How Do You Find the Area Between a Curve and the X-Axis When the Function Is Negative?

You negate the definite integral when the function is negative throughout the interval. Equivalently, use the absolute value: $A=-\int_a^b f(x)\,dx=\int_a^b |f(x)|\,dx$, because geometric area cannot be negative.

Why the Sign Changes

A definite integral measures signed area, not total geometric area. Regions above the x-axis contribute positively, while regions below the x-axis contribute negatively. Therefore, if $f(x)\leq 0$ throughout ([a,b]), its integral is non-positive and the area is the negative of that integral.

For example, find the area between $f(x)=x^2-4$ and the x-axis from $x=-1$ to $x=1$. Since throughout this interval, first write

$$A=-\int_{-1}^{1}(x^2-4)\,dx.$$

Now evaluate the signed integral:

$$\int_{-1}^{1}(x^2-4)\,dx =\left[\frac{x^3}{3}-4x\right]_{-1}^{1} =-\frac{22}{3}.$$

Hence,

$$A=-\left(-\frac{22}{3}\right)=\frac{22}{3}.$$

If a curve crosses the x-axis, find every root in the interval and split the integral at each point where the function changes sign. This prevents positive and negative contributions from cancelling.

For a sign-changing example, suppose $f(x)$ has one root $x=c$ inside ([a,b]), with $f(x)\geq 0$ on ([a,c]) and $f(x)\leq 0$ on ([c,b]). Then

$$A=\int_a^c f(x)\,dx-\int_c^b f(x)\,dx.$$

This is equivalent to $\int_a^b |f(x)|\,dx$, but splitting makes the method and boundaries clearer to an examiner.

The common misconception is that a negative definite integral represents negative area. It does not. The sign records the curve's position relative to the x-axis; area is the magnitude of each region.

IB Exam Technique

This is IB Mathematics: Analysis and Approaches SL 5.11, relevant to both Paper 1 and Paper 2. Examiners expect the correct integral expression before its evaluation. Show the roots used as boundaries, split at sign changes, and add positive areas. Do not integrate straight across multiple regions, because cancellation can produce displacement-like signed area rather than the requested total area.

On Paper 2, a GDC may locate non-exact roots, but you must still show the integral setup and give the area to the required accuracy.