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# Properties of Definite Integrals

In this section, we study some fundamental properties of definite integrals which are very useful in evaluating integrals

#### Property I $\displaystyle \int _{ a }^{ b }{ f\left( x \right) } dx=\int _{ a }^{ b }{ f\left( t \right) } dt$

i.e. integration is independent of change of variable.

Proof   Let $\displaystyle \phi \left( x \right)$ be a primitive of f(x) .Then $\displaystyle \frac { d }{ dx } \left\{ \phi \left( x \right) \right\} =f\left( x \right) \\ \Rightarrow \frac { d }{ dx } \left\{ \phi \left( t \right) \right\} =f\left( t \right)$ $\displaystyle Hence\int _{ a }^{ b }{ f\left( x \right) } dx={ \left[ \phi \left( x \right) \right] }_{ a }^{ b }=\phi \left( b \right) -\phi \left( a \right) \qquad .....\left( i \right) \\ and\quad \int _{ a }^{ b }{ f\left( t \right) } dt={ \left[ \phi \left( t \right) \right] }_{ a }^{ b }=\phi \left( b \right) -\phi \left( a \right) \qquad .....\left( ii \right)$

From (i) and (ii), we have $\displaystyle \int _{ a }^{ b }{ f\left( x \right) } dx=\int _{ a }^{ b }{ f\left( t \right) } dt$

#### Property II $\displaystyle \int _{ a }^{ b }{ f\left( x \right) } dx=-\int _{ a }^{ b }{ f\left( x \right) } dx$

i.e., if the limits of a definite integral are interchanged then its value changes by minus sign only.

Proof   Let $\displaystyle \phi \left( x \right)$ be a primitive of f(x) .Then $\displaystyle \int _{ a }^{ b }{ f\left( x \right) } dx=\phi \left( b \right) -\phi \left( a \right) \\ and\quad -\int _{ b }^{ a }{ f\left( x \right) } =-\left[ \phi \left( b \right) -\phi \left( a \right) \right] \\ =\phi \left( b \right) -\phi \left( a \right) \\ \therefore \int _{ a }^{ b }{ f\left( x \right) } dx=-\int _{ b }^{ a }{ f\left( x \right) } dx$

#### Property III $\displaystyle \int _{ a }^{ b }{ f\left( x \right) } dx=\int _{ a }^{ c }{ f\left( x \right) } dx+\int _{ c }^{ b }{ f\left( x \right) } dx\quad where\quad a

Proof   Let $\displaystyle \phi \left( x \right)$ be a primitive of f(x) .Then $\displaystyle \int _{ a }^{ b }{ f\left( x \right) } dx=\phi \left( b \right) -\phi \left( a \right) ....\left( i \right) \\ and\quad \int _{ a }^{ c }{ f\left( x \right) } dx+\int _{ c }^{ b }{ f\left( x \right) } dx=\left[ \phi \left( c \right) -\phi \left( a \right) \right] +\left[ \phi \left( b \right) -\phi \left( c \right) \right] =\phi \left( b \right) -\phi \left( a \right) ....\left( ii \right)$

From (i) and (ii), we have $\displaystyle \int _{ a }^{ b }{ f\left( x \right) } dx=\int _{ a }^{ c }{ f\left( x \right) } dx+\int _{ c }^{ b }{ f\left( x \right) } dx$

#### Generalization

The above property can be generalized into following form $\displaystyle \int _{ a }^{ b }{ f\left( x \right) } dx=\int _{ a }^{ { c }_{ 1 } }{ f\left( x \right) } dx+\int _{ a }^{ { c }_{ 2 } }{ f\left( x \right) } dx+......\int _{ c }^{ b }{ f\left( x \right) } dx\\ where\quad a<{ c }_{ 1 }<{ c }_{ 2 }<{ c }_{ 3 }....{ c }_{ n-1 }<{ c }_{ n }

#### Property IV $\displaystyle \int _{ 0 }^{ a }{ f\left( x \right) } dx=\int _{ 0 }^{ a }{ f\left( a-x \right) } dx$

Proof   On RHS put a-x=t, so that dx=-dt. Also when x=0, t=a and when x=a , t=0 $\displaystyle \therefore \int _{ 0 }^{ a }{ f\left( a-x \right) } dx=\int _{ a }^{ 0 }{ f\left( t \right) } \left( -dt \right) =\int _{ a }^{ 0 }{ f\left( t \right) } dt=\int _{ 0 }^{ a }{ f\left( t \right) } dt\qquad \left[ by\quad prop(ii) \right] \\ =\int _{ 0 }^{ a }{ f\left( x \right) } \qquad \left[ by\quad prop(i) \right]$

NOTE  This property is useful to calculate a definite integral without first finding corresponding indefinite integrals which may be difficult or sometime even possible to find.

#### Property V $\displaystyle \int _{ -a }^{ a }{ f\left( x \right) } dx\begin{matrix} =2\int _{ 0 }^{ a }{ f\left( x \right) } dx,\quad if\quad f\left( x \right) is\quad an\quad even\quad function \\ =0,\quad if\quad f\left( x \right) is\quad an\quad odd\quad function \end{matrix}$

Proof   From property III, we get $\displaystyle \int _{ -a }^{ a }{ f\left( x \right) } dx=\int _{ -a }^{ 0 }{ f\left( x \right) } dx+\int _{ 0 }^{ a }{ f\left( x \right) } dx\qquad ......\left( i \right)$ $\displaystyle Now,\quad \int _{ -a }^{ 0 }{ f\left( x \right) } dx=\int _{ a }^{ 0 }{ f\left( -t \right) } \left( -dt \right) \quad where\quad t=-x\\ =-\int _{ a }^{ 0 }{ f\left( -t \right) } \left( -dt \right) =\int _{ 0 }^{ a }{ f\left( -t \right) } \left( -dt \right) \qquad \left[ by\quad Prop.II \right] \\ =\int _{ 0 }^{ a }{ f\left( -x \right) } dx\qquad \left[ by\quad Prop.I \right]$ $\displaystyle =\begin{matrix} \int _{ 0 }^{ a }{ f\left( x \right) } dx,\quad if\quad f\left( x \right) \quad is\quad an\quad even\quad function \\ -\int _{ 0 }^{ a }{ f\left( x \right) } dx,\quad if\quad f\left( x \right) \quad is\quad an\quad odd\quad function \end{matrix}\qquad ......\left( ii \right)$

Thus when f(x) is an even function $\displaystyle \left( i \right) \quad and\quad \left( ii \right) \Rightarrow \int _{ -a }^{ a }{ f\left( x \right) } dx=\int _{ 0 }^{ a }{ f\left( x \right) } dx+\int _{ 0 }^{ a }{ f\left( x \right) } dx\\ =2\int _{ 0 }^{ a }{ f\left( x \right) } dx$

and, when f(x) is an odd function $\displaystyle \left( i \right) \quad and\quad \left( ii \right) \Rightarrow \int _{ -a }^{ a }{ f\left( x \right) } dx=-\int _{ 0 }^{ a }{ f\left( x \right) } dx+\int _{ 0 }^{ a }{ f\left( x \right) } dx \\=0$ #### Property VI $\displaystyle \int _{ 0 }^{ 2a }{ f\left( x \right) } dx\begin{matrix} =2\int _{ 0 }^{ a }{ f\left( x \right) } dx,\quad if\quad f\left( 2a-x \right) =f\left( x \right) \\ =0,\quad if\quad f\left( 2a-x \right) =-f\left( x \right) \end{matrix}$

Proof   From property III, we get $\displaystyle \int _{ 0 }^{ 2a }{ f\left( x \right) } dx=\int _{ 0 }^{ a }{ f\left( x \right) } dx+\int _{ 0 }^{ 2a }{ f\left( x \right) } dx\qquad ......\left( i \right)$

Consider the integral $\displaystyle \int _{ a }^{ 2a }{ f\left( x \right) } dx$ . Putting x= 2 a -t so, that d x=-d t . Also, when x=a,  t=a and when x=2 a , t=0 $\displaystyle \therefore \int _{ 0 }^{ 2a }{ f\left( x \right) } dx\\ =\int _{ a }^{ 0 }{ f\left( 2a-t \right) } \left( -dt \right) \\ =-\int _{ a }^{ 0 }{ f\left( 2a-t \right) } dt\\ =\int _{ 0 }^{ a }{ f\left( 2a-t \right) } dt\qquad \left[ by\quad Prop.II \right] \\ =\int _{ 0 }^{ a }{ f\left( 2a-x \right) } dx\qquad \left[ by\quad Prop.I \right]$ $\displaystyle =\begin{matrix} \int _{ 0 }^{ a }{ f\left( x \right) } dx,\quad if\quad f\left( 2a-x \right) =f\left( x \right) \\ -\int _{ 0 }^{ a }{ f\left( x \right) } dx,\quad if\quad f\left( 2a-x \right) =-f\left( x \right) \end{matrix}\qquad ......\left( ii \right)$

Thus, when f(2a-x ) =f(x) $\displaystyle \left( i \right) \quad and\quad \left( ii \right) \Rightarrow \int _{ 0 }^{ 2a }{ f\left( x \right) } dx\\ =\int _{ 0 }^{ a }{ f\left( x \right) } dx+\int _{ 0 }^{ a }{ f\left( x \right) } dx\\ =2\int _{ 0 }^{ a }{ f\left( x \right) } dx$

and when f(2a -x) = -f(x) when $\displaystyle\left( i \right) \quad and\quad \left( ii \right) \Rightarrow \int _{ 0 }^{ 2a }{ f\left( x \right) } dx\\ =\int _{ 0 }^{ a }{ f\left( x \right) } dx-\int _{ 0 }^{ a }{ f\left( x \right) } dx\\ =0$ If f(2 a-x) = f(x) then the graph of f(x) is symmetrical about x=a so, $\displaystyle \int _{ 0 }^{ a }{ f\left( x \right) } dx=\int _{ a }^{ 2a }{ f\left( x \right) } dx$

In case $\displaystyle f\left( 2a-x \right) =-f\left( x \right)$

then graph of f(x) is shown in figure $\displaystyle \int _{ 0 }^{ a }{ f\left( x \right) } dx=\int _{ a }^{ 2a }{ f\left( x \right) } dx$

#### Property VII $\displaystyle \int _{ a }^{ b }{ f\left( x \right) } dx=\int _{ a }^{ b }{ f\left( a+b-x \right) } dx$

Proof

Putting x=a+b-t, dx=-dt, when x=a,t=b, when x=b, t=a we get $\displaystyle \int _{ a }^{ b }{ f\left( x \right) } dx\\ =\int _{ a }^{ b }{ f\left( a+b-t \right) } \left( -dt \right) \\ =-\int _{ a }^{ b }{ f\left( a+b-t \right) } dt\\ =\int _{ a }^{ b }{ f\left( a+b-t \right) } dt\qquad \left[ by\quad Prop.II \right] \\ =\int _{ a }^{ b }{ f\left( a+b-x \right) } dx\qquad \left[ by\quad Prop.I \right]$

April 18, 2019
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