Question

integrate |x - 1| dx from -2 to 6

Answer 1

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

Given integral is

$\red{\rm :\longmapsto\:\displaystyle\int_{-2}^6\tt |x - 1| \: dx}$

can be rewritten as

$\rm \: = \: \: \displaystyle\int_{-2}^1\tt |x - 1| \: dx \: + \: \displaystyle\int_{1}^6\tt |x - 1| \: dx$

We know,

By definition of Modulus function

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

So,

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

that is

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

So, using this, the above integral can be rewritten as

$\rm \: = \: \: \displaystyle\int_{-2}^1\tt (1 - x) \: dx \: + \: \displaystyle\int_{1}^6\tt (x - 1) \: dx$

We know,

$\boxed{ \rm{ \displaystyle\int\tt {x}^{n} dx = \frac{ {x}^{n + 1} }{n + 1} + c}}$

So, using this identity, we get

$\rm \: = \: \: \bigg(x - \dfrac{ {x}^{2} }{2} \bigg)_{-2}^1 + \bigg(\dfrac{ {x}^{2} }{2} - x\bigg)_{1}^6$

$\rm \: = \: \: \bigg(1 - \dfrac{1}{2} \bigg) - \bigg( - 2 - 2\bigg) +\bigg(18 - 6 \bigg) - \bigg( \dfrac{1}{2} - 1 \bigg)$

$\rm \: = \: \: \dfrac{1}{2} + 4 + 12 + \dfrac{1}{2}$

$\rm \: = \: \: 17$

Hence,

$\red{\rm :\longmapsto\:\displaystyle\int_{-2}^6\tt |x - 1| \: dx = 17}$

Additional Information :-

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

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

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

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

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

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

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