∫1−1x|x|dx= [MP PET 1990; Pb. CET 2004]
Let f(x)=x|x|. Then f(−x)=−x|−x|=−x|x|=−f(x) Therefore ∫1−1x|x|dx=0 (By the property of definite integral).
∫10e2 In xdx= [MP PET 1990]