integration問題

Solving y = 4-x and y = 3x, x=1 and y=3
Solving y = 4 - x and 3y = x, x=3 and y=1
Therefore, area bounded by the three straight lines
= ∫(3x - x/3) dx (from 0 to 1) + ∫(4 - x - x/3) dx (from 1 to 3)
= [3x2/2- x2/6]10 + [4x - 2x2/3]31
= 4


2008-01-11 23:43:58 補充：
The three graphs are shown as follows:http://i207.photobucket.com/albums/bb173/kkkpopup/int_graph1.pngFrom that, y=3x is above 3y=x for x lying between 0 and 1y=4-x is above 3y=x for x lying between 1 and 3