Newton-Leibnitz's Formula

Newton Leibnitz's Theorem is one of the important parts of Calculus, which applies to measuring the change in the function at a certain point. Mathematically, it forms a powerful tool by which slopes of functions are determined, the maximum and minimum of functions found, and problems on motion, growth, and decay, to name a few. These integration concepts have been broadly applied in branches of mathematics, physics, engineering, economics, and biology.

This Story also Contains

1. Newton Leibnitz's Theorem
2. Solved Examples Based on Newton Leibnitz's Theorem
3. Summary

Newton Leibnitz's Theorem

Newton Leibnitz's Theorem states that If the functions u(x) and v(x) are defined and f(t) is a continuous function, then

$\frac{d}{d x}\left[\int_{\mathbf{u}(\mathbf{x})}^{\mathbf{v}(\mathbf{x})} \mathbf{f}(\mathbf{t}) \mathrm{dt}\right]=\mathbf{f}(\mathbf{v}(\mathbf{x})) \cdot \frac{\mathrm{d}}{\mathrm{dx}}\{\mathbf{v}(\mathbf{x})\}-\mathbf{f}(\mathbf{u}(\mathbf{x})) \cdot \frac{d}{d x}\{\mathbf{u}(\mathbf{x})\}$

It gives us the condition to differentiate the definite integral of which limits are functions of a different variable. When the limits of the integral are completely different functions compared to the function of the integral sign, then the Newton-Leibniz method is applicable. This is used to find the derivative of the integration. It is only useful in the case of definite integral. It can be used to find the differentiation of the nth order.

Definite integration calculates the area under a curve between two specific points on the x-axis.

Let f be a function of x defined on the closed interval [a, b]. F be another function such that $\frac{d}{d x}(F(x))=f(x)$ for all x in the domain of f, then $\int_a^b f(x) d x=[F(x)+c]_a^b=F(b)-F(a)$is called the definite integral of the function f(x) over the interval [a, b], where a is called the lower limit of the integral and b is called the upper limit of the integral.

Proof:

$\begin{array}{ll}\text { Let } & \frac{d}{d x}\{F(x)\}=f(x) \\ \Rightarrow & \int_{u(x)}^{v(x)} f(t) d t=F(v(x))-F(u(x)) \\ \Rightarrow & \frac{d}{d x}\left[\int_{u(x)}^{v(x)} f(t) d t\right]=\frac{d}{d x}(F(v(x))-F(u(x))) \\ \Rightarrow & \frac{d}{d x}\left[\int_{u(x)}^{v(x)} f(t) d t\right]=F^{\prime}(v(x)) \frac{d}{d x}\{v(x)\}-F^{\prime}(u(x)) \frac{d}{d x}\{u(x)) \\ \Rightarrow & \frac{d}{d x}\left[\int_{u(x)}^{v(x)} f(t) d t\right]=f(v(x)) \frac{d}{d x}\{v(x)\}-f(u(x)) \frac{d}{d x}\{u(x)\}\end{array}$

$\frac{d}{d t}\left(\int_{f(t)}^{\phi(t))} F(x) d x\right)=F(\phi(t)) \phi^{\prime}(t)-F(f(t)) f^{\prime}(t)$

Recommended Video Based on Newton Leibnitz's Theorem

Solved Examples Based on Newton Leibnitz's Theorem

Example 1: Let $F: R \rightarrow R$ be a differentiable function having $f(2)=6, f^{\prime}(2)=\left(\frac{1}{48}\right)$. Then $\lim _{x \rightarrow 2} \int_6^{f(x)} \frac{4 t^3}{x-2} d t$ equals

1) 18

2) 24

3) 36

4) 12

Solution

As we learnt in

NEWTON LEIBNITZ THEOREM -

$\frac{d}{d t}\left(\int_{f(t)}^{\phi(t))} F(x) d x\right)=F(\phi(t)) \phi^{\prime}(t)-F(f(t)) f^{\prime}(t)$

$f(2)=6 ; f^{\prime}(2)=1 / 48$

$\lim _{x \rightarrow 2} \int_6^{f(x)} \frac{4 t^3 d t}{x-2}=\lim _{x \rightarrow 2} \frac{4(f(x))^3 \times f^{\prime}(x)}{1}$

$\Rightarrow 4(f(2))^3 \times f^{\prime}(2)$

$\Rightarrow 4 \times 6^3 \times \frac{1}{48}=\frac{4 \times 216}{48}=18$

Hence, the answer is the option 1.

Example 2: If $f(x)=\int_0^x t(\sin x-\sin t) d t$

then

1) $f^{\prime \prime \prime}(x)+f^{\prime \prime}(x)=\sin x$
2) $f^{\prime \prime \prime}(x)+f^{\prime \prime}(x)-f^{\prime}(x)=\cos x$
3) $f^{\prime \prime \prime}(x)+f^{\prime}(x)=\cos x-2 x \sin x$
4) $f^{\prime \prime \prime}(x)-f^{\prime \prime}(x)=\cos x-2 x \sin x$

Solution

As we have learned

NEWTON LEIBNITZ THEOREM -

$\frac{d}{d t}\left(\int_{f(t)}^{\phi(t))} F(x) d x\right)=F(\phi(t)) \phi^{\prime}(t)-F(f(t)) f^{\prime}(t)$

$f(x)=\int_0^x t \sin x d t-\int_0^x t \sin t d t$

$=\frac{x^2}{2} \sin x-\int_0^x t \sin t$

$f^{\prime}(x)=x \sin x+\frac{x^2}{2} \cos x-x \sin x$

$f^{\prime}(x)=\frac{x^2}{2} \cos x$

$f^{\prime \prime}(x)=x \cos x-\frac{x^2}{2} \sin x$

$f^{\prime \prime \prime}(x)=-x \sin x+\cos x-x \sin x-\frac{x^2}{2} \cos x$

Thus $f^{\prime \prime \prime}(x)+f^{\prime}(x)=\cos x-2 x \sin x$

Hence, the answer is the option 3.

Example 3: Let $f:(0, \infty) \rightarrow R$ and $F(x)=\int_1^x f(t) d t$. If $F\left(x^2\right)=x^2(1+x)$ then $f(4)$ equals :

1) 5/4

2) 7

3) 4

4) 2

Solution

As we learned

NEWTON LEIBNITZ THEOREM -

$\frac{d}{d t}\left(\int_{f(t)}^{\phi(t))} F(x) d x\right)=F(\phi(t)) \phi^{\prime}(t)-F(f(t)) f^{\prime}(t)$

$F^{\prime}(x)=f(x)$

$F(x)=x(1+\sqrt{x})=x+x^{\frac{3}{2}}$

$F^{\prime}(x)=f(x)=1+\frac{3}{2} \sqrt{x}$

$\therefore f(4)=4$

Hence, the answer is the option (3).

Example 4: $\lim _{x \rightarrow 0} \frac{\int_0^{x^2} \cos t^2 d t}{x \sin x}=?$ :

1) 1

2) 2

3) 0

4) 0.5

Solution

As we learned

NEWTON LEIBNITZ THEOREM -

$\frac{d}{d t}\left(\int_{f(t)}^{\phi(t))} F(x) d x\right)=F(\phi(t)) \phi^{\prime}(t)-F(f(t)) f^{\prime}(t)$

Limit= $\lim _{x \rightarrow 0} \frac{\frac{d}{d x} \int_0^{x^2} \cos t^2 d t}{\frac{d}{d x}(x \sin x)}=\lim _{x \rightarrow 0} \frac{\cos \left(x^2\right)^2 \cdot \frac{d\left(x^2\right)}{d x}}{\sin x+x \cos x}$

$=\lim _{x \rightarrow 0} \frac{2 x \cos x^4}{\sin x+x \cos x}=\lim _{x \rightarrow 0} \frac{2 x \cos x^4}{x \cdot\left(\frac{\sin x}{x}\right)+x \cos x}=\lim _{x \rightarrow 0} \frac{2 \cos x^4}{1+\cos x}=1$

Hence, the answer is the option (1).

Example 5: if $\int_{\sin x}^1 t^2 f(t) d t=1-\sin x, x \in\left(0, \frac{\Pi}{2}\right)$ then $f\left(\frac{1}{\sqrt{3}}\right)=$ :

1) 3
2) $\frac{1}{3}$
3) $\frac{1}{\sqrt{3}}$
4) $\sqrt{3}$

Solution

As we learned

NEWTON LEIBNITZ THEOREM -

$\frac{d}{d t}\left(\int_{f(t)}^{\phi(t))} F(x) d x\right)=F(\phi(t)) \phi^{\prime}(t)-F(f(t)) f^{\prime}(t)$

On differentiating both sides, we get

$\Rightarrow-\sin ^2 x f(\sin x) \cos x=-\cos x$

$\begin{aligned} & \Rightarrow f(\sin x)=\operatorname{cosec}^2 x \\ & \Rightarrow f(x)=\frac{1}{x^2} \\ & \Rightarrow f\left(\frac{1}{\sqrt{3}}\right)=3\end{aligned}$

Hence, the answer is the option (1).

Summary

Newton Lebinzn's theorem is a powerful tool in calculus that allows us to calculate the area under a curve between two specific points. It provides a deeper understanding of mathematical ideas paramount for later developments in many scientific and engineering disciplines.