Math Doubts

Equality property of Definite integrals

Formula

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

Introduction

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.

Example

$\displaystyle \int_{1}^{2}{x^2\,}dx$ 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 $[1, 2]$.

Proof

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

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