Calculus review: integrals

An integral is very similar to a sum.  If a curve is described by a function, the integral of that function between two x values gives you the area under it on a graph.  There are many approximation schemes for integrals that just compute them as sums, such as Midpoint Rule, Trapezoid Rule, or Simpson's Rule.  Integration is also the inverse operation of differentiation: taking the derivative and then the integral of a function gets you back the original function.

Here, we are going to assume some familiarity with integration and remind the rusty among you of how to do a bit more . . .

The most basic functions can be integrated just by remembering the derivative formulae and inverting.  For instance,
since $$\frac{d x^n}{dx}=nx^{n-1},$$ if we want to integrate $$x^m,$$ we can let $$m=n-1$$ and see that   $$\frac{1}{n}\frac{d x^n}{dx}=x^{n-1}$$ and replace n-1 with m to see that
$$\int x^m=\frac{1}{m+1}x^{m+1}.$$  We can then check by differentiating: we get $$x^m,$$ so all's well that ends well!

$$e^x,\;\sin x,\;\rm{and}\;\cos x$$ can all be handled using this principle too.

For more complicated functions, there are five major techniques, which we'll look at in increasing order of complexity.
1) Integration by substitution
2) Integration by parts
3) Trig substitution
4) Partial fractions
5) Gaussian integrals
6) Completing the square in Gaussian integrals

1) Integration by substitution
Given a composition of functions, it is often convenient to define a new variable for the inner function.  For instance, let's compute
$$\int_a^b \big[ 1+x\big]^2dx$$
by defining $$u=1+x,$$ $$du=dx.$$
We then have
$$\int_{1+a}^{1+b} u^2 du=\frac{u^3}{3}\bigg|_{1+a}^{1+b}$$
Notice a few things:
a) we have replaced the limits of integration: if x runs from a to b, u=1+x runs from 1+a to 1+b.
b) we have replaced dx with dx=du: in general, we need to differentiate u and solve algrebraically for dx in terms of du, and then replace dx with its value in terms of du.

Let's do a harder example:
$$\int_a^b \big[1+2x \big]^{1/2}dx.$$
Let $$u=1+2x,\;du=2dx$$ so that
$$dx=\frac{du}{2}.$$
We then find
$$\int_{1+2a}^{1+2b} \frac{u^{1/2}}{2}du=\frac{u^{3/2}}{2}\bigg|_{1+2a}^{1+2b}.$$

To do these without making a mistake, it's good to go step by step.  First, write out clearly what u equals and then du = . . . dx, and solve carefully for dx.  Then make these replacements in the integral.  After that, make sure to change the bounds of integration to have the right range for the new variable!

2) Integration by parts
Often an integrand is of the form of a product of two functions, one of which is has a simple integral.  How do we handle these cases?  Write the integral as
$$\int u dv$$ where dv is the function that has a simple integral.  We then have the formula
$$\int_a^b u dv=uv\bigg|_a^b-\int_a^b v du.$$  Try differentiating this formula to prove it is the correct integral of u dv:
$$\frac{d}{dx}\big[uv\big|_a^b-\int_a^b v du\big]=u \frac{dv}{dx}+v\frac{du}{dx}-v\frac{du}{dx}.$$

Let's try an example.  Conveniently, this example will turn out to give us a useful formula for the integral of natural log.

$$\int \ln x dx=\int 1 \cdot \ln x dx.$$

We don't know how to integrate $$\ln x,$$ so we better not have it be our dv.  But we do know how to integrate $$1dx:$$ that's just x!  So we set
$$u=\ln x,\;dv=1dx.$$

We then have
$$\int \ln x dx=x \ln x - \int \frac{x}{x}dx=x\ln x -x.$$

This is great because we now have a general technique for integrating any power of x against natural log.  Let's try x:

$$\int x \ln x dx=\int \ln x dx=\frac{x^2}{2} \ln x - \int \frac{x^2}{2x}dx=\frac{x^2}{2}\ln x -\frac{x^2}{4}.$$

Try deriving the formula for $$\int x^n \ln x dx$$ yourself!

Another integral commonly evaluated using integration by parts is $$\int x^n e^x dx,$$ where n is a positive whole number.  Here things get a little trickier.  We want to eventually lower the integrand to involve only $$e^x$$ and no powers of x, so we need to do something that lowers the power of x each time we do it.  This suggests letting $$u=x^n,$$ because $$du=nx^{n-1}.$$  We then have

$$\int x^n e^x=x^n e^x-\int nx^{n-1}e^xdx.$$  We can then integrate the integral by parts again to reduce the power of x by one more, and keep doing so until we get the integrand to be $$e^x,$$ which we know how to integrate.