Periodic Function Integration
Definite Integration & Area
17
JEE Qs
8%
Hard
60
min
Master the splitting of integration limits using periodicity to reduce complex integrals to simpler forms over a single period.
🧮 Key Formulas
✅ Key Points for JEE
- 1The first step is always to correctly identify the fundamental period (T) of the given periodic function.
- 2When limits of integration are multiples of the period (e.g., 0 to nT), the integral can be simplified to n times the integral over one period (e.g., 0 to T).
- 3If limits are not exact multiples of the period (e.g., a to b), split the integral into parts covering full periods and remaining fractional parts, then apply periodicity.
- 4The integral of a periodic function over any interval whose length is equal to its period (T) will always yield the same value.
- 5Many problems combine periodic function properties with other definite integration properties, especially King's Rule (Integral from a to b of f(x) dx = Integral from a to b of f(a+b-x) dx) and odd/even function properties.
⚠️ Common Mistakes
- ✕Incorrectly identifying the fundamental period of functions (e.g., mistaking 2π for the period of sin²x or |sinx|).
- ✕Applying periodicity properties incorrectly when the limits of integration are not exact multiples or ranges of the period.
- ✕Forgetting to handle the 'remaining' interval when the integration limits are not perfectly aligned with multiples of the period.
- ✕Ignoring the conditions under which the function is periodic or continuous, especially for piecewise functions.
📝 Practice Questions
See allQ1. Let f(x) = ∫t0 (1) 253 (2) 154 (3) 125 (4) 157 →
Q11.Let the area enclosed between the curves |y| = 1 −x2 and x2 + y2 = 1 be α. If 9α = βπ + γ; β, γ are integers, then the value of |β −γ| equals. (1) 27 (2) 33 (3) 15 (4) 18
Q21.If 24 ∫ 0 4 (sin 4x − 12π + [2 sin x])dx = 2π + α, where [⋅] denotes the greatest integer function, then α is equal to _______.
Q6. Let for f(x) = 7 tan8 x + 7 tan6 x −3 tan4 x −3 tan2 x, I1 = ∫π/40 f(x)dx and I2 = ∫π/40 xf(x)dx. Then 7I1 + 12I2 is equal to : (1) 2 (2) 1 (3) 2π (4) π
Q13.The area of the region, inside the circle (x −2√3)2 + y2 = 12 and outside the parabola y2 = 2√3x is : (1) 3π + 8 (2) 6π −16 (3) 3π −8 (4) 6π −8
Q7. The area of the region enclosed by the curves y = x2 −4x + 4 and y2 = 16 −8x is : (1) 8 (2) 4 3 3 (3) 8 (4) 5 x ∈R. Then the numbers of local maximum and local minimum points of f ,
NCERT Chapters
- Class 11 Maths Ch 2: Relations and Functions
- Class 12 Maths Ch 7: Integrals