Area Under the x-axis

Video lesson · from 52:221 worked example

One lesson video covers all of Integration, so it opens at 52:22 for this topic — not from the beginning.

Theory

If the area is below the x-axis, the definite integral will yield a negative value. Since area must be positive, we can calculate it as the area between two curves, where the upper curve is $y=0$ (the x-axis) and the lower curve is the function.

\int_a^b (0 - f(x)) \,dx = - \int_a^b f(x) \,dx

⚠️ Common Examiner Traps

The negative is information, not a mistake: a negative integral tells you the region lies below the axis. Interpret it rather than quietly dropping the sign.
Never add a negative area to a positive one: work out each piece separately and add the magnitudes.
Find every root in the interval: each crossing is a place the integral must be split.
Total area versus net value: read carefully — "the area enclosed" wants the sum of the magnitudes; a plain definite integral may legitimately be negative.

Worked examples

Example 1

Calculate the shaded area shown.

The limits are $x=1$ to $x=4$ . The curve $y=x^2-4x$ is entirely below the x-axis.

\begin{aligned} \text{Area} &= - \int_1^4 (x^2 - 4x) \,dx \\ &= - \left[ \frac{x^3}{3} - 2x^2 \right]_1^4 \\ &= - \left( \left( \frac{64}{3} - 32 \right) - \left( \frac{1}{3} - 2 \right) \right) \\ &= - \left( - \frac{32}{3} - \left(- \frac{5}{3} \right) \right) \\ &= - \left( - \frac{27}{3} \right) \\ &= 9 \end{aligned}

The area is 9 square units.