Find the integral of the function f(x) = (2x^2 -5)/x from x=1 to x= 2.

Given the curve f(x) =
(2x^2-5)/x

We need to find the integral of f(x) between 1
and 2.

Let F(x) = Int
f(x).

Then the definite integral is
:

A = F(2) - F(1)

Let us
determine F(x).

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

We will
simplify.

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

==> F(x) = Int 2x dx - 5*Int
dx/x

==> F(x) = 2x^2/2 - 5*ln x +
C

==> F(x) = x^2 - 5lnx +
C

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

==> F(2) = 4-5ln2 +
C

==> A = 4-5ln2 - 1 = 3-5ln2 = -0.47 (
approx.)

Then the definite integral is 3-5ln2
= -0.47.