Katie,

Think of "the definite integral from 2 to 6 of f(x-4)dx" as being from x=2 to x=6. Let y = x-4. Then dy = dx (because dy/dx = 1) and the integral runs from

y = x-4 = 2-4 = -2
to
y = x-4 = 6-4 = 2

Does this help?

Note that the integral from x=a to x=b of f(x)dx is the same as the integral from y=a to y=b of f(y)dy or the integral from theta=a to theta=b of f(theta)d theta.

I don't understand how to solve definite integrals when the function has something
more than just x inside the parenthesis such as f(4-x).

In general this is called "integration by substitution" - look it up in your text.

-Good hunting!
RD