how do you deal with an absolute value inside an integral? like in this case: ∫ ⎮(x^2) -4x+3⎮ and the upper bound is 4, lower bound is 0

Question

anonymous · Answer

because bounds are >=0: [0,4] it did seems that absolute value means much in this case...

anonymous · Answer

$$=\int\limits_{ }^{}(x ^{2}-4x+3) dx=x ^{3}/3 - 2x ^{2}+3x +cons$$

anonymous · Answer

i think I'm wrong... let me check something

anonymous · Answer

yep!
expression inside the integral could be re-written as:
(x-3)(x-1)

anonymous · Answer

this expression >=0 if x=[3,4] & x=[0,1]
right?

anonymous · Answer

and it's <=0 (negative) if x=[1,3]

anonymous · Answer

then your integral should be broken in following parts:
$$=\int\limits_{0}^{1}(x ^{2}-4x+3)dx -\int\limits_{1}^{3}(x ^{2}-4x+3)dx +\int\limits_{3}^{4}(x ^{2}-4x+3)dx=...$$
can you finish it now?

anonymous · Answer

ya! thanks!

anonymous · Answer

i hope it's right... I think :)