For proving the difference rule of integration, assume $f(x)$, $g(x)$, $p(x)$ and $q(x)$ are four functions in $x$, and $c_1$ and $c_2$ are constants. We can now start proving the difference rule of indefinite integration in the functions by using the mathematical relationship between integration and difference.

Now, add the constant $c_1$ to the function $p(x)$, and their sum is written as an expression $p(x)+c_1$. Assume, the function $f(x)$ is the derivative of the mathematical expression $p(x)+c_1$. It can be expressed in the following mathematical form by the differentiation.

$\dfrac{d}{dx}{\, \Big(p(x)+c_1\Big)} \,=\, f(x)$

As per the integral calculus, the function $p(x)$ is called as an anti-derivative or primitive of function $f(x)$ and $c_1$ is called the integral constant. It is also written mathematically in the form of indefinite integration.

$\implies$ $\displaystyle \int{f(x) \,}dx \,=\, p(x)+c_1$

$\,\,\, \therefore \,\,\,\,\,\,$ $p(x)+c_1 \,=\, \displaystyle \int{f(x) \,}dx$

In the same way, add the constant $c_2$ to the function $q(x)$. Now, the addition of them is expressed as $q(x)+c_2$. Let the derivative of expression $q(x)+c_2$ is represented by the function $g(x)$. It is written mathematically by the differential calculus.

$\dfrac{d}{dx}{\, \Big(q(x)+c_2\Big)} \,=\, g(x)$

According to the integral calculus, the function $q(x)$ is called the integral of $g(x)$ with respect to $x$, and $c_2$ is called the constant of integration. Mathematically, the relationship between them can be written by the integration.

$\implies$ $\displaystyle \int{g(x) \,}dx \,=\, q(x)+c_2$

$\,\,\, \therefore \,\,\,\,\,\,$ $q(x)+c_1 \,=\, \displaystyle \int{g(x) \,}dx$

Now, subtract the expression $q(x)+c_2$ from the expression $p(x)+c_1$ and then evaluate its derivative in mathematical form.

$\dfrac{d}{dx}{\, \Big[\Big(p(x)+c_1\Big)-\Big(q(x)+c_2\Big)\Big]}$

According to the difference rule of differentiation, the derivative of difference of the functions is equal to the difference of their derivatives.

$\implies$ $\dfrac{d}{dx}{\, \Big[\Big(p(x)+c_1\Big)-\Big(q(x)+c_2\Big)\Big]}$ $\,=\,$ $\dfrac{d}{dx}{\,\Big(p(x)+c_1\Big)}$ $-$ $\dfrac{d}{dx}{\, \Big(q(x)+c_2\Big)}$

It is taken that the derivatives of expressions $p(x)+c_1$ and $q(x)+c_2$ are $f(x)$ and $g(x)$ respectively in the first step.

$\implies$ $\dfrac{d}{dx}{\, \Big[\Big(p(x)+c_1\Big)-\Big(q(x)+c_2\Big)\Big]}$ $\,=\,$ $f(x)-g(x)$

Express the differential equation in integral form by using the mathematical relationship between them.

$\implies$ $\displaystyle \int{\Big(f(x)-g(x)\Big)\,}dx$ $\,=\,$ $\Big(p(x)+c_1\Big)$ $-$ $\Big(q(x)+c_2\Big)$

According to the results of the first step, we know that

$(1)\,\,\,$ $p(x)+c_1 \,=\, \displaystyle \int{f(x)\,}dx$

$(2)\,\,\,$ $q(x)+c_2 \,=\, \displaystyle \int{g(x)\,}dx$

Now, the expressions $p(x)+c_1$ and $q(x)+c_2$ can be replaced by their equivalent expressions.

$\,\,\,\therefore\,\,\,\,\,\,$ $\displaystyle \int{\Big(f(x)-g(x)\Big)\,}dx$ $\,=\,$ $\displaystyle \int{f(x) \,}dx$ $-$ $\displaystyle \int{g(x) \,}dx$

Therefore, it is proved successfully that the indefinite integral of difference of functions is equal to difference of their integrals, and it is called as the difference rule of integration.

Latest Math Topics

Dec 13, 2023

Jul 20, 2023

Jun 26, 2023

Latest Math Problems

Jan 30, 2024

Oct 15, 2023

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved