Picture: stacking thin strips under the curve: Imagine the region between a curve y = f(x) and the x-axis, sliced into thin vertical strips of width dx and height f(x).
Adding up all those strips from x = a to x = b gives the area — and that sum is exactly the definite integral.
So whenever a question says "find the area under the curve / the area of the shaded region" between two x-values, you are being asked for an integral. In AI, type it straight into the GDC.
IB-style question — area of a garden bed
A flower bed has one edge modelled by y = 0.5x² between x = 0 and x = 4 metres, with the other edge along the x-axis (x in metres, y in metres).
Find the area of the flower bed.
Step by step
- The area under the curve is the definite integral of y from 0 to 4.
- Integrate (or just type it into the GDC).
Final answer
The flower bed has area 32/3 ≈ 10.7 m². (Units are m² because both axes are in metres.)
Always check whether the curve dips below the axis: An integral counts area below the x-axis as negative. If the region crosses the axis, the plain integral gives a signed total, not the true geometric area.
For a true area where the curve goes below the axis, split at the x-intercepts and add the absolute values — or, on the GDC, integrate |f(x)|.
Picture: the strip height is (top − bottom): When a region sits between two curves, each thin vertical strip runs from the lower curve up to the upper curve, so its height is (top − bottom).
Adding the strips gives the area as the integral of (upper − lower) between the x-values where the curves meet. This automatically handles curves that dip below the x-axis, because what matters is the gap between them, not their height above the axis.
IB-style question — area between two cost models
Two pricing models give cost per unit f(x) = 6 − 0.5x² and g(x) = 2 (in dollars), where x is thousands of units, 0 ≤ x ≤ 3.
(a) Find where the two models give the same cost.
(b) Find the area of the region between the two curves.
Step by step
- (a) Set the curves equal to find the intersection.
- Take the positive root (x ≥ 0).
- (b) On 0 ≤ x ≤ 2.83 the curve f is above g, so integrate (top − bottom).
- Simplify the integrand and evaluate (use the GDC).
Final answer
(a) The models match at x = 2√2 ≈ 2.83 (thousand units). (b) The enclosed area ≈ 7.54. Always integrate top minus bottom between the intersection points.
Find the limits first — they are usually the intersections: Most "area between curves" questions do not hand you the limits. Solve f(x) = g(x) (on the GDC, find the intersection points) and use those x-values as a and b.
If you are unsure which curve is on top, just evaluate both at one x-value in the interval — the larger one is the "top".