Prove that 7/3=

In order to solve the problem, we'll have to determine the
extreme values of f(x), over the interval [0,2].

To
calculate the maximum or minimum values of f(x), we'll have to verify if the function is
increasing or decreasing over the given interval.

To verify
the monotony of the function, we'll calculate the first derivative of the
function.

f'(x) =
[(x^4+5)/(x^4+2)]'

Since the function is a ratio, we'll
apply the quotient rule:

(u/v)' = (u'v -
uv')/(v^2)

u = x^4+5 => u' =
4x^3

v = x^4+2 => v' =
4x^3

u'v = 4x^3(x^4+2)

uv' =
4x^3(x^4+5)

u'v - uv' = 4x^3(x^4+2) -
4x^3(x^4+5)

We'll factorize by
4x^3:

u'v - uv' =
4x^3(x^4+2-x^4-5)

We'll eliminate like
terms:

u'v - uv' =
-12x^3

f'(x)
=  -12x^3/(x^4+2)^2

Since the denominator is always
positive, f'(x)<0 for any value of x.

So, the
function is decreasing over [0,2].

0=<x=<2
=> f(2)=<f(x)=<f(0)

So, the function
has a maximum for x = 0 and a minimum for x = 2.

We'll
calculate f(0):

f(0)  =
(0^4+5)/(0^4+2)

f(0) =
5/2

f(2) = (16+5)/(16+2)

f(2)
= 21/18

f(2) = 7/6

We'll
express the
inequality:

 f(2)=<f(x)=<f(0)

7/6
=< f(x)=<5/2

Int (7/6)dx =< Int
f(x)dx=< Int (5/2)dx

7/6 =< Int
f(x)dx=< 5/2

We'll multiply by 2 the
inequality:

7/3 =< Int f(x)dx=<
5 q.e.d.