First you have to know HowToDifferentiate. Having learned how to differentiate, your goal in integrating some function f(x) is to find another function F(x) such that F'(x) = f(x). You can then write that the integral of f(x) is F(x)+C (any constant C works), and compute definite integrals with the rule Integrala to b f(x) dx = F(b) - F(a).
How do you find this magic F(x)? Some possibilities:
Memorize some standard integral formulas. Some useful ones for AlgorithmAnalysis are:
a f(x) [a constant]
f(ax) [a constant]
xn [n constant not equal to -1]
ax [a constant]
ax/ln a [follows from ax = ex ln a]
x ln x - x
- Guess but verify. Guess F(x) and compute F'(x) to see if it's f(x). May be time-consuming unless you are good at guessing, and can put enough constants in F(x) to let you adjust F'(x) to equal f(x). Example: if f(x) = 2/x, you may remember the 1/x formula and try F(x) = a ln bx. Then F'(x) = ab/(bx) = a/x and you can set a = 2, quietly forget you ever put in b, and astound your friends (who also forgot the a f(x) rule) by announcing that the integral is 2 ln x. Sometimes if the answer comes out wrong you can see how to fudge F(x) to make it work: if for f(x) = ln x you guess F(x) = x ln x, then F'(x) = ln x + 1 and you can notice that you need to add a -x term (the integral of -1) to get rid of the 1.
- There's a complicated technique called "integration by parts" which is the integral version of the duv = u dv + v du formula, but it doesn't work as often as one might like. The rule is that Integral u dv = uv - Integral v du; an example is Integral ln x dx = x ln x - Integral x d(ln x) = x ln x - Integral x (1/x) dx = x ln x - Integral 1 dx = x ln x - x. You probably shouldn't bother memorizing this unless you need to pass AP Calculus again, although you can rederive it from the duv = u dv + v du formula.
Use a computer algebra system like Mathematica, Maple, or Maxima. Mathematica's integration routine is available on-line at http://integrals.wolfram.com.
- Look your function up in a big book of integrals. This is actually less effective that using Mathematica, but may continue to work during power failures.