Newton Leibniz Theorem in Calculus Explained Clearly

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.