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

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

Latest Math Problems

A best free mathematics education website for students, teachers and researchers.

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Learn how to solve the maths problems in different methods with understandable steps.

Copyright © 2012 - 2021 Math Doubts, All Rights Reserved