Prove the inequality 1/3=

We'll prove the inequality, using the following
property:

If the function could be integrated over the range [a;b]
and a < f(x) < b, then Int adx < Int f(x)dx < Int
bdx.

First, we need to determine the monotony of the function over
the range [1 ; 2].

We'll calculate the 1st derivative of f(x) using
quotient rule.

f'(x) = [(3x-2)'*(x+2)
-(3x-2)*(x+2)']/(x+2)^2

f'(x) = [3(x+2) - 3x +
2]/(x+2)^2

f'(x) = (3x + 6 - 3x +
2)/(x+2)^2

We'll eliminate like
termS:

f'(x) = (8)/(x+2)^2

We notice
that f'(x) is positive over the range [1 ; 2], then the function is increasing over the range [1
; 2].

That means that f(1) < f(x) < f(2)
(1)

We'll calculate f(1) = 1/3

We'll
calculate f(2) = 1

We'll re-write the inequality
(1):

1/3 < f(x) < 1

We'll
integrate:

(1/3)*Int dx < Int f(x)dx < Int
dx

We'll apply
Leibniz-Newton:

(1/3)*(2-1)< Int f(x)dx <
(2-1)

(1/3)*< Int f(x)dx <
1

We notice that the given inequality (1/3)*<
Int f(x)dx < 1 is verified, over the range [1 ; 2].