We have,
#|x-1|=-(x-1)=1-x,if (x-1)<=0,i.e.,if x<=1............(1)#
, &, #|x-1=x-1, if (x-1)>=0, i.e., if x>1..........(2)#.
So, we split #[0,5]# as #[0,1]uu[1,5]#. Therefore,
#I=int_0^5|x-1|dx=int_0^1|x-1|dx+int_1^5|x-1|dx=I_1+I_2#,
say, where, #I_1=int_0^1|x-1|dx, and, I_2=int_1^5|x-1|dx#.
Now, in #I_1, 0<=x<=1, &, so, |x-1|=1-x, by (1)#.
#:. I_1=int_0^1(1-x)dx=[x-x^2/2]_0^1=(1-1/2)-0=1/2#.
Next, in #I_2, 1<=x<=5, so that, |x-1|=x-1, by (2)#.
#:. I_2=int_1^5(x-1)dx=[x^2/2-x]_1^5=(25/2-5)-(1/2-1)=15/2+1/2=8#
Finally, we have, #I=I_1+I_2=1/2+8=17/2#.
Eventually, #I# can also be visualised as the Area bounded by
#y=|x-1|, X#-axis, #x=0, &, x=5# [refer to the graph ].
This Area consists of 2 right-triangles# : (Delta_1)# having sides of
lengths #1,1,sqrt2#; & hence Area#=1/2*1*1=1/2#, &, # : (Delta_2)#
with sides #4,4,4sqrt2# and Area#=1/2*4*4=8#
Thus, total Area#=1/2+8=17/2#, as before!
Enjoy Maths.!
