Area Under Curves
Prepared for IIT‑JEE (Main + Advanced) – Mathematics
1. Simple Explanation
The area under a curve (y=f(x)) between two points (x=a) and (x=b) (with (a<b)) is the amount of plane space the graph of (f) occupies above the (x)-axis and below the curve.
Mathematically it is defined as the definite integral
[ \text{Area} = \int_{a}^{b} f(x),dx ]
If the curve dips below the (x)-axis, the integral still gives the net area (positive above, negative below). To get the exact geometric area one must take the absolute value or split the integral at the points where (f(x)=0).
2. Key Points
| Concept | What It Means | Formula / Procedure |
|---|---|---|
| Definite Integral | Limit of Riemann sums; sum of infinitely many infinitesimal rectangles | (\displaystyle \int_{a}^{b} f(x),dx = \lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)\Delta x) |
| Area Above the (x)-axis | All (f(x) \ge 0) on ([a,b]) | (A = \int_a^b f(x),dx) |
| Area Below the (x)-axis | All (f(x) \le 0) on ([a,b]) | (A = -\int_a^b f(x),dx) |
| General Area | Curve crosses (x)-axis | Split at zeros: (A = \int_a^{c_1} |
| Area Between Two Curves | Upper curve (y = u(x)), lower curve (y = l(x)) | (A = \int_a^b [u(x)-l(x)],dx) |
| Vertical / Horizontal Strips | Use (dx) (vertical) if integrating w.r.t. (x); use (dy) (horizontal) if integrating w.r.t. (y) | (A = \int_{x_1}^{x_2} f(x),dx) or (A = \int_{y_1}^{y_2} g(y),dy) |
| Symmetry | If (f) is even, area from (-a) to (a) is (2\int_0^a f(x)dx). If odd, area is zero. | Use symmetry to reduce calculation. |
| Trapezoidal / Simpson’s Rules | Numerical approximation when antiderivative is hard | (\displaystyle \int_a^b f(x)dx \approx \frac{b-a}{2}[f(a)+f(b)]) (Trapezoid) |
| Area of a Polygon/Region given by Functions | Combine multiple integrals or use geometry formula | Apply piecewise integration. |
3. Worked Example 1 – Simple Curved Area
Problem: Find the area under the curve (y = x^2) from (x = 0) to (x = 2).
Solution
-
Set up the integral
[ A = \int_{0}^{2} x^2 , dx ]
-
Find the antiderivative
[ \int x^2 dx = \frac{x^3}{3} ]
-
Evaluate between limits
[ A = \left. \frac{x^3}{3} \right|_{0}^{2} = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} ]
Answer: The area is (\displaystyle \frac{8}{3}) square units.
4. Worked Example 2 – Area Between Two Curves
Problem: Find the area enclosed between the curves
(y = x^2) and (y = 4x).
Solution
-
Find intersection points
Set (x^2 = 4x \Rightarrow x^2 - 4x = 0 \Rightarrow x(x-4)=0).
So (x = 0) and (x = 4). -
Determine which curve is on top
For (x\in(0,4)):
(4x - x^2 = x(4-x) > 0).
Thus (y=4x) is above (y=x^2). -
Set up the integral
[ A = \int_{0}^{4}\bigl(4x - x^2\bigr),dx ]
-
Integrate
[ \int(4x - x^2)dx = 2x^2 - \frac{x^3}{3} ]
-
Evaluate
[ A = \left[2x^2 - \frac{x^3}{3}\right]_{0}^{4} = \left(2(4)^2 - \frac{(4)^3}{3}\right) - 0 = \left(32 - \frac{64}{3}\right) = \frac{96-64}{3} = \frac{32}{3} ]
Answer: The area between the curves is (\displaystyle \frac{32}{3}) square units.