## Newton Leibniz Theorem: Brief Introduction

Newton Leibniz Theorem has a wide range of applications in calculus used to determine the definite integrals of a function whose limits are functions of any independent variable. In this article, we will be discussing the proof of the theorem in detail.

Integration of The Curve

## History of Isaac Newton

Isaac Newton

Name: Isaac Newton

Born: 4 January 1643

Died: 31 March 1727

Field: Mathematics

Nationality: British

## Statement of Newton Leibniz Theorem

$\dfrac{d}{d t}\left(\int_{a(t)}^{b(t)} f(x, t) d x\right)={\int_{a}^{b}\delta_{t}f(x, t) )dx} + f(b,t).\dfrac{db}{dt} - f(a,t).\dfrac{da}{dt}$

where $a(t)$ and $b(t)$ are definite limits of the integral.

## Proof of Newton Leibniz Rule for Integration

$I =\int_{a(t)}^{b(t)} f(x, t) d x$

Taking derivative,

$\dfrac{dI}{d t}=\dfrac{d}{d t}\left(\int_{a(t)}^{b(t)} f(x, t) d x\right)$

Now,

$\dfrac{dI}{d x}=\mathop{\lim}\limits_{\delta {t} \rightarrow 0} \dfrac{\int_{a+\delta {a}}^{b+\delta {b}} f(x, t+\delta {t}) dx-\int_{a}^{b} f(x, t) d x}{\delta {t}}$

$\dfrac{dI}{d x}=\mathop{\lim}\limits_{\delta {t} \rightarrow 0} \dfrac{1}{\delta {t}}({\int_{a+\delta {a}}^{b+\delta {b}} f(x, t+\delta {t}) dx-\int_{a}^{b} f(x, t) d x})$

$\dfrac{dI}{d x}=\mathop{\lim}\limits_{\delta {t} \rightarrow 0} \dfrac{1}{\delta {t}}({\int_{a+\delta {a}}^{a}} f(x, t+\delta {t}) dx + {\int_{a}^{b} f(x, t+\delta {t}) dx} + {\int_{b}^{b+\delta {b}}} f(x, t+\delta {t}) dx -\int_{a}^{b} f(x, t) d x)$

$\dfrac{dI}{d x}=\mathop{\lim}\limits_{\delta {t} \rightarrow 0} \dfrac{1}{\delta {t}}(-{\int_{a}^{a+\delta {a}} }f(x, t+\delta {t}) dx + {\int_{a}^{b}( f(x, t+\delta {t}) - f(x, t) )dx }+ {\int_{b}^{b+\delta {b}} f(x, t+\delta {t}) dx})$

We know

$\int_{a}^{b} f(x) d x = F(b)-F(a)$

From the slope definition,

$F(b)-F(a) = F’(c)(b-a)$

$ {\int_{b}^{b+\delta {b}} f(x, t+\delta {t}) dx}=F(x,t+\delta {t})-F(x,t)=f(c,t)(b+\delta {b}-b)=\delta {b}.f(c,t)$

$\mathop{\lim}\limits_{h \rightarrow 0} f(c,t).\delta {b} \rightarrow f(b,t).\delta {b}$

Now solving expression,

$\dfrac{dI}{d x}=\mathop{\lim}\limits_{\delta {t} \rightarrow 0} \dfrac{1}{\delta {t}}( {\int_{a}^{b}( f(x, t+\delta {t}) - f(x, t) )dx + f(b,t).\delta {b} - f(a,t).\delta {a}})$

$\dfrac{dI}{d x}=\mathop{\lim}\limits_{\delta {t} \rightarrow 0}( {\int_{a}^{b}\dfrac{( f(x, t+\delta {t}) - f(x, t) )dx}{\delta {t}} + f(b,t).\dfrac{\delta {b}}{\delta {t}} - f(a,t).\dfrac{\delta {a}}{\delta {t}}})$

$\dfrac{dI}{d x}= {\int_{a}^{b}\delta_{t}f(x, t) )dx} + f(b,t).\dfrac{db}{dt} - f(a,t).\dfrac{da}{dt}$

## Limitations of Newton Leibniz Theorem

Newton Leibniz Theorem can not be applied in the case of indefinite integral. It holds good only in the case of Definite Integral.

Newton Leibniz Theorem is used to find definite integrals; however, it does not use the basic properties of definite integrals.

## Applications of Newton Leibniz Theorem

The theorem can be used to find the integration as well as differentiation. We can find the first order, second order, and even $n^{th}$ order derivative.

The Newton Leibniz Theorem is used to find the definite integral when limits are themselves functions of any independent variable.

## Solved Examples

1. Find $\dfrac{d}{d x} \int_{1}^{x^{4}} \sec d t$.

Ans: Let $u=x^{4}$. Then,

$\dfrac{d}{d x} \int_{1}^{x^{4}} \sec t d t \\\Rightarrow \dfrac{d}{d x} \int_{1}^{u} \operatorname{sectdt} \\\Rightarrow \dfrac{d}{d u}\left(\int_{1}^{u} \operatorname{sect} d t\right. \\\Rightarrow \sec u \dfrac{d u}{d x} \\\Rightarrow \sec \left(x^{4}\right) \cdot 4 x^{3}$

2. Find the derivative of $F(x)=\int_{\pi / 2}^{x^{3}} \cos t d t$.

Ans: $F^{\prime}(x)=\dfrac{d F}{d u} \dfrac{d u}{d x}\\\Rightarrow\dfrac{d}{d u}\left[\int_{\pi / 2}^{u} \operatorname{costdt}\right] \dfrac{d u}{d x} \\\Rightarrow(\cos u)\left(3 x^{2}\right) \\\Rightarrow\left(\cos x^{3}\right)\left(3 x^{2}\right) \\\left.F(x)=\int_{\pi / 2}^{x^{3}} \operatorname{costdt}=\sin t\right \\\Rightarrow \sin x^{3}-\sin \dfrac{\pi}{2} \\\Rightarrow\left(\sin x^{3}\right)-1 \\ \Rightarrow F^{\prime}(x) = \left(\cos x^{3}\right)\left(3 x^{2}\right)$

3. Evaluate $\int_{-\infty}^{0} x e^{x} d x$.

Ans: $\int_{-\infty}^{0} x e^{x} d x=\mathop{\lim}\limits_{t \rightarrow-\infty} \int_{t}^{0} x e^{x} d x$

We integrate by parts with $u=x, d v=e^{x} d x$

$\text { so that } d u=d x, v=e^{x} $

$\int_{t}^{0} x e^{x} d x=\left.x e^{x}\right|_{t} ^{0}-\int_{t}^{0} e^{x} d x $

$\Rightarrow-t e^{t}-1+e^{t}$

We know that $e^{t} \rightarrow 0$

as $t \rightarrow-\infty$, and by l'Hopital's Rule, we have

$\mathop{\lim}\limits_{t \rightarrow-\infty} e^{t} $

$\Rightarrow \mathop{\lim}\limits_{t \rightarrow-\infty} \dfrac{t}{e^{-t}} $

$\Rightarrow \mathop{\lim}\limits_{t \rightarrow-\infty} \dfrac{t}{e^{-t}} $

Using l'Hopital's Rule,

$\Rightarrow \mathop{\lim}\limits_{t \rightarrow-\infty}\left(-e^{t}\right)=0 $

$\text { Therefore } \int_{-\infty}^{0} x e^{x} d x $

$\Rightarrow \mathop{\lim}\limits_{t \rightarrow-\infty}\left(-t e^{t}-1+e^{t}\right) $

$\Rightarrow-0-1+0=-1$

## Important Formulas to Remember

Newton Leibniz Formula:

$\dfrac{d}{d t}\left(\int_{a(t)}^{b(t)} f(x, t) d x\right)={\int_{a}^{b}\delta_{t}f(x, t) )dx} + f(b,t).\dfrac{db}{dt} - f(a,t).\dfrac{da}{dt}$

## Important Points to Remember

Newton Leibniz Theorem is applicable in the case of definite integrals.

It can be used to find differentiation as well as integration.

## Conclusion

We have a different set of formulas in Calculus that are used to find the integral of functions. However, the normal formulas and their properties do not apply in the case of functions where limits are themselves functions, so we need Newton Leibniz Theorem. The theorem is a unique theorem in its character and sense and is very helpful in solving problems.

## FAQs on Newton Leibniz Theorem

**1. What are indefinite integrals? Can we apply Newton Leibniz Theorem to the case of Indefinite Integrals?**

The integrals that are expressed without limits are called indefinite integrals. As these integrals take antiderivatives of other functions and have no definite limits, so these functions contain a constant in their value. Also, Indefinite Integrals of a function are never unique as they contain a constant will that can be anything and vary from person to person. Newton Leibniz Theorem is defined only for Definite Integrals, i.e., it holds good only in the case of definite integrals and so it cannot be applied in the case of Indefinite Integrals.

**2. What do you understand about the Leibniz Theorem? Is it the same as Newton's Leibniz Theorem?**

Leibniz Theorem also known as the Leibniz rule is defined for derivatives of the antiderivative and antiderivatives of the function are the functions that could have given the function as a derivative. According to the Leibniz rule, if two functions u(x) and v(x) are differentiable n times, then their product u(x). v(x) is also differentiable n times.

No, the Newton Leibniz theorem is not the same as Leibniz Rule. The former is used to calculate the definite integrals whereas the latter is used to find the derivatives of the product of functions.

**3. Can we apply Newton Leibniz Theorem for normal Definite integrals?**

Yes, we can apply Newton Leibniz Theorem in the case of normal definite integrals, i.e., the integrals in which limits are constant rather than functions of any independent variable. Newton Leibniz Theorem holds good in both types of definite integrals. Newton Leibniz Theorem is only not applicable in the case of indefinite integrals. Also, the properties of normal definite integrals are not applicable in the case of Newton Leibniz theorem. The theorem is considered a special case in the theory of definite integrals.