AI Tutor
Imagine integrating f(x)=x3f(x) = x^3 from βˆ’2-2 to 22 and getting exactly 00, even though there is clearly colored space between the curve and the xx-axis.
A standard definite integral evaluates net signed area, meaning regions above the xx-axis count as positive while regions below count as negative.
To find the true geometric areaβ€”the actual physical amount of surface enclosedβ€”we must prevent these positive and negative areas from canceling each other out.
How do we mathematically force regions under the axis to count as positive space?

The Split-and-Sum Strategy

The geometric area is formally defined as ∫ab∣f(x)βˆ£β€‰dx\int_a^b |f(x)| \, dx, which flips all negative regions above the horizontal axis.
Because we cannot easily antidifferentiate an absolute value directly, we locate every root (where f(x)=0f(x) = 0) inside the interval [a,b][a, b] and slice the integral into separate sub-intervals.
On any sub-interval where the function stays entirely below the xx-axis, the integral yields a negative number, so taking its absolute value guarantees a positive geometric area.
Formally, if roots partition [a,b][a, b] into sub-intervals with endpoints a=c0<c1<β‹―<ck+1=ba = c_0 < c_1 < \dots < c_{k+1} = b, the total geometric area is the summation of magnitudes: Area=βˆ‘i=0k∣∫cici+1f(x) dx∣\text{Area} = \sum_{i=0}^{k} \left| \int_{c_i}^{c_{i+1}} f(x)\,dx \right|.
Let's put this procedure into action on a curve that dips below and climbs back above the axis.

Worked Example: A Cubic Crossing the Axis