Calculus/Integration techniques

Approximation Techniques

Simpson's Rule

For a given function, f(x), the area below the curve can be approximated using a cubic equation (many books quote quadratic). An even number of strips must be chosen for the formula to work correctly. The formula is:

$A=\frac{h}{3}[y_0+y_n+4(y_1+y_3+...+y_{n-1})+2(y_2+y_4+...+y_{n-2})]$ where h is the width of the strip

Other Numerical Approximation Techniques

Because from first principles, $f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$

Rearranging, $f(x+h)\approx f(x)+h{f}^{\prime }(x)$

This approximation is more accurate as h approaches 0.

Recognizing Derivatives and Reversing Derivative Rules

If we recognize a function $g(x)$ as being the derivative of a function $f(x)$, then we can easily express the antiderivative of $g(x)$:

$\int g(x)dx=f(x)+C.$

For example, since

$\frac{d}{dx}\sin x=\cos x$

we can conclude that

$\int \cos xdx=\sin x+C.$

Similarly, since we know $e^x$ is its own derivative,

$\int e^{x}dx=e^x+C.$

The power rule for derivatives can be reversed to give us a way to handle integrals of powers of $x$. Since

$\frac{d}{dx}{x}^n=n{x}^{n-1}$,

we can conclude that

$\int n{x}^{n-1}dx=x^n+C,$

or, a little more usefully,

$\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$.

Integration by Substitution

For many integrals, a substitution can be used to transform the integrand and make possible the finding of an antiderivative. There are a variety of such substitutions, each depending on the form of the integrand.

Integrating with the derivative present

If a component of the integrand can be viewed as the derivative of another component of the integrand, a substitution can be made to simplify the integrand.

For example, in the integral

$\int 3x^2(x^3+1{)}^{5}dx$

we see that $3x^2$ is the derivative of $x^3+1$. Letting

$u=x^3+1$

we have

$\frac{du}{dx}=3x^2$

or, in order to apply it to the integral,

$du=3x^{2}dx$.

With this we may write $\int 3x^2(x^3+1{)}^{5}dx=\int u^{5}du=\frac{1}{6}{u}^6+C=\frac{1}{6}(x^3+1{)}^6+C.$

Note that it was not necessary that we had exactly the derivative of $u$ in our integrand. It would have been sufficient to have any constant multiple of the derivative.

For instance, to treat the integral

$\int x^4\sin (x^5)dx$

we may let $u=x^5$. Then

$du=5x^{4}dx$

and so

$\frac{1}{5}du=x^{4}dx$

the right-hand side of which is a factor of our integrand. Thus,

$\int x^4\sin (x^5)dx=\int \frac{1}{5}\sin udu=-\frac{1}{5}\cos u+C=-\frac{1}{5}\cos x^5+C.$

In general, the integral of a power of a function times that function's derivative may be integrated in this way. Since $\frac{d[g(x)]}{dx}=g^{\prime }(x)$,

we have $dx=\frac{d[g(x)]}{g^{\prime }(x)}.$

Therefore, $\int g^{\prime }(x)[g(x){]}^{n}dx$	$=\int g^{\prime }(x)[g(x){]}^n\frac{d[g(x)]}{g^{\prime }(x)}$
	$=\int [g(x){]}^{n}d[g(x)]$
	$=\frac{[g(x){]}^{n+1}}{n+1}$

Integration by Parts

If y = u v where u and v are functions of x,

Then $y^{\prime }=(uv{)}^{\prime }=v^{\prime }u+u^{\prime }v$

Rearranging, $u{v}^{\prime }=(uv{)}^{\prime }-v{u}^{\prime }$

Therefore, $\int u{v}^{\prime }dx=\int (uv{)}^{\prime }dx-\int v{u}^{\prime }dx$

Therefore, $\int u{v}^{\prime }dx=uv-\int v{u}^{\prime }dx$, or

$\int udv=uv-\int vdu$.

This is the integration by parts formula. It is very useful in many integrals involving products of functions, as well as others.

For instance, to treat

$\int x\sin xdx$

we choose $u=x$ and $dv=\sin xdx$. With these choices, we have $du=dx$ and $v=-\cos x$, and we have

$\int x\sin xdx=-x\cos x-\int (-\cos x)dx=-x\cos x+\int \cos xdx=-x\cos x+\sin x+C.$

Note that the choice of $u$ and $dv$ was critical. Had we chosen the reverse, so that $u=\sin x$ and $dv=xdx$, the result would have been

$\frac{1}{2}{x}^2\sin x-\int \frac{1}{2}{x}^2\cos xdx.$

The resulting integral is no easier to work with than the original; we might say that this application of integration by parts took us in the wrong direction.

So the choice is important. One general guideline to help us make that choice is, if possible, to choose $u$ to be the factor of the integrand which becomes simpler when we differentiate it. In the last example, we see that $\sin x$ does not become simpler when we differentiate it: $\cos x$ is no simpler than $\sin x$.

An important feature of the integration by parts method is that we often need to apply it more than once. For instance, to integrate

$\int x^2{e}^{x}dx$,

we start by choosing $u=x^2$ and $dv=e^x$ to get

$\int x^2{e}^{x}dx=x^2{e}^x-2\int x{e}^{x}dx.$

Note that we still have an integral to take care of, and we do this by applying integration by parts again, with $u=x$ and $dv=e^{x}dx$, which gives us

$\int x^2{e}^{x}dx=x^2{e}^x-2\int x{e}^{x}dx=x^2{e}^x-2(x{e}^x-e^x)+C=x^2{e}^x-2x{e}^x-e^x+C.$

So, two applications of integration by parts were necessary, owing to the power of $x^2$ in the integrand.

Note that any power of x does become simpler when we differentiate it, so when we see an integral of the form

$\int x^{n}f(x)dx$

one of our first thoughts ought to be to consider using integration by parts with $u=x^n$. Of course, in order for it to work, we need to be able to write down an antiderivative for $dv$.

Integration by Partial Fractions

It is possible to convert a rational function into a sum of simpler rational expressions.

In all cases, we begin with an integrand which is a rational function in which the degree of the numerator is less than the degree of the denominator. To apply the method to integrands for which the numerator has degree that is not less than the degree of the denominator, one must manipulate the integrand into that form. For instance, to evaluate

$\int \frac{x^3+x+2}{x^2+3}dx$

we begin by writing

$\frac{x^3+x+2}{x^2+3}=x-\frac{2x}{x^2+3}$

and then apply the partial fractions method to the rational function.

We then factor the denominator completely. The fundamental theorem of algebra tells us that we can always factor a polynomial into linear and irreducible quadratic factors. This gives us four cases to consider.

When the denominator factors as a product of distinct linear factors

When the denominator is a product of distinct linear factors, we can write the integrand as a sum of rational functions of the form

$\frac{A}{ax+b}$

with one term for each of the factors of the denominator.

For example, to evaluate the integral

$\int \frac{x}{x^2+7x+10}dx$

we first factor the denominator

$\frac{x}{(x^2+7x+10)}=\frac{x}{(x+2)(x+5)}$.

Then we write

$\frac{x}{(x+2)(x+5)}=\frac{A}{x+2}+\frac{B}{x+5}$

and solve for A and B. This equation can be transformed to

$x=A(x+5)+B(x+2).$

Letting $x=-5$, we conclude that $B=5/3$. Letting $x=-2$, we conclude that $A=-2/3$.

Thus,

$\int \frac{x}{x^2+7x+10}dx=\int \frac{x}{(x+2)(x+5)}dx=\int (-\frac{2/3}{x+2}+\frac{5/3}{x+5})dx=-\frac{2}{3}\ln |x+2|+\frac{5}{3}\ln |x+5|+C.$

When the denominator factors into linear factors, at least one repeated

$\frac{f(x)}{(ax+b{)}^n(cx+d{)}^m}=\frac{A}{(ax+b{)}^n}+\frac{B}{(ax+b{)}^{n-1}}+...+\frac{D}{ax+b}+\frac{E}{(cx+d{)}^m}+\frac{F}{(cx+d{)}^{m-1}}+...+\frac{H}{cx+d}$

When the denominator's factors include distinct, irreducible quadratic factors

$\frac{f(x)}{(a{x}^2+bx+c)(dx+e)}=\frac{Ax+B}{a{x}^2+bx+c}+\frac{C}{dx+e}$

When the denominator's factors include repeated irreducible quadratic factors

Therefore, $\frac{f(x)}{(a{x}^2+bx+c{)}^n(dx+e{)}^m}$	$=\frac{Ax+B}{(a{x}^2+bx+c{)}^n}+\frac{Cx+D}{(a{x}^2+bx+c{)}^{n-1}}+...+\frac{Gx+H}{a{x}^2+bx+c}$
	$+\frac{I}{(dx+e{)}^m}+\frac{J}{(dx+e{)}^{m-1}}+...+\frac{L}{dx+e}$