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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
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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?
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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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