Evaluate the following Integral:

Let

Let us plot the given interval,

Interval (0, 5) can now be divided into sub-intervals: (1, 2), (2, 3), (3, 4) and (4, 5)

In (1, 2),

y = |x – 1| = (x+1)

y = |x – 2| = -(x – 2)

y = |x – 3| = -(x – 3)

y = |x – 4| = -(x – 4)

In(2, 3),

y = |x – 1| = (x – 1)

y = |x – 2| = (x – 2)

y = |x – 3| = -(x – 3)

y = |x – 4| = -(x – 4)

In (3, 4),

y = |x – 1| = (x – 1)

y = |x – 2| = (x – 2)

y = |x – 3| = (x – 3)

y = |x – 4| = -(x – 4)

In (4, 5)

y = |x – 1| = (x – 1)

y = |x – 2| = (x – 2)

y = |x – 3| = (x – 3)

y = |x – 4| = (x – 4)

Now,

I = 18 square units.