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Integration doubt

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first draw function y=[3x]

                           it will y=0 when x=[0 ,1/3)

                                    y=1 when x=[1/3,2/3)

                                    y=2 when  x=[2/3,1)

                                    y=3 when x= [3,4/3)

we need to find algebric area till x=1

so area =0*(1/3)+1*(2/3-1/3)+2*(1-2/3)

=0+1/3+2/3=1
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