Q.

The value of 12∫03|x2-3x+2|dx is ________ .            [2023]


Ans.

(22)

Let I=12∫03|x2-3x+2|dx

=12∫03|(x-2)(x-1)|dx

=12×[∫01(x2-3x+2)dx-∫12(x2-3x+2)dx+∫23(x2-3x+2)dx]

=12×[(x33-3x22+2x)01+(x33-3x22+2x)23-(x33-3x22+2x)12]=22