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**(*on [*x*) ≤ M*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 .

If *f* ( *x* ) < 2*x*, then

We can find by looking at the graph and getting the area of the triangle:

We conclude that

If * *and* **, *then

**Be Careful:** The integral of a product is *not* necessarily the product of the integrals.

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