INSTRUCTOR: Find the area of a region bounded above by the curve y equals x cubed and below by y equals 0 over the interval 0 to 3. Now, before we set this up as a double integral, I'd like to just note, this is actually a question you might see at the end of a Calculus 1 course or possibly in a Calculus 2 course at the beginning. So we're actually going to see as we go, one of the steps looks hopefully very familiar to us from previous courses.

Now, because we're explicitly given bounds in the y direction, y equals x cubed and y equals 0, and also bounds in the x direction, interval 0 to 3, we can set this up as the integral from 0 to 3 with respect to x of the integral from y equals 0 to y equals x cubed with respect to y of the function 1. Integrating this first with respect to y and evaluating at those endpoints, this becomes the integral from 0 to 3 with respect to x of the function x cubed.

And pausing here for a moment, this is the function we might see in our earlier calculus courses. Now, taking the antiderivative with respect to x, this will be the function x to the fourth over 4, evaluated from 0 to 3, which is equal to 81 over 4. So it says, this area is 18/4 units squared.