# Prove that 7/3=< definite integral of f(x)=<5 when x=0 to x=2. f(x)=(x^4+5)/(x^4+2)

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### 2 Answers

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.**

To prove that 7/3 < = I = <5, Where I = Integral of f(x) = (x^4+5)/(x^4+2). from x= 0 to x = 2.

Soution:

We know that f(x) = (x^4+5)/(x^4+2) has the highest value 5/2 when x = 0 and least value (2^4+5)/(2^4+2) = 21/18 = 7/6 .when x = 2.

Therefore, (7/6 < = (x^4+5)/(x^4+2) < (5/2) , for all x in the interval (0,2).

So let Integral (x^4+5)/ (x^4+2) dx from 0 to 2 = I.

Then in the interval (0 , 2).

(7/6) < (x^4+5)/(x^2+2) < 5/2.

Therefore ,

(7/6) INtegral dx < = I = < (5/2) Integral dx , from x= 0 to x = 3.

(7/6) {(x atx =2 - ) - (x at x = 0) < I < (5/2)* {(x at x=2)-(x at x= 0)}

(7/6)*2 <= I = < (5/2)*2

7/3 <= I = < (5/2).