Question

Find the value of integration|(1 - x)|.dx limit 0,4​

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle\int_{0}^4 \rm |1 - x| \: dx$

can be rewritten as

$\rm \:  =  \:\:\displaystyle\int_{0}^1 \rm |1 - x| \: dx + \displaystyle\int_{1}^4 \rm |1 - x| \: dx$

We know,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\: |x| = \begin{cases} &\sf{ - x \: \: if \: x < 0} \\ &\sf{x \: \: if \: x \geqslant 0 \: \: \: \: \: \: \: \: } \end{cases}\end{gathered}\end{gathered}$

So,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\: |1 - x| = \begin{cases} &\sf{ 1- x \: \: if \: x < 1} \\ &\sf{x - 1 \: \: if \: x \geqslant 1 \: \: \: \: \: \: \: \: } \end{cases}\end{gathered}\end{gathered}$

So, above integral can be rewritten as

$\rm \:  =  \:\:\displaystyle\int_{0}^1 \rm (1 - x) \: dx + \displaystyle\int_{1}^4 \rm (x - 1) \: dx$

$\rm \:  =  \:  \: \bigg(x - \dfrac{ {x}^{2} }{2} \bigg)_{0}^1 + \bigg(\dfrac{ {x}^{2} }{2} - x\bigg)_{1}^4$

$\rm \:  =  \:  \: \bigg(1 - \dfrac{1}{2} \bigg) + \bigg(\dfrac{16}{2} - 4 \bigg) - \bigg(\dfrac{1}{2} - 1 \bigg)$

$\rm \:  =  \:  \: \bigg(\dfrac{1}{2} \bigg) + \bigg(8- 4 \bigg) - \bigg( - \dfrac{1}{2} \bigg)$

$\rm \:  =  \:  \: \bigg(\dfrac{1}{2} \bigg) + \bigg(4\bigg) + \bigg( \dfrac{1}{2} \bigg)$

$\rm \:  =  \:  \: 5$

Additional Information :-

$\underbrace{\boxed{ \tt{\displaystyle\int_{a}^b \rm f(x)dx \: = \: \displaystyle\int_{a}^b \rm f(y)dy }}}$

$\underbrace{\boxed{ \tt{\displaystyle\int_{a}^b \rm f(x)dx \: = - \: \displaystyle\int_{b}^a \rm f(x)dx }}}$

$\underbrace{\boxed{ \tt{\displaystyle\int_{a}^b \rm f(x)dx \: = \: \displaystyle\int_{a}^b \rm f(a + b - x)dx}}}$

$\underbrace{\boxed{ \tt{\displaystyle\int_{0}^a \rm f(x)dx \: = \: \displaystyle\int_{0}^a \rm f(a - x)dx}}}$

$\underbrace{\boxed{ \tt{\displaystyle\int_{ - b}^b \rm f(x)dx = 2\displaystyle\int_{0}^b \rm f(x)dx \: if \: f( - x) = f(x) }}}$

$\underbrace{\boxed{ \tt{\displaystyle\int_{ - b}^b \rm f(x) dx \: = 0 \: if \: f( - x) = f(x) }}}$

$\underbrace{\boxed{ \tt{\displaystyle\int_{0}^{2a} \rm f(x)dx = 2\displaystyle\int_{0}^a \rm f(x)dx \: \: if \: f(2a - x) = f(x) }}}$

$\underbrace{\boxed{ \tt{\displaystyle\int_{0}^{2a} \rm f(x)dx = 0 \: \: if \: f(2a - x) = - \: f(x) }}}$