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.

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