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Sum rule of Integration


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

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


Let $f(x)$ and $g(x)$ represent two functions in $x$, then the indefinite integral of each function with respect to $x$ can be written in mathematical form as follows.

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

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

The sum of functions is written as $f(x)+g(x)$ in mathematics. The integral of sum of the functions with respect $x$ is written in the following mathematical form in integral calculus.

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

According to integral calculus, the integral of sum of two or more functions is equal to the sum of their integrals. The following equation expresses this integral property and it is called as the sum rule of integration.

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

The sum rule of indefinite integration can also be extended to sum of infinite number of functions.

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


Evaluate $\displaystyle \int{(3+4x)\,}dx$

Now, use the integral sum rule to evaluate the integral of sum of the functions.

$\implies$ $\displaystyle \int{(3+4x)\,}dx$ $\,=\,$ $\displaystyle \int{3\,}dx$ $+$ $\displaystyle \int{4x\,}dx$

$\implies$ $\displaystyle \int{(3+4x)\,}dx$ $\,=\,$ $3x+c_1$ $+$ $4\displaystyle \int{x\,}dx$

$\implies$ $\displaystyle \int{(3+4x)\,}dx$ $\,=\,$ $3x+c_1$ $+$ $4 \times \Big(\dfrac{x^2}{2}+c_2\Big)$

$\implies$ $\displaystyle \int{(3+4x)\,}dx$ $\,=\,$ $3x+c_1$ $+$ $4 \times \dfrac{x^2}{2}+4 \times c_2$

$\implies$ $\displaystyle \int{(3+4x)\,}dx$ $\,=\,$ $3x+c_1$ $+$ $\dfrac{4 \times x^2}{2}+4c_2$

$\implies$ $\displaystyle \int{(3+4x)\,}dx$ $\,=\,$ $3x+c_1$ $+$ $\require{cancel} \dfrac{\cancel{4} \times x^2}{\cancel{2}}+4c_2$

$\implies$ $\displaystyle \int{(3+4x)\,}dx$ $\,=\,$ $3x+c_1$ $+$ $2 \times x^2+4c_2$

$\implies$ $\displaystyle \int{(3+4x)\,}dx$ $\,=\,$ $3x+c_1$ $+$ $2x^2+4c_2$

$\implies$ $\displaystyle \int{(3+4x)\,}dx$ $\,=\,$ $3x+2x^2+c_1+4c_2$

$\implies$ $\displaystyle \int{(3+4x)\,}dx$ $\,=\,$ $3x+2x^2+c$


Learn how to prove the addition law of integration in integral calculus.

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