Q1. Find the area under y = x² from x = 0 to x = 3.
A = ∫₀³ x² dx = [x³/3] from 0 to 3
A = 3³/3 − 0 = 27/3
A = 9 square units
Application of Integrals — Class 12
A definite integral is literally an area. Learn to set up the integral for the region between a curve and the x-axis, or between two curves.
Area under a curveThe area bounded by y = f(x), the x-axis, and x = a to x = b is A = ∫ₐᵇ f(x) dx (assuming f(x) ≥ 0).
Area between two curvesIf f(x) ≥ g(x) on [a,b], the area between them is ∫ₐᵇ [f(x) − g(x)] dx.
Solved examples
Formula sheet
| Quantity | Formula |
|---|---|
| Area under a curve | A = ∫ₐᵇ f(x) dx |
| Area between two curves | A = ∫ₐᵇ [f(x) − g(x)] dx |
| Area of a circle x²+y²=r² | πr² (via integration) |
Common mistakes & exam wins
- Sketch the curve(s) first — it tells you the correct limits and which function is "on top".
- If part of the curve dips below the x-axis, that area comes out negative — take its absolute value.
- For area between curves, always subtract the LOWER function from the UPPER one.
Frequently asked questions
How do you find the area under a curve using integration?
Evaluate the definite integral of the function between the given x-limits: A = ∫ₐᵇ f(x) dx.
How do you find the area between two curves?
Integrate the difference of the two functions (upper minus lower) over the interval where they bound the region.
What if the area comes out negative?
It means the curve lies below the x-axis in that region; take the absolute value for the actual area.