How do you find the area below the x axis?

Signed area versus geometric area

A definite integral computes net signed area. When a function f(x) is above the x-axis, the function values are positive, so multiplying the height by the width dx gives positive area elements. When a curve dips below the x-axis, f(x) is negative. Multiplying a negative height by dx gives a negative value. A standard definite integral from a to b counts this region as negative.

Geometric area, by contrast, cannot be negative. If you need the physical or geometric area bounded by the curve and the x-axis, you must convert negative sections into positive quantities. Integrating without checking for regions below the axis causes positive and negative areas to cancel each other out, giving an incorrect total area.

Steps to find the total area

Step 1: Set f(x) = 0 and solve for x to find every intercept inside your interval [a, b]. These roots mark where the curve crosses the axis.

Step 2: Split the original interval into subintervals using those roots as split points. On each subinterval, the curve stays entirely above or entirely below the x-axis.

Step 3: Integrate f(x) across each subinterval separately. If a subinterval produces a negative value, take its absolute value or multiply that integral by negative one. Add the positive values together to get the total area.

A worked example

Consider f(x) = x^2 - 4 on the interval from x = 0 to x = 3. Setting x^2 - 4 = 0 gives roots at x = 2 and x = -2. Within [0, 3], the curve crosses the x-axis at x = 2. Between 0 and 2, the curve is below the axis; between 2 and 3, it is above.

First, integrate from 0 to 2: [x^3 / 3 - 4x] from 0 to 2 gives 8/3 - 8 = -16/3. Since this area lies below the axis, its geometric area is |-16/3| = 16/3. Next, integrate from 2 to 3: [x^3 / 3 - 4x] from 2 to 3 gives (9 - 12) - (8/3 - 8) = -3 - (-16/3) = 7/3. Adding both parts yields 16/3 + 7/3 = 23/3 square units.