# Prove the inequality 1/3=<integral of (3x-2)/(x+2)=<1the limits of integration are 1 and 2.

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3(x-2)/(x+2) = 3(x+2)/(x+2)-8/(x+2) = 3-8/x+2

Therefore F(x) = Int(3x-2)/(x+2) dx = Int 3dx - Int 8dx/x+2 = 3x-ln(x+2).

=> Int 3(x-2)/(x+2) dx between 1 to 2 = F(2) - F(1) = 3(2-1) - 8ln(4/3) = 0.6985 appr which is between 1/3 and 1.

So **1/3 <= Integral of (3x-2)/(x+2) < = 1.**

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].**