altogether.

Another theorem that is commonly used is "integration by parts".

Example 7.2.4: Applying the Substitution Rule

In order for integration by parts to be useful, three conditions should be satisfied:

Theorem 7.2.5: Integration by PartsSuppose fandgare two continuously differentiable functions. LetG(x) = f(x) g(x). Thenf(x) g'(x) dx = ( G(b) - G(a) ) - f'(x) g(x) dx

- The integrand must be a product of two expressions (one of which could be
1)- You must know the antiderivative of one of the two expressions
- The derivative of the other expression should become easier

Integration by parts can also be used to prove more interesting facts that we might need later.

Example 7.2.6: Applying Integration by PartsEvaluate the following integrals:

Integration by parts can be used to provide an effective means to compute the value of an integral numerically. Of course, Riemann sums can be used to approximate an integral, but convergence is usually slow. A much faster convergence scheme is based on the

Example 7.2.7: Integration by Parts and LimitsSuppose f:[a, b]is a continuously differentiable function. Show that:R

f(x) sin(nx) dx = 0f(x) cos(nx) dx = 0

Now we can state and prove the Trapezoid Rule.

Theorem 7.2.8: Mean Value Theorem for IntegrationIf fandgare continuous functions defined on[a, b]so thatg(x) 0, then there exists a numberc [a, b]withf(x) g(x) dx = f(c) g(x) dx

The Trapezoid Rule is useful because the error

Proposition 7.2.9: Trapezoid RuleLet fbe a twice continuously differentiable function defined on[a, b]and setIfK = sup{ |f''(x) |, x [a, b] }h = (b - a) / n, wherenis a positive integer, thenwhere|R(n)| < K/12 (b-a) h^{2}.

Our final integration technique uses

Example 7.2.10: Application of the Trapezoid RuleCompare the numeric approximations to the integral obtained by using (a) a left Riemann sum and (b) the Trapezoid Rule, using a partition of size 5 and of size 100.sin(x) cos(x) dx

This theorem sounds complicated, so we should restate it in a more useful form:

Theorem 7.2.11: Partial Fraction DecompositionSuppose p(x)is a polynomial of degreensuch thatp(x) =, where eachqis polynomial of degree_{j}jthat is irreducible over. IfRs(x)is another polynomial of degree less thannwith no factors in common withp(x), then the rational functions(x) / p(x)can be written as a finite sumwhere eachris a polynomial of degree less than_{j}j.

Several of the assumptions are technical:

p(x)is a polynomial that can be factored into polynomialsq_{j}- each
qcan not be factored any further (with real coefficients)_{j}s(x)is another polynomial whose factors, if any, do not cancel any of theq, and whose degree is less than the degree of_{j}p- Then
s(x)/p(x)can be written as a finite sum of simple rational functions whose denominators consist of theq's and whose numerators have degree at most_{j}j-1- finding the
ramounts to solving a system of linear equations, as the examples will show_{j}

- If the degree of
*s*is greater than or equal to the degree of*p*, then you can split*s(x) / p(x)*using long division, and the remainder will be of the form where the theorem applies. - If
*s*does have factors in common with*p*, first cancel out those factors, then apply the theorem to the resulting rational function

Example 7.2.12: Integrating Rational FunctionsTo compute the following integrals, use partial fraction decomposition to write the rational function as a sum of simple rational functions.