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= [ 2/1+1 . x pangkat 1+1 ]
= [x pangkat 2]
=[ (2) pangkat 2] -[0 pangkat 0]
=4 satuan luas
C. integral batas bawah 0 batas atas 2 x kuadrat dx
= [1/2+1 . x pangkat 2+1]
=1/3x pangkat 3
=[1/3 (2)pangkat 3]-[1/3(2) pangkat 0]
=8/3 satuan luas.