JEE Mains Maths · Definite Integration

Integral Equations and Leibniz's Rule

Integrals whose limits depend on x, differentiated by Leibniz's rule; equations in which a definite integral is an unknown constant; and equations of the form ∫ f(λx) dλ = a f(x), solved by power functions.

All Definite Integration notes JEE Mains Maths strategy

Why this matters

Twenty-eight PYQs. Two thirds give an equation containing an integral with a variable limit: differentiating removes the integral and leaves a differential equation. The rest hide constants inside integrals or scale the variable. Three ideas cover the page.

Concept 1 of 3: Differentiating an integral with variable limits

\frac{d}{dx}\int_{u(x)}^{v(x)}f(t)\,dt=f(v)\,v'-f(u)\,u'

. Differentiating an equation that contains such an integral removes the integral and leaves an ordinary equation in

f

, usually a first-order linear differential equation. Putting

x

equal to the lower limit, where the integral is 0, gives the constant. For a limit of an integral over a power of

(x-a)

, use L'Hôpital's rule with this derivative.

Definition

$\frac{d}{dx}\int_{u(x)}^{v(x)}f(t)\,dt=f(v(x))v'(x)-f(u(x))u'(x)$ .
At the lower limit the integral is 0: this fixes a constant.
$\int_0^x(x-t)f(t)\,dt$ : differentiate once to $\int_0^xf$ , twice to $f(x)$ .

Leibniz's rule

\frac{d}{dx}\int_{u(x)}^{v(x)}f(t)\,dt=f\big(v(x)\big)v'(x)-f\big(u(x)\big)u'(x)

Worked example

F(x)=\int_0^{x^2}\sqrt{1+t}\,dt

. Find

F'(1)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 28 Jan 2025 · Q66Moderate

Example 1 · Definite Integration · Integral Equations and Leibniz's Rule

Let for some function

y = f(x),\int_{0}^{x} tf(t)dt =x^{2}f(x)

x > 0

and

f(2) = 3

. Then

f(6)

is equal to :

The chain factor

An upper limit

x^2

brings a factor

2x

; a lower limit contributes with a minus sign. Forgetting either is the usual slip.

Drill 18 more on differentiating an integral with variable limits

Concept 2 of 3: Definite integrals as unknown constants

A definite integral with fixed limits is a number. In

f(x)=x+\int_0^1(x-t)f(t)\,dt

, take the

x

out:

f(x)=x+Ax-B

with

A=\int_0^1f

and

B=\int_0^1tf

. Substitute this form back into the definitions of

A

and

B

; that gives linear equations for the constants.

Definition

Only the part independent of $x$ is constant: split $\sin(x+y)=\sin x\cos y+\cos x\sin y$ first.
Name each integral: $A,B,\dots$ ; write $f$ in terms of them.
Substitute back and solve the linear system.

The form it forces

f(x)=g(x)+A\,h(x),\qquad A=\int_a^b k(t)f(t)\,dt

Worked example

f(x)=1+x\int_0^1f(t)\,dt

. Find

f(x)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 24 · Q79Moderate

Example 2 · Definite Integration · Integral Equations and Leibniz's Rule

The function

f(x)

, that satisfies the condition

f(x) =x+\int_{0}^{\pi/2} \sin x\cdot \cos yf(y)dy

, is :

x inside the integral is not a constant

\int_0^1(x-t)f(t)\,dt

the

x

must come out first: it is

x\int f-\int tf

. Naming the whole integral a constant is wrong.

Drill 5 more on definite integrals as unknown constants

Concept 3 of 3: Integrals of f(λx): power-function solutions

Putting

s=\lambda x

turns

\int_0^1f(\lambda x)\,d\lambda

into

\frac1x\int_0^xf(s)\,ds

. An equation '

\int_0^1f(\lambda x)\,d\lambda=a\,f(x)

' is solved by a power

f=cx^k

, because then the left side is

\frac{f(x)}{k+1}

. The given values of

f

fix

k

and

c

Definition

$\int_0^1f(\lambda x)\,d\lambda=\frac1x\int_0^xf(s)\,ds$ .
$f=cx^k\Rightarrow\int_0^1f(\lambda x)\,d\lambda=\frac{f(x)}{k+1}$ , so $a=\frac1{k+1}$ .
Two values of $f$ fix $c$ and $k$ .

Power functions

f(x)=cx^k\ \Rightarrow\ \int_0^1 f(\lambda x)\,d\lambda=\frac{f(x)}{k+1}

Worked example

\int_0^1f(\lambda x)\,d\lambda=\frac13f(x)

and

f(1)=2

. Find

f(3)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 27 July 2022 · Q178Moderate

Example 3 · Definite Integration · Integral Equations and Leibniz's Rule

Let

f

a

differentiable function satisfying

f(x) =\frac{2}{\sqrt{3}}\int_{0}^{\sqrt{3}} f\left( \frac{\lambda^{2}x}{3} \right)d\lambda,x > 0

and

f(1) =\sqrt{3}

. If

y = f(x)

passes through the point

(\alpha,6)

, then

\alpha

is equal to

Confirm the guess

A power function is a guess. Check that it satisfies the equation for every

x

, and that the value of

a

it forces matches the one given.

Drill 2 more on integrals of f(λx): power-function solutions

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Differentiating an integral with variable limits
Leibniz's rule
$\frac{d}{dx}\int_{u(x)}^{v(x)}f(t)\,dt=f\big(v(x)\big)v'(x)-f\big(u(x)\big)u'(x)$
Definite integrals as unknown constants
The form it forces
$f(x)=g(x)+A\,h(x),\qquad A=\int_a^b k(t)f(t)\,dt$
Integrals of f(λx): power-function solutions
Power functions
$f(x)=cx^k\ \Rightarrow\ \int_0^1 f(\lambda x)\,d\lambda=\frac{f(x)}{k+1}$

Watch out for (3)

The chain factor
→ Differentiating an integral with variable limits
x inside the integral is not a constant
→ Definite integrals as unknown constants
Confirm the guess
→ Integrals of f(λx): power-function solutions

Test yourself on Definite Integration

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Definite Integration chapter test Drill all 28 questions