The value of the integral over [0, ] where the area is above the x-axis, is 2. The value of the integral over [ ,2 ] where the area is below the x-axis, is -2. Therefore the net area over [0, 2 ] is 0.

The net area is square units, which is the sum of the two areas that lie above the x-axis minus the area that lies below the x-axis.

1. The curve is decreasing on (0, 4), so the area function is concave downward on the interval.
2. The curve is positive when x < 2 and negative when x > 2, so the area function has a maximum at x = 2.
3. The curve appears to have equal areas above and below the x-axis on the interval, so the area function is 0 when x = 4.

A Net Area Function

The total area between f(x) = sin x and the x-axis between x = 0 and x = 2 is 4 square units.

The TI-89 is not able to integrate , as indicated by the result being the same as the command.

However, the graph of y = x³ – 3x² – x + 3 on the interval [0, 4] has two regions above the x-axis and one below.

The window is [0, 4] x [-5, 15].

After finding the zeros of the curve (x = 1 and x = 3), integrate the curve over the separate intervals (0, 1), (1, 3), and (3, 4) then add their absolute values.

The total area is square units.

The area between the curves is , or approximately 11.6821 square units.

The exact answer is , which is approximately 6.12573 units.

The result is the same as the decimal approximation found in Question 17.3.1.

One possible net area function is shown below.