1. ## Integration

$\displaystyle \int_{-2}^{2}|x+1|d(x)$In this case can I break the mod value and write like this $\displaystyle \int_{-2}^{2}xd(x)+\int_{-2}^{2}1d(x)$

2. ## Re: Integration

No. On the interval [-2,1) |x+1| = -x-1, while on the interval [-1,2], |x+1| = x+1. So what you WANT is:

$\displaystyle -\int_{-2}^{-1}x\ dx - \int_{-2}^{-1}1\ dx + \int_{-1}^2 x\ dx + \int_{-1}^2 1\ dx$

3. ## Re: Integration

Originally Posted by Deveno
No. On the interval [-2,1) |x+1| = -x-1,
TYPO! Deveno meant "On the interval [-2, -1].

while on the interval [-1,2], |x+1| = x+1. So what you WANT is:

$\displaystyle -\int_{-2}^{-1}x\ dx - \int_{-2}^{-1}1\ dx + \int_{-1}^2 x\ dx + \int_{-1}^2 1\ dx$

4. ## Re: Integration

Hello, srirahulan!

$\displaystyle \displaystyle\int_{\text{-}2}^{2}|x+1|\,dx$

$\displaystyle \text{Graph the function: }\,y \:=\:|x+1|\,\text{ on }[\text{-}2,2]$

Code:
                  |
|          .*
|        .*.:
|      .*...:
|    .*.....:
|  .*.......:
|.*.........:
*.         .*...........:
:.*.     .*.|...........:
:...*. .*...|...........:
---+-----*-----+-----+-----+---
-2    -1     |     1     2
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$\displaystyle \displaystyle\text{We want: }\:\int^{\text{-}1}_{\text{-}2}(\text{-}x-1)\,dx\:+\:\int^2_{\text{-}1}(x+1)\,dx$

5. ## Re: Integration

Notice that, per Soroban's graph, this can be done with any "integration". The answer is the sum of the areas of two triangles.

6. ## Re: Integration

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