The geometric area enclosed by a curve and an axis is always positive. If the curve crosses the axis, split the region at each intercept and add the absolute values of the definite integrals. This distinction is essential because a definite integral measures accumulated signed change, whereas the question asks for physical area.
A definite integral gives signed area. For a curve :
Therefore, gives the net signed area, not necessarily the total geometric area. The required area is
which is normally evaluated by splitting the integral where .
Before integrating, solve within the stated interval. These roots divide the interval into sections where the sign of is constant, so each section can be handled correctly.
| Boundary axis | Area expression |
|---|---|
| -axis | Integrate with respect to : $A=\int |
| -axis | Write and integrate with respect to : $A=\int |
For a region bounded by the -axis, horizontal slices are usually simpler. Express the curve as , identify any -values where , and split there.
Find the total area between and the -axis for .
First find the intercepts:
The function is positive on and , but negative on . Hence,
Evaluating gives
By contrast, integrating without splitting gives , which is the net signed area. Using a GDC, store full precision and round only the final answer.
This is AHL 5.12 content. In an IB response, show the intercepts, split the integral at every sign change, write the complete integral expression before using the GDC, and give a positive final area with square units. If the command term is calculate or determine, a bare calculator value may not secure method credit; the set-up must be visible. The common misconception is reporting a negative integral as a negative area: integrals can be negative, but geometric areas cannot.
