Rishabh Author: Rishabh Kumar (IIT + ISI | Global Math Mentor)
Published: October 2025
Category: Calculus | Applications of Integration
🔹 What Does “Area Under a Curve” Mean?
In mathematics, integration can be thought of as the process of adding up infinitely many small quantities.
When applied to a function y=f(x)y = f(x)y=f(x), the area under the curve between two points x=ax = ax=a and x=bx = bx=b is given by: A=∫abf(x) dxA = \int_a^b f(x) \, dxA=∫abf(x)dx
This represents the accumulated value — geometrically, the region bounded by the curve, the x-axis, and the vertical lines x=ax = ax=a and x=bx = bx=b.
Simply put: Differentiation breaks things down. Integration builds them up.
🔹 Step-by-Step Example 1 — Simple Polynomial
Find the area under the curve y=x2y = x^2y=x2 from x=0x = 0x=0 to x=3x = 3x=3. A=∫03×2 dx=[x33]03=273=9A = \int_0^3 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^3 = \frac{27}{3} = 9A=∫03x2dx=[3×3]03=327=9
✅ Area = 9 square units
🔹 Step-by-Step Example 2 — Area Between Curve and x-Axis
If a curve dips below the x-axis, integration gives a signed area (negative values).
To find the total area, take absolute values or split the region at intercepts.
Example:
Find the total area enclosed by y=x2−4xy = x^2 – 4xy=x2−4x and the x-axis.
1️⃣ Find intercepts: x(x−4)=0⇒x=0,4x(x – 4) = 0 \Rightarrow x = 0, 4x(x−4)=0⇒x=0,4
2️⃣ Integrate: A=∫04∣x2−4x∣ dxA = \int_0^4 |x^2 – 4x| \, dxA=∫04∣x2−4x∣dx
Since y<0y < 0y<0 between 0 and 4, A=−∫04(x2−4x) dx=−[x33−2×2]04=−(643−32)=323A = -\int_0^4 (x^2 – 4x) \, dx = -\left[\frac{x^3}{3} – 2x^2\right]_0^4 = -\left(\frac{64}{3} – 32\right) = \frac{32}{3}A=−∫04(x2−4x)dx=−[3×3−2×2]04=−(364−32)=332
✅ Total Area = 323 sq. units\frac{32}{3} \, \text{sq. units}332sq. units
🔹 Example 3 — Area Between Two Curves
If two curves bound a region, y1=f(x)y_1 = f(x)y1=f(x) and y2=g(x)y_2 = g(x)y2=g(x),
then the enclosed area is: A=∫ab∣f(x)−g(x)∣ dxA = \int_a^b |f(x) – g(x)| \, dxA=∫ab∣f(x)−g(x)∣dx
Example:
Find the area enclosed between y=xy = xy=x and y=x2y = x^2y=x2 from x=0x = 0x=0 to x=1x = 1x=1. A=∫01(x−x2) dx=[x22−x33]01=12−13=16A = \int_0^1 (x – x^2) \, dx = \left[\frac{x^2}{2} – \frac{x^3}{3}\right]_0^1 = \frac{1}{2} – \frac{1}{3} = \frac{1}{6}A=∫01(x−x2)dx=[2×2−3×3]01=21−31=61
✅ Area = 16 sq. units\frac{1}{6} \, \text{sq. units}61sq. units
🔹 The Geometric Insight
Each infinitesimal strip of width dxdxdx under the curve forms a rectangle of height f(x)f(x)f(x).
By summing infinitely many such rectangles, we get the exact area — a perfect demonstration of the limit concept in calculus.
🔹 Common Pitfalls
- ❌ Forgetting that areas below the x-axis are negative.
- ❌ Not sketching the region before integrating.
- ❌ Mixing up limits of integration or axis of rotation.
- ❌ Ignoring absolute area in exam problems.
🔹 Real-World Applications
- Physics: work done by a variable force (area under F–x graph)
- Economics: total cost, profit, or demand (area under rate curves)
- Probability: area under probability density functions
- Engineering: flow rate, cross-sectional area, energy consumption
🔹 Advanced Example — IB/STEP Level
Find the area bounded by y=e−xy = e^{-x}y=e−x, the x-axis, and x=0x = 0x=0 to x=2x = 2x=2. A=∫02e−x dx=[−e−x]02=1−e−2A = \int_0^2 e^{-x} \, dx = [-e^{-x}]_0^2 = 1 – e^{-2}A=∫02e−xdx=[−e−x]02=1−e−2
✅ Area = 1−e−2 sq. units1 – e^{-2} \, \text{sq. units}1−e−2sq. units
🌟 Why It Matters
The “area under the curve” bridges algebra, geometry, and calculus — forming the foundation of physics, data analysis, and optimization.
It teaches students not only how to compute, but also how to model continuous change — a hallmark of higher mathematical thinking.
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✅ Interpret integrals geometrically
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