The problem (1):-
Since 1−x2>0 when −1<x<1 and then ∣∣1−x2∣∣=1−x2
But 1−x2<0 when 1<x or x<−1, thus ∣∣1−x2∣∣=x2−1
Thus
∫−22∣∣1−x2∣∣dx=∫−2−1∣∣1−x2∣∣dx+∫−11∣∣1−x2∣∣dx+∫12∣∣1−x2∣∣dx=∫−2−1(x2−1)dx+∫−11(1−x2)dx+∫12(x2−1)dx=(3x3−x)−2−1+(x−3x3)−11+(3x3−x)12=(3−1+1)−(3−8+2)+(3−1+1)=4
The problem (2):-
Since 2x+3>0 when x>2−3 , then ∣2x+3∣=2x+3
But 2x+3<0 when x<2−3 , thus ∣2x+3∣=−2x−3
Therefore we can find the value of the following integral
∫−22∣2x+3∣dx=∫−22−3∣2x+3∣dx+∫−32∣2x+3∣dx=∫−22−3(−2x−3)dx+∫2−32(2x+3)dx=(−x2−3x)−22−3+(x2+3x)2−32=(−49+29)−(−4+6)+(4+6)−(49−29)=225
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