$\displaystyle \int_{-2}^{2}|x+1|d(x)$In this case can I break the mod value and write like this $\displaystyle \int_{-2}^{2}xd(x)+\int_{-2}^{2}1d(x)$
Hello, srirahulan!
$\displaystyle \displaystyle\int_{\text{-}2}^{2}|x+1|\,dx$
$\displaystyle \text{Graph the function: }\,y \:=\:|x+1|\,\text{ on }[\text{-}2,2]$
$\displaystyle \displaystyle\text{We want: }\:\int^{\text{-}1}_{\text{-}2}(\text{-}x-1)\,dx\:+\:\int^2_{\text{-}1}(x+1)\,dx$Code:| | .* | .*.: | .*...: | .*.....: | .*.......: |.*.........: *. .*...........: :.*. .*.|...........: :...*. .*...|...........: ---+-----*-----+-----+-----+--- -2 -1 | 1 2 |
Indeed...using just "plain geometry" (ok, a lame pun), it is apparent that the integral:
$\displaystyle \int_{-2}^2 x+1\ dx = 4$, while:
$\displaystyle \int_{-2}^2 |x+1|\ dx = 5$.
The reason being, the triangle with base extending from (-2,0) to (-1,0) in the former integral has a "peak" at (-2,-1) and lies BELOW the x-axis (so has "negative area", that is the integral is negative in this region), while in the latter integral lies above the x-axis (with "peak" at (-2,1)). This triangular area has a base of 1, and a height of 1, and so has area 1/2, making a "swing" of a difference of 1 between the two integrals.
And yes, Halls, that WAS a typo. :P