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what is newton -leibnitz's rule how to apply it in a question ?explain with 3-4 example

pradyot mayank , 16 Years ago
Grade 12
anser 2 Answers
Badiuddin askIITians.ismu Expert

Dear pradyot

Leibniz Integral Rule

The Leibniz integral rule gives a formula for differentiation of a definite integral whose limits are functions of the differential variable,

5849-972_7662_908735170818779a3a888ae684ae94d3.png

 

where the partial derivative of f indicates that inside the integral only the variation of ƒ ( x, α ) with α is considered in taking the derivative.

Example

Here, we consider the integration of

\textbf I\;=\;\int_0^{\frac{\pi}{2}}\,\frac{1}{\left(a\,\cos^2\,x+b\,\sin^2\,x\right)^2}\;dx,\,

where both a,\,b\,>\,0, by differentiating under the integral sign.

Let us first find \textbf J\;=\;\int_0^{\frac{\pi}{2}}\,\frac{1}{a\,\cos^2\,x+b\,\sin^2\,x}\;dx.\,

Dividing both the numerator and the denominator by \cos^2\,x yields

\begin{align} \textbf J\; &=\;\int_0^{\frac{\pi}{2}}\,\frac{\sec^2\,x}{a\,+b\,\tan^2\,x}\;dx \\ &=\,\frac{1}{b}\,\int_0^{\frac{\pi}{2}}\,\frac{1}{\left(\sqrt{\,\frac{a}{b}\,}\right)^2+\tan^2\,x}\;d(\tan\,x)\, \\ &=\,\frac{1}{\sqrt{\,a\,b\,}}\,\left(\tan^{-1}\left(\sqrt{\,\frac{b}{a}\,}\,\tan\,x\right)\right)\,\bigg|_0^{\frac{\pi}{2}}\;=\;\frac{\pi}{2\,\sqrt{\,a\,b\,}}. \end{align}

The limits of integration being independent of a,\, \textbf J\;=\;\int_0^{\frac{\pi}{2}}\,\frac{1}{a\,\cos^2\,x+b\,\sin^2\,x}\;dx\, gives us

\frac{\partial\,\textbf J}{\partial\,a}\;=\;-\,\int_0^{\frac{\pi}{2}}\,\frac{\cos^2\,x\;dx}{\left(a\,\cos^2\,x+b\,\sin^2\,x\right)^2}\,

whereas \textbf J\;=\;\frac{\pi}{2\,\sqrt{\,a\,b\,}} gives us

\frac{\partial\,\textbf J}{\partial\,a}\;=\;-\frac{\pi}{4\,\sqrt{\,a^3\,b\,}}.\,

Equating these two relations then yields

\,\int_0^{\frac{\pi}{2}}\,\frac{\cos^2\,x\;dx}{\left(a\,\cos^2\,x+b\,\sin^2\,x\right)^2}\;=\;\frac{\pi}{4\,\sqrt{\,a^3\,b\,}}.\,

In a similar fashion, pursuing \frac{\partial\,\textbf J}{\partial\,b}\, yields

\,\int_0^{\frac{\pi}{2}}\,\frac{\sin^2\,x\;dx}{\left(a\,\cos^2\,x+b\,\sin^2\,x\right)^2}\;=\;\frac{\pi}{4\,\sqrt{\,a\,b^3\,}}.\,

Adding the two results then produces

\textbf I\;=\;\int_0^{\frac{\pi}{2}}\,\frac{1}{\left(a\,\cos^2\,x+b\,\sin^2\,x\right)^2}\;dx\;=\;\frac{\pi}{4\,\sqrt{\,a\,b\,}}\left(\frac{1}{a}+\frac{1}{b}\right),\,

which is the value of the integral \textbf I.\,

 

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Badiuddin

Last Activity: 16 Years ago
Kushagra Madhukar
Dear student,
Please find the answer to your problem below.
 
The Newton – Leibnitz’s law can be stated as follows
Example-
 
Thanks and regards,
Kushagra
Last Activity: 5 Years ago
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