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

The functions are in same form but they are expressed in different variables. In such case, their definite integrals of a function in different variables over a closed interval are equal and this property is called the equality property of definite integrals.

Let $f(x)$ be a function in terms of $x$ and its definite integral with respect to $x$ over an interval $[a, b]$ is written in the following mathematical form.

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

Suppose, $f(y)$ be a function in same form but it is defined in terms of a variable $y$. The definite integral of the function $f(y)$ with respect to $y$ over same interval is written mathematically as follows.

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

The definite integral of $f(x)$ with respect to $x$ is equal to the definite integral of $f(y)$ with respect to $y$ over the interval $[a, b]$.

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

This mathematical property is called the equality property of definite integrals.

$\displaystyle \int_{1}^{2}{x^2\,}dx$ and $\displaystyle \int_{1}^{2}{t^2\,}dt$

Let us evaluate the definite integrals of both functions over a closed interval $[1, 2]$ to understand the equality property of the definite integrals.

$=\,\,\,$ $\bigg[\dfrac{x^{2+1}}{2+1}+c\bigg]_{1}^{2}$

$=\,\,\,$ $\bigg[\dfrac{x^3}{3}+c\bigg]_{1}^{2}$

$=\,\,\,$ $\bigg[\dfrac{2^3}{3}+c\bigg]$ $-$ $\bigg[\dfrac{1^3}{3}+c\bigg]$

$=\,\,\,$ $\bigg[\dfrac{8}{3}+c\bigg]$ $-$ $\bigg[\dfrac{1}{3}+c\bigg]$

$=\,\,\,$ $\dfrac{8}{3}$ $+$ $c$ $-$ $\dfrac{1}{3}$ $-$ $c$

$=\,\,\,$ $\dfrac{8}{3}$ $-$ $\dfrac{1}{3}$ $+$ $c$ $-$ $c$

$=\,\,\,$ $\dfrac{8-1}{3}$ $+$ $\cancel{c}$ $-$ $\cancel{c}$

$=\,\,\,$ $\dfrac{7}{3}$

$\displaystyle \int_{1}^{2}{t^2\,}dt$

$=\,\,\,$ $\bigg[\dfrac{t^{2+1}}{2+1}+c\bigg]_{1}^{2}$

$=\,\,\,$ $\bigg[\dfrac{t^3}{3}+c\bigg]_{1}^{2}$

$=\,\,\,$ $\bigg[\dfrac{2^3}{3}+c\bigg]$ $-$ $\bigg[\dfrac{1^3}{3}+c\bigg]$

$=\,\,\,$ $\bigg[\dfrac{8}{3}+c\bigg]$ $-$ $\bigg[\dfrac{1}{3}+c\bigg]$

$=\,\,\,$ $\dfrac{8}{3}$ $+$ $c$ $-$ $\dfrac{1}{3}$ $-$ $c$

$=\,\,\,$ $\dfrac{8}{3}$ $-$ $\dfrac{1}{3}$ $+$ $c$ $-$ $c$

$=\,\,\,$ $\dfrac{8-1}{3}$ $+$ $\cancel{c}$ $-$ $\cancel{c}$

$=\,\,\,$ $\dfrac{7}{3}$

$\therefore\,\,\,$ $\displaystyle \int_{1}^{2}{t^2\,}dt$ $\,=\,$ $\dfrac{7}{3}$ $\,=\,$ $\displaystyle \int_{1}^{2}{x^2\,}dx$

It can be understood that the definite integral of $x$ square with respect to $x$ is equal to the definite integral of square of $t$ with respect to $t$ over an interval $[1, 2]$.

Learn how to prove the equality property of definite integrals mathematically in calculus.

Latest Math Topics

Jul 24, 2024

Dec 13, 2023

Jul 20, 2023

Latest Math Problems

Jan 30, 2024

Oct 15, 2023

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved