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

*print*Print*list*Cite

The area between the curve f(x) = (x^2-5x)/x^3, the x-axis and x=1 and x= 2 can be found by determining the definite integral of f(x) = (x^2-5x)/x^3 between x = 1 and x = 2

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

Int [ (x^2-5x)/x^3 dx]

=> Int [ 1/x - 5/x^2 dx]

=> ln x + 5/x + C

To find the definite integral we determine the value between f(2) and f(1).

f(2) = ln 2 + 5/2 + C

f(1) = ln 1 + 5 + C

f(2) - f(1) = ln 2 - 5/2

**The enclosed area is ln 2 - 5/2**

**1.81 square units**

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.**