When on the interval , then = the area under the graph of f from a to b. To see what happens when over part of the interval consider the function . First consider the interval [ ] and follow the following steps with Maple:

> plot(6-2*x,x=-2..3);

The area of this triangular region is given by

> Int(6-2*x,x=-2..3);

To evaluate this we can write

> int(6-2*x,x=-2..3);

or recognize that this is the area of a triangle with base 5 and height 10. Now consider what happens when we look at the curve over [ ].

> plot(6-2*x,x=3..7);

Again we see a triangular region (the one between the curve and the x-axis) but this time it is below the axis. It has area 16, but when we evaluate

> int(6-2*x,x=3..7);

We get negative 16. When we put the two together and ask for we get

> int(6-2*x,x=-2..7);

which is the sum of 25 and -16. This is what we mean by the definite integral as a signed area.

Submission:

Follow the same process for the integral . That is plot the function given there over the two intervals where the function is positive and negative respectively, evaluate the definite integrals and finally show that the definite integral over the entire interval is the same as the sum of your two earlier integrals.

Submission worksheet: