When the integral and the area disagree.

• Where the curve lies below the x-axis, f(x) < 0, so ∫ₐᵇ f(x) dx comes out negative. Area is never negative, so take the absolute value: the area is |∫ₐᵇ f(x) dx|.
• If the curve crosses the axis inside [a, b], one integral over the whole interval lets the pieces cancel. Split the interval at each crossing point, integrate each piece and add the absolute values: A = |A₁| + A₂ + ….
• Worked example: y = 3x + 2 meets the x-axis at x = −2/3. On [−1, −2/3] the line is below the axis and the integral is −1/6; on [−2/3, 1] it is 25/6. The area between x = −1 and x = 1 is 1/6 + 25/6 = 13/3 (a single integral would give 4).
• Worked example: y = cos x on [0, 2π] is above the axis on [0, π/2], below on [π/2, 3π/2] and above on [3π/2, 2π]. The pieces have areas 1, 2 and 1, so the area is 4, although ∫₀^(2π) cos x dx = 0.
• Worked example: y = x² − 4 on [0, 3] is below the axis on [0, 2] (area 16/3) and above on [2, 3] (area 7/3). The area is 23/3; the single integral gives −3.
• Worked example: y = x(x − 2) on [0, 3] gives ∫₀³ x(x − 2) dx = 0, yet the region has area 4/3 + 4/3 = 8/3. A zero integral does not mean zero area.
• Always find where f(x) = 0 inside the interval before integrating; a quick sketch shows which pieces are below the axis.

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