Mathematics 75 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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?

OpenStudy (anonymous):

ya! thanks!

OpenStudy (anonymous):

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

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