Question

The value of integral -1 to 1 (x-[x])dx is

Answer 1

$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

$(1). \: \boxed{ \blue{ \bf \: \int \:x \: dx \: = \: \dfrac{ {x}^{2} }{2} + c }}$

$\large\underline\purple{\bold{Solution :- }}$

$\rm :\implies\:Let \: I \: = \: \int_{ - 1}^1 \: (x - [x] \: )dx$

$\rm :\implies\:I \: = \: \int_{ - 1}^0 \: (x - [x] \: )dx + \int_{ 0}^1 \: (x - [x] \: )dx$

We know that

$\boxed{ \pink{\begin{gathered}\begin{gathered}\bf \:[x] \: = \begin{cases} &\sf{ - 1 \: \: \: \: \: \: \: when \: x \: \in \: [ - 1,0)} \\ &\sf{0 \: \: \: \: \: when \: x \: \in \: [0,1)} \end{cases}\end{gathered}\end{gathered}}}$

Therefore,

$\rm :\implies\:I \: = \int_{ - 1}^0 \: (x - ( - 1)\: )dx + \int_{ 0}^1 \: (x - 0 \: )dx$

$\rm :\implies\:I \: = \int_{ - 1}^0 \: (x + 1)\: )dx + \int_{ 0}^1 \: (x \: )dx$

$\rm :\implies\:I = \bigg[\dfrac{ {x}^{2} }{2} + x \bigg]_{ - 1}^0 + \bigg[\dfrac{ {x}^{2} }{2} \bigg]_0^1$

$\rm :\implies\:I = (0 + 0) - (\dfrac{1}{2} - 1) + \dfrac{1}{2} - 0$

$\rm :\implies\:I = \dfrac{1}{2} + \dfrac{1}{2}$

$\rm :\implies\:I = 1$

Hence,

$\large\boxed{ \blue{ \bf \: \: \int_{ - 1}^1 \: (x - [x] \: )dx = 1}}$