Applications of Integrals
Learn Applications of Integrals with notes, examples, and practice questions.
Chapter Overview — Why Integrate in Real Life?
Area is the most visual application of integration — if you can shade the region on a graph, you are halfway to the correct integral.
In the Integrals chapter you learned how to find ∫ f(x) dx. This chapter answers: where does that number actually matter?
Definite integrals measure geometric quantities — mainly area under (or between) curves. Physics and economics use the same idea: total quantity = accumulation of an instantaneous rate.
Every area problem reduces to: sketch the region → identify bounds → set up ∫ (top curve − bottom curve) dx or ∫ x dy when convenient.
Area Under a Curve y = f(x)
Area under curve
A = ∫ₐᵇ y dx = ∫ₐᵇ f(x) dx (when f(x) ≥ 0)
Signed area
∫ₐᵇ f(x) dx can be negative where f(x) < 0
Total geometric area
Split at zeros; add |∫| of each sub-interval
For f(x) ≥ 0 on [a, b], the area of the region bounded by y = f(x), the x-axis, and the lines x = a and x = b equals ∫ₐᵇ f(x) dx.
If f(x) dips below the x-axis on part of the interval, split the integral at points where f(x) = 0. Take absolute value of each piece, or integrate separately and add magnitudes.
Area Between Two Curves
Between curves (x-form)
A = ∫ₐᵇ [f(x) − g(x)] dx, f(x) ≥ g(x) on [a,b]
Between curves (y-form)
A = ∫ᶜᵈ [x_right(y) − x_left(y)] dy
Intersection first
Solve f(x) = g(x) to find limits a and b
When two curves y = f(x) (upper) and y = g(x) (lower) meet at x = a and x = b, the enclosed area is the integral of the vertical strip height (upper − lower).
Always subtract the lower function from the upper. If unsure, test a point between the intersection x-values.
Area with Respect to the y-Axis
Area with y as variable
A = ∫ᶜᵈ x dy (when x ≥ 0 and curve is x = h(y))
Between two x = … curves
A = ∫ᶜᵈ [x_upper(y) − x_lower(y)] dy
Sometimes the curve is given as x = h(y). Then horizontal strips are easier: width in y-direction is dy, length of strip is x.
Use this when the region is better described by left/right boundaries in terms of y — for example, area bounded by a parabola x = y² and the line x = 4.
Step-by-Step Problem Strategy
- Draw a neat sketch — mark intersection points and which curve is on top
- Find limits: solve f(x) = g(x) or read from the question
- Write integrand = (upper function) − (lower function)
- Integrate and evaluate using Fundamental Theorem
- Include units if the question is applied (m², cm², etc.)
- Check: is the answer positive? Area cannot be negative
Common CBSE Patterns
- Area bounded by parabola y² = 4ax and its latus rectum
- Area between line y = mx + c and a curve (circle, ellipse, parabola)
- Area enclosed by two circles or a circle and a line
- Area in first quadrant only — limits often 0 to intersection point
- Symmetry: if region is symmetric about y-axis, compute one side × 2
Connection to Physics (Quick Insight)
If v(t) is velocity, then ∫ₜ₁ᵗ² v(t) dt = displacement. If v(t) ≥ 0, the same integral gives distance travelled.
The mathematics is identical to area — only the labels change. This is why mastering area problems strengthens applied integration.
Solved Examples
Step-by-step solutions — read each step before checking the final answer.
Example 1: Area Under a Parabola
Find the area bounded by y = x², x = 0, x = 2, and the x-axis.
- 1Region: parabola above x-axis from x = 0 to x = 2.
- 2A = ∫₀² x² dx = [x³/3]₀².
- 3= 8/3 − 0 = 8/3 square units.
Answer
8/3 sq. units
Example 2: Area Between Line and Curve
Find the area between y = x and y = x² in the first quadrant.
- 1Intersection: x = x² → x = 0 or x = 1. Limits: 0 to 1.
- 2On [0,1], line y = x is above parabola y = x².
- 3A = ∫₀¹ (x − x²) dx = [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6.
Answer
1/6 sq. units
Example 3: Area Using y as Variable
Find the area bounded by x = y² and the line x = 4.
- 1Intersection: y² = 4 → y = ±2. By symmetry, A = 2 ∫₀² 4 dy − 2 ∫₀² y² dy.
- 2Or single integral: A = ∫₋₂² (4 − y²) dy.
- 3= [4y − y³/3]₋₂² = (8 − 8/3) − (−8 + 8/3) = 32/3.
Answer
32/3 sq. units
Example 4: Circle and Line
Find the area enclosed by y = √(4 − x²) and y = 1.
- 1Solve √(4−x²) = 1 → 4 − x² = 1 → x = ±√3.
- 2Top curve: semicircle; bottom: y = 1. Limits: −√3 to √3.
- 3A = ∫₋√₃^√₃ [√(4−x²) − 1] dx (standard integral or geometry + calculus).
Answer
Evaluate using ∫√(a²−x²) dx formula — practice this standard form
Example 5: Below x-Axis Piece
Find the total area between y = sin x and the x-axis from x = 0 to x = 2π.
- 1sin x ≥ 0 on [0, π] and sin x ≤ 0 on [π, 2π].
- 2A = ∫₀^π sin x dx + |∫_π^2π sin x dx|.
- 3= [−cos x]₀^π + |[−cos x]_π^2π| = 2 + 2 = 4.
Answer
4 sq. units
Key Points to Remember
- ✓ Area = ∫ (upper curve − lower curve) over correct limits
- ✓ Always find intersection points before setting up the integral
- ✓ Sketch the region — prevents wrong limits and wrong order of subtraction
- ✓ Use ∫ x dy when horizontal strips are simpler
- ✓ Split the integral if the curve crosses the axis
- ✓ Symmetry can halve your calculation effort
Exam Tips
- • Draw the figure even if not asked — examiners award diagram marks
- • State limits clearly after solving f(x) = g(x)
- • Write 'Area =' before the integral for clarity
- • Memorise ∫√(a² − x²) dx — it appears in circle/ellipse area problems
- • Check first-quadrant-only conditions in conic section questions
- • Final answer must include square units (or unit²)
Formula Cheat Sheet
Quick reference — NCERT board exam formulas
Area & Volume
- Area under curve:A = ∫ₐᵇ f(x) dx
- Between curves:A = ∫ₐᵇ |f(x) − g(x)| dx
- Circle area = πr² ; Triangle = ½ × base × height
- Cylinder volume = πr²h
- Cone volume = (1/3)πr²h
- Sphere volume = (4/3)πr³
Practice MCQs — Applications of Integrals
Test your understanding with topic-wise multiple choice questions. Explanations appear after each answer.
Area under y = f(x) from x = a to x = b (f ≥ 0) is: