what is the integral 2 to 5 of abs(x-3)dx . I know you have to split them up but not so sure how and what that does to the problem. also this problem... integral -1 to 2 of (x-2 absx)dx Thanks!!

Question

ganeshie8 · Answer

yes we need to split the integral such that the function never becomes negative :

$$
\large    
|x-3| = \left\{
  \begin{array}{lr}
    3-x  & : x < 3\
    x-3 & : x \ge 3
  \end{array}
\right.
$$

ganeshie8 · Answer

$$\large \int \limits_2^5 |x-3| dx =  \int \limits_2^3( 3-x) dx +  \int \limits_3^5 (x-3) dx  $$

ganeshie8 · Answer

see if that makes more or less sense...

ganeshie8 · Answer

we have used the definition of |x| function :

$$
\large    
|x| = \left\{
  \begin{array}{lr}
    x  & : x < 0\
    -x & : x \ge 0
  \end{array}
\right.
$$

anonymous · Answer

for the first one, where did you get integral 2 to 3 and 3 to 5 from? and for the second one.. what would the integral look like? $$\int\limits_{-1}^{0} (x-2x) + \int\limits_{0}^{2} (x-2x)dx$$ ?

aravindg · Answer

For first one he just split the integral where there is change in how function is defined. 

x-3=0 means  x=3

So 2 to 5 interval was split as sum of two integrals 2 to 3 then 3 to 5

agreene · Answer

The second one kinda depends on what it is, your notation is a bit ambiguous, is it:
$$\int_{-1}^2(x-2)|x|dx \space \space \text{or} \space \space \space \int_{-1}^2 x-2|x| dx$$

anonymous · Answer

sorry, it is $$\int\limits_{1}^{2} (x-2\left| x \right|)dx $$

agreene · Answer

in that case, you can do a lot of splitting:
$$\int_{-1}^2x-2|x|dx =\int_{-1}^2xdx-\int_{-1}^22|x|dx=\int_{-1}^2xdx-2\int_{-1}^2|x|dx$$
you will still have to fix the continuity issue for x=-1