The area under a curve between two x-values is the definite integral of the function. This area can be positive if above the x-axis and negative if below. To find the total area under a curve, the curve is broken into sections where the function is either above or below zero and the integral is evaluated over each section adding or subtracting areas as appropriate. The example problem demonstrates finding the total area under the curve defined by y=x^2-x-2 between -2 and 3 by breaking it into three sections and evaluating the integral over each.

1. Area Under the Curve

2. What is “area under the curve?”The definite integral can be used to find the area between a graph curve and the ‘x’ axis, between two given ‘x’ values. This area is called the ‘area under the curve’ regardless of whether it is above or below the ‘x’ axis.

3. When the area is above the x axis, the area is positive. When the area is below the x axis, the area is negative.

4. Example problemy = x² - x – 2We need to create three separate integrals

5. Finding the integralsThe zeros of the function f(x) that lie between -2 and 3 form the boundaries of the separate area segments, and these will be our integrals-2 < x < -1 -1 < x < 22 < x < 3

6. In Integral Form:

7. Integrate the equation using the general power rule:1/3x^3 – 1/2 x^2 – 2x

8. Plug in beginning points and endpoints for each of three intervals and subtract the beginning point from the end point.

9. Plug in numbers and solve1/3(-1)^3 – ½ (-1)^2 – 2(-1) – (1/3(-2)^3 – ½ (-2)^2 – 2(-2) = 1.83333 1/3(2)^3 – ½ (2)^2 – 2(2) – (1/3(-1)^3 – ½ (-1)^2 – 2(-1)) = 4.51/3(3)^3 – ½ (3)^2 – 2(3) – (1/3(2)^3 – ½ (2)^2 – 2(2)) = 1.83333

10. The total shaded area will be A = A1 + A2 + A3Total area = 1.83333 + 4.5 + 1.83333 = 8.16666The net area takes into account the negative area:1.83333 - 4.5 + 1.83333 = - 0.83333