Caluculate the area bounded between the curve f(x) = (x^2-5x)/x^3 and x=1 and x= 2

f(x)- (x^2-5x)/x^3

Let us
simplify.

==> f(x)= x^2/x^3 -
5x/x^3

==> f(x)= 1/x
-5/x^2

==> f(x)= 1/x -
5x^-2

==> Now we will find the
integral.

==> Int f(x)= Int 1/x - 5x^-2
dx

==> Int f(x) = ln x - 5x^-1/-1  +
C

==> Int f(x) = lnx +
5/x

Now we will find Int
f(2)

==> F(2) = ln2 + 5/2 = 2.5 + ln
2

==> F(1) = ln1 + 5 =
5

Then the bounded area is F(2) -F(1) = 2.5 +ln2 - 5 =
-1.81

==> Then the area is 1.81 square
units.