Problem

Evaluate

Solution

This integral can be split up into the sum of two integrals (Property 4, p. 331):

Let's graph these functions separately and try to evaluate their integrals by looking at the areas under the graphs.

This region has the same amount of area above the x axis as it does below the x axis, so these areas cancel each other out, and the value of the integral is zero.

This region is a triangle of base 2π, height 10π. Its area is

The sum of these values is the answer.