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

The integral of difference of functions equals to difference of their integrals is called the difference rule of integration.

When $f(x)$ and $g(x)$ represent two functions in $x$, the indefinite integral of every function with respect to $x$ is written in the following mathematical form.

$(1)\,\,\,$ $\displaystyle \int{f(x)\,}dx$

$(2)\,\,\,$ $\displaystyle \int{g(x)\,}dx$

The difference of two functions is written as $f(x)-g(x)$ in mathematical form. The indefinite integral of difference of the functions with respect $x$ is written as follows.

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

As per integral calculus, the integral of difference of any two functions is equal to the difference of their integrals. The property can be expressed as equation in mathematical form and it is called as the difference rule of integration.

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

Evaluate $\displaystyle \int{(1-2x)\,}dx$

Now, use the integral difference rule for evaluating the integration of difference of the functions.

$=\,\,\,$ $\displaystyle \int{1\,}dx-\int{2x\,}dx$

$=\,\,\,$ $\displaystyle \int{1\,}dx-2\int{x\,}dx$

$=\,\,\,$ $x+c_1-2\Big(\dfrac{x^2}{2}+c_2\Big)$

$=\,\,\,$ $x+c_1-2 \times \dfrac{x^2}{2}-2 \times c_2$

$=\,\,\,$ $x+c_1-\dfrac{2x^2}{2}-2c_2$

$=\,\,\,$ $\require{cancel} x+c_1-\dfrac{\cancel{2}x^2}{\cancel{2}}-2c_2$

$=\,\,\,$ $x+c_1-x^2-2c_2$

$=\,\,\,$ $x-x^2+c_1-2c_2$

$=\,\,\,$ $x-x^2+c$

Learn how to derive the difference rule of indefinite integration in integral calculus.

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