Newton Leibniz Formula for Differentiating definite integrations of variable limits

Leibnitz rule is a very important formula for differentiating a given definite integration having functions of variable limits. This rule is also known as Fundamental Theorem of Calculus.

Conditions for applying Newton Leibniz Formula for Differentiation

Let the given definite integration is as below,

\int_{g(x)}^{h(x)}f(t)dt

Where f is a continuous function on [a, b] and g(x) and h(x) are differentiable functions of x whose values lie in [a, b] .

 

Method for applying Newton Leibniz Formula for Differentiation

Now, we have to evaluate the above integration.

Let,

P(x)= \int_{g(x)}^{h(x)}f(t)dt

Differentiating w.r.t to x , using Newton Leibniz Formula

\frac {dP(x)}{dx}= f(h(x))\times h'(x)–f(g(x))\times g'(x)

where h'(x)=\frac{dh(x)}{dx} and g'(x)=\frac{dg(x)}{dx}

 

Some More Questions to Read:

• IIT JEE Advanced 2025, Paper-1, Marks (+4, 0); Limits Question on Leibniz Rule

• IIT JEE Advanced 2025 Paper-2 Limits Question based on L’Hospital’s Rule

Proof of Newton Leibniz Formula for Differentiation

Let \frac {dF(x)}{dx}= f(x)   \Longrightarrow   \int f(x) dx = F(x) + C  

Now, Let P(x)= \int_{g(x)}^{h(x)}f(t)dt

\therefore P(x)=\left| F(t) \right|_{g(x))}^{h(x))}

P(x)= F(g(x)) – F(h(x))

Now, differentiating P(x) w.r.t x , we get

\frac {dP(x)}{dx} = F'(g(x)) g'(x) – F'(h(x)) h'(x) ……..(1)

\because \frac {dF(x)}{dx}= f(x) \Longrightarrow F'(x) = f(x)

\therefore equation (1) converts into

\frac {dP(x)}{dx} = f(g(x)) g'(x) – f(h(x)) h'(x)

Hence,

\frac {d}{dx} \int_{g(x)}^{h(x)}f(t)dt = f(g(x)) g'(x) – f(h(x)) h'(x)

This is the required proof of Newton Leibniz Formula for Differentiation

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Example on Newton Leibniz Formula for Differentiation

Question:

Evaluate \lim\limits_{x\to 0^{+}} \int_{0}^{x^{2}} \frac {sin \sqrt{t}} {x^{3}}  

Solution:

First write the given question correctly, as below 

\lim\limits_{x\to 0^{+}} \frac {\int_{0}^{x^{2}} {sin \sqrt{t}}} {x^{3}}  

Since the limit is in \frac {0}{0} form therefore, Applying L’Hopital’s Rule, we get

\lim\limits_{x\to 0^{+}} \frac {sin \sqrt{x^{2}}\times2x-sin \sqrt{0^{2}}\times0} {3 x^{2}}  

hence we get

\lim\limits_{x\to 0^{+}} \frac {sin{x} \times2x} {3 x^{2}}  

\lim\limits_{x\to 0^{+}} \frac {sin{x}} {x} \times {\frac{2x} {3 x}}= \frac {2}{3}

So, the answer is \frac {2}{3}

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‘Modified’ Newton Leibniz Formula

Modified Newton Leibniz Formula for Differentiating definite integrations of variable limits

Let the given definite integration is as below,

\int_{g(x)}^{h(x)}f(x, t)dt

Where f is a continuous function on [a, b] and g(x) and h(x) are differentiable functions of x whose values lie in [a, b] .

 

Method for applying Modified Newton Leibniz Formula for Differentiation

Now, we have to evaluate the above integration.

Let,

P(x)= \int_{g(x)}^{h(x)}f(x, t)dt

Differentiating w.r.t to x , using Newton Leibniz Formula

\frac {dP(x)}{dx}= \int_{g(x)}^{h(x)} \frac{\partial f(x, t) }{\partial x}dt + f(h(x))\times h'(x)–f(g(x))\times g'(x)

where \frac{\partial f(x, t) }{\partial x} is partial derivative of f(x, t) w.r.t x taking t as a constant,   h'(x)=\frac{dh(x)}{dx} and g'(x)=\frac{dg(x)}{dx}

 

 

Example on Modified Newton Leibniz Formula for Differentiation

Question:

Evaluate  

Solution:

 

 

Some More Questions to Read:

• JEE Mains 2025 Question on Limits Held on January 29, 2025; Shift-1

• JEE Mains 2019; A.M, G.M Inequality based question.

 

Some Important Concept to Read:

• L’ Hospital Rule complete Content 

 

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