Proof of Equality property of Definite integrals

According to the equality property of definite integrals, the definite integrals of a function in different variables over a closed interval are equal.

$\displaystyle \int_{a}^{b}{f(x)\,}dx$ $\,=\,$ $\displaystyle \int_{a}^{b}{f(y)\,}dy$

Now, let’s learn how to prove the equality property of definite integrals mathematically in calculus.

Find the definite integral of a function

Let $f(x)$ be a function in terms of $x$ and assume that its indefinite integral is $g(x)$ plus constant of integration $c$. So, the definite integral of the function $f(x)$ with respect to $x$ over the interval $[a, b]$ is written in the following mathematical form in calculus.

$\displaystyle \int_{a}^{b}{f(x)\,}dx$ $\,=\,$ $\big[g(x)+c\big]_{a}^{b}$

Now, calculate the definite integral of the function $f(x)$ with respect to $x$ over the interval of $[a, b]$.

$=\,\,\,$ $\big[g(b)+c\big]$ $-$ $\big[g(a)+c\big]$

$=\,\,\,$ $g(b)$ $+$ $c$ $-$ $g(a)$ $-$ $c$

Perform the subtraction of the expressions to find their difference mathematically.

$=\,\,\,$ $g(b)$ $-$ $g(a)$ $+$ $c$ $-$ $c$

$=\,\,\,$ $g(b)$ $-$ $g(a)$ $+$ $\cancel{c}$ $-$ $\cancel{c}$

$=\,\,\,$ $g(b)$ $-$ $g(a)$

$\therefore\,\,\,$ $\displaystyle \int_{a}^{b}{f(x)\,}dx$ $\,=\,$ $g(b)$ $-$ $g(a)$

Evaluate the definite integral of function

Let $f(y)$ be a function in terms of $y$ and it is same as the function $f(x)$. In other words, the expressions of the function $f(x)$ and $f(y)$ are same but they are defined in two variables $x$ and $y$. Hence, the definite integral of the function will be $g(y)$ plus integral constant. Due to similar formation of both functions, the integral constant for the function $f(y)$ is also same.

Therefore, the definite integral of $f(y)$ with respect to $y$ over the interval $[a, b]$ is expressed mathematically as follows.

$\displaystyle \int_{a}^{b}{f(y)\,}dy$ $\,=\,$ $\Big[g(y)+c\Big]_{a}^{b}$

Now, find the definite integral of the function $f(y)$ with respect to $y$ over the interval of $[a, b]$.

$=\,\,\,$ $\Big[g(b)+c\Big]$ $-$ $\Big[g(a)+c\Big]$

It is time to find the subtraction of the expressions to get the difference between them in mathematical form.

$=\,\,\,$ $g(b)$ $+$ $c$ $-$ $g(a)$ $-$ $c$

$=\,\,\,$ $g(b)$ $-$ $g(a)$ $+$ $c$ $-$ $c$

$=\,\,\,$ $g(b)$ $-$ $g(a)$ $+$ $\cancel{c}$ $-$ $\cancel{c}$

$=\,\,\,$ $g(b)$ $-$ $g(a)$

$\therefore\,\,\,$ $\displaystyle \int_{a}^{b}{f(y)\,}dy$ $\,=\,$ $g(b)$ $-$ $g(a)$

Compare the definite integrals of functions

Now, compare the definite integrals of both functions to understand the relation between them.

$(1).\,\,\,$ $\displaystyle \int_{a}^{b}{f(x)\,}dx$ $\,=\,$ $g(b)$ $-$ $g(a)$

$(2).\,\,\,$ $\displaystyle \int_{a}^{b}{f(y)\,}dy$ $\,=\,$ $g(b)$ $-$ $g(a)$

It can be easily understood by comparison that the definite integrals of both functions are same.

$\therefore\,\,\,$ $\displaystyle \int_{a}^{b}{f(x)\,}dx$ $\,=\,$ $\displaystyle \int_{a}^{b}{f(y)\,}dy$