Now another function is going to join the party. Assume both f and g are integrable functions.
- The integral of a sum is the sum of the integrals. In symbols,
Also, the integral of a difference is the difference of the integrals:
We've taken whatever weighted area was between g and the x-axis, and stuck that on top of f. If g is negative, in some places we may actually be subtracting area from f. The integral of (f + g) is the integral of f plus the integral of g.
- If f (x) is smaller than g(x), the integral of f is smaller than the integral of g. In symbols, if
f (x) < g(x) for all x in [a, b]
then
- If both f and g are positive, it's clear that there's more area between g and the x-axis than between f and the x-axis:
- If both f and g are negative, the weighted area between g and the x-axis is closer to 0 than the weighted area between f and the x-axis, so we still have
- If f is negative and g is positive, then
is negative and 
is positive. - If f and g are sometimes negative and sometimes positive, we can split the interval [a, b] into sub-intervals on which one of the previous conditions holds, and go from there.
- Let m and M be constants. If m ≤ f(x) ≤ M on [a, b], then
This is using the last property. We're comparing f to the constant functions m and M. If m ≤ f (x)on [a, b], then we know from the previous property that
Since 
, we have half the inequality explained.
Similarly, if f (x) ≤ M on [a,b] then we know from the previous property that
and we know that .
Sample Problem
If f (x) < 2x, then
We can find 
by looking at the graph and getting the area of the triangle:
We conclude that
Sample Problem
If 
and 
, then
Be Careful: The integral of a product is not necessarily the product of the integrals.