Leibniz Rule — Differentiation under integral sign
Definite Integration & Area
17
JEE Qs
8%
Hard
60
min
Master the correct application of both parts of the Leibniz rule – differentiating the integrand with respect to x (partial derivative) and handling the variable limits using the chain rule with utmost care for signs and substitutions.
🧮 Key Formulas
✅ Key Points for JEE
- 1The Leibniz Rule allows differentiation of definite integrals where the limits of integration are functions of the differentiation variable (x) and/or the integrand itself depends on x.
- 2Always identify if the integrand f(x, t) contains the differentiation variable 'x'; if not, the partial derivative term ∫ (∂/∂x f(x, t)) dt simplifies to zero, leading to the simpler form of the rule.
- 3Remember to apply the chain rule correctly to the limits: substitute the limit into the integrand and multiply by the derivative of the limit function.
- 4When computing the partial derivative (∂/∂x f(x, t)), treat the integration variable 't' as a constant and differentiate only with respect to 'x'.
- 5This rule is frequently combined with L'Hopital's Rule to evaluate limits involving definite integrals, especially those resulting in indeterminate forms like 0/0 or ∞/∞.
⚠️ Common Mistakes
- ✕Forgetting to differentiate the limits of integration (g'(x) and h'(x)) or applying incorrect signs in the subtraction part.
- ✕Failing to recognize or incorrectly applying the partial derivative term when the integrand f(x, t) depends on both x and t.
- ✕Confusing the variable of integration (t) with the variable of differentiation (x), leading to incorrect substitution or differentiation.
📝 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 12 Maths Ch 5: Continuity and Differentiability
- Class 12 Maths Ch 7: Integrals