Concept understanding — Area of a Bounded Plane Region by Integration
Building on the geometric meaning of ∫abf(x)dx (Remarks under the limit-of-a-sum definition), this topic packages the vertical/horizontal-strip argument into ready-to-use area formulas.
Bounded by a curve, the x-axis, and x=a,x=b (vertical strips, height =∣y∣, width Δx):
If y=f(x)≥0 throughout [a,b] (curve above the x-axis): A=∫abydx.
If y=f(x)≤0 throughout [a,b] (curve below the x-axis): A=−∫abydx=∫abydx.
If f changes sign on [a,b]: split at the zeros c1,c2,… into subintervals of constant sign, apply the two rules above on each piece, and add the absolute values — A=∫ac1f+∫c1c2f+⋯ (never just ∫abfdx, which can cancel positive against negative area).
Bounded by a curve, the y-axis, and y=c,y=d (horizontal strips, mirror image): A=∫cdxdy if the curve lies to the right of the y-axis (x≥0); A=−∫cdxdy if to the left; split-and-add-absolute-values if it crosses.
Bounded between two curves. If f(x)≥g(x) on [a,b] (an "upper" curve U and "lower" curve L), the region between them and the ordinates x=a,x=b has area
A=∫ab[f(x)−g(x)]dx=∫ab(yU−yL)dx.
The y-axis mirror (curves x=f(y)≥x=g(y), a "right" curve R and "left" curve L) gives A=∫cd(xR−xL)dy.
General working rule (no need to identify upper/lower by name): draw an arbitrary vertical line cutting the region; call the y-value where it enters the region yENTRY and where it exitsyEXIT (both read off the bounding curves' equations); then
A=∫ab[yEXIT−yENTRY]dx
(and the horizontal-strip mirror A=∫cd[xEXIT−xENTRY]dy). This is the most robust approach when a region's boundary switches which curve is "on top" partway through, or when it is more natural to sketch and read off entry/exit points than to name f,g globally.
Tip
Always sketch the region first — the sketch immediately tells you whether to integrate in x or y, where the curves cross (the limits of integration), and whether a piecewise split is needed. When a region has symmetry (about an axis), compute the area of one symmetric piece and multiply, rather than integrating over the whole region directly.
y=x2−x−2=(x−2)(x+1) is ≥0 outside [−1,2] and ≤0 inside [−1,2] — split [−3,3] at x=−1,2 into three pieces and add absolute values.
✓Final answer
Area =15 sq units.
Factor the parabola to find its zeros inside [−3,3], split into three constant-sign pieces, and add absolute values.
Step 1. Factor the curve and find its zeros.2+x−x2+y=0⇒y=x2−x−2=(x−2)(x+1), zero at x=−1,2; both lie inside [−3,3].
Step 2. Determine the sign of y on each piece. Since the parabola opens upward, y≥0 outside the roots and y≤0 between them: at x=−3, y=(−5)(−2)=10>0; at x=0, y=(−2)(1)=−2<0; at x=3, y=(1)(4)=4>0. So