Curves that cross the axis

Area is always positive, but an integral is not. Where the curve is below the x-axis, y is negative, so ∫ y dx comes out negative there.

If a curve is partly above and partly below the axis, a single integral lets the parts cancel, and gives the wrong area.

Method:

• Find where the curve crosses the x-axis (solve y = 0).
• Split the integral at those points.
• Work out each part and take its size (drop any minus sign).
• Add the sizes.

Example: y = x² – 1 from x = 0 to x = 2. It crosses at x = 1.

• ∫₀¹ (x² – 1) dx = 1/3 – 1 = –2/3, so this part has area 2/3 (below the axis).
• ∫₁² (x² – 1) dx = (8/3 – 2) – (1/3 – 1) = 4/3 (above the axis).
• Total area = 2/3 + 4/3 = 2.

A single integral from 0 to 2 would give –2/3 + 4/3 = 2/3. That is a correct integral but the wrong area.

For modulus functions such as |x – 2|, the graph is a V shape. Split where the inside is zero; often each piece is just a triangle, and ½ × base × height gives the answer as a check.

A sketch again helps: shade the region, and mark which parts are under the axis.