Area as a Definite Integral

The area under y=f(x) from x=a to x=b (above the x-axis) is ∫ₐᵇ f(x)dx. If f(x)<0 on part of [a,b], the integral gives signed area — below-axis regions subtract. To find total area (always positive), integrate |f(x)|: split the interval at each zero of f.

Area Between Two Curves

Integrating with Respect to y

Sometimes a region is more naturally described by horizontal strips than vertical ones. If the boundary curves can be written as x = f(y) and x = g(y), integrate with respect to y: A = ∫ᶜᵈ [right − left]dy = ∫ᶜᵈ [f(y) − g(y)]dy where c and d are the y-values of the intersection points.

When to use this: if the top and bottom boundaries change (e.g., the region is bounded above by one curve and below by another that changes at an x-value requiring case-by-case analysis), switching to horizontal integration with respect to y often eliminates the need to split the integral.

Applications: Why Area Calculations Matter

Frequently Asked Questions
What does negative area mean?▼
Negative signed area means the curve is below the x-axis in that region, contributing negatively to the integral. It does not mean the geometric area is negative. To find total geometric area, compute ∫|f(x)|dx by splitting at x-intercepts.
How do I set up area between curves problems?▼
Step 1: Find intersection points (bounds). Step 2: Determine which curve is on top. Step 3: Set up ∫[top−bottom]dx. Step 4: Evaluate. Always sketch the region first — it prevents sign errors and incorrect bounds.
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References & Further Reading
  • Stewart, J. (2015). Calculus, §5.1–5.3, §6.1. Cengage.
  • Apostol, T. (1967). Calculus, Vol. 1, Ch. 1–2. Wiley.
  • Strang, G. (1991). Calculus, Ch. 5. Wellesley-Cambridge.

Setting Up the Integral

The area between a curve y = f(x) and the x-axis from x = a to x = b is ∫ₐᵇ |f(x)| dx. The absolute value matters: if f(x) is negative over part of [a,b], that region is below the x-axis, and you must integrate f and −f separately and add the magnitudes.

📋 Area between sin(x) and x-axis on [0, 2π]
Splitsin(x) ≥ 0 on [0,π], sin(x) ≤ 0 on [π, 2π]
Area₁∫₀^π sin(x)dx = [−cos x]₀^π = 1+1 = 2
Area₂|∫_π^(2π) sin(x)dx| = |[−cos x]_π^(2π)| = |−1−1| = 2
Total area= 2 + 2 = 4 (not ∫₀^(2π) sin x dx = 0)

Area Between Two Curves

The area between y = f(x) (top curve) and y = g(x) (bottom curve) is ∫ₐᵇ [f(x) − g(x)] dx, where a and b are the x-coordinates where the curves intersect.

📋 Area between y = x and y = x² on [0,1]
Intersectionsx = x² → x=0, x=1. On (0,1): x > x² (line above parabola) ✓
Integral∫₀¹ (x − x²)dx = [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6
📋 Area between y = 4 − x² and y = x + 2
Intersections4−x² = x+2 → x²+x−2=0 → (x+2)(x−1)=0 → x=−2, x=1
Top curveAt x=0: 4−0=4 vs 0+2=2. Parabola is on top.
Integral∫₋₂¹ (4−x²−x−2)dx = ∫₋₂¹ (2−x−x²)dx = [2x−x²/2−x³/3]₋₂¹ = 9/2

The 4-Step Method

📋 Standard approach for any area problem
Step 1Sketch the region (identify which curve is on top)
Step 2Find intersection points (these give limits of integration)
Step 3Set up ∫[top − bottom]dx, checking for sign changes
Step 4Evaluate using FTC and simplify
AM
Dr. Aisha Malik, PhD Mathematics
Senior Lecturer in Applied Mathematics · 12 years teaching calculus at university level

Dr. Malik holds a PhD in Applied Mathematics from the University of Edinburgh and has taught calculus to over 4,000 students at both undergraduate and postgraduate level. Her research focuses on numerical methods for differential equations. She has reviewed this article for mathematical accuracy and pedagogical clarity.

Technically reviewed by: Prof. James Chen, Stanford Mathematics Department