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.

Practice: