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• Trigonometry
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Proof of Sum rule of Integration

Let $f(x)$, $g(x)$, $u(x)$ and $v(x)$ represent four functions in $x$, and also $c_1$ and $c_2$ are constants. Now, let’s derive the formula for integral sum rule in mathematical form by the relation between differentiation and integration in integral calculus.

Relation between Derivative & Integrals of functions

Add a constant $c_1$ to the function $u(x)$, then the sum of them can be written as expression $u(x)+c_1$. Let $f(x)$ be the derivative of the mathematical expression $u(x)+c_1$. The relationship between them can be written in mathematical form by the differentiation.

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

According to the integral calculus, the function $u(x)$ is called a primitive or anti-derivative of function $f(x)$ and $c_1$ is called the constant of integration. It can be written in mathematical form by the indefinite integration.

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

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

Similarly, add a constant $c_2$ to the function $v(x)$. Now, the addition of them is written as an expression $v(x)+c_2$. Assume, the derivative of expression $v(x)+c_2$ is equal to $g(x)$. It can be written mathematically by the differential calculus.

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

As per integral calculus, the function $v(x)$ is called the integral of $g(x)$ with respect to $x$ and $c_2$ is called a constant of integration. The relationship between them can be written mathematically by the integration.

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

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

Relation between Derivative & Integrals of sum

Now, add both mathematical expressions and evaluate its derivative.

$\implies$ $\dfrac{d}{dx}{\, \Big[\Big(u(x)+c_1+v(x)+c_2\Big)\Big]}$ $\,=\,$ $\dfrac{d}{dx}{\, \Big[\Big(u(x)+c_1\Big)+\Big(v(x)+c_2\Big)\Big]}$

The derivative of sum is equal to sum of their derivatives as per sum rule of differentiation.

$\implies$ $\dfrac{d}{dx}{\, \Big[\Big(u(x)+c_1+v(x)+c_2\Big)\Big]}$ $\,=\,$ $\dfrac{d}{dx}{\,\Big(u(x)+c_1\Big)}$ $+$ $\dfrac{d}{dx}{\, \Big(v(x)+c_2\Big)}$

In the first step, we have assumed that the derivatives of the expressions $u(x)+c_1$ and $v(x)+c_2$ are $f(x)$ and $g(x)$ respectively.

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

As per the relationship between differentiation and integration, the mathematical expression in differential form can be written integral form.

$\implies$ $\displaystyle \int{\Big(f(x)+g(x)\Big)\,}dx$ $\,=\,$ $u(x)+c_1$ $+$ $v(x)+c_2$

Proving the Addition rule of Integration

In first step, we have derived that

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

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

Now, replace the values of expressions $u(x)+c_1$ and $v(x)+c_2$ in the above equation.

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

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