# Apply the property of monothony of definite integral to proove that integral of f(x) = (x-1)(x-3)(x-5) if x= 2 to x = 3 is positive.

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f(x) = (x-1)(x-2)(x-3) is a continuous finction between 2 and 3

f(x) = - ve for all x for x < 1.

f(x) = postive for 1 < x< 3 as f (x) = (x-1)(x-3)(x-5) = (+ve)(-ve)(-ve),

f' (x) = {(x-1)(x-3)(x-5)}' =(x-1)(x-3)+(x-3)(x-5)+(x-5)(x-1) = {3x^2-(4+8+6)x+3+15+6} = 3x^2 -18x +24 = 3(x^2-6x+8) = 3(x-4)(x-2). f'(x) = 0 at x =2. and f"(x) = 3(x^2-6x+8) = 3(2x-6), At x=2, f"(2) = 2(2*2-6) = -4 is negative..

So f(2) i local maximum. And f(2) = 0.

f(x) <f (2) , f(x) > 0 in 2 < x <3 and f(3) = 0.

Therefore Integral f(x) = integral of a positive f(x) for 2 <x <3 must be positive.

To prove that the integral of f(x) is positive, we'll prove first that f(x) is positive over the interval [2,3].

We'll check this by substituting x by the extreme values of the interval.

For x = 2

f(2) = (2-1)(2-3)(2-5)

f(2) = 1*(-1)*(-3)

f(2) = 3 > 0

For x = 3

f(3) = (3-1)(3-3)(3-5)

f(3) = 0

According to the rule, if the function is positive over the given interval, that means that the definite integral of the function is also positive, over the given interval, [2,3].

We'll calculate the definite integral of f(x).

For this reason, we'll remove the brackets first:

f(x) = (x-1)(x-3)(x-5)

f(x) = (x^2 - 4x + 3)(x-5)

f(x) = x^3 - 5x^2 - 4x^2 + 20x + 3x - 15

We'll combine like terms:

f(x) = x^3 - 9x^2 + 23x - 15

Int f(x)dx = Int (x^3 - 9x^2 + 23x - 15)dx

Int f(x)dx = Int (x^3)dx - 9Int(x^2)dx + 23Intxdx - 15Int dx

Int f(x)dx = x^4/4 - 9x^3/3 + 23x^2/2 - 15x

Int f(x)dx = F(3) - F(2)

Int f(x)dx = (3^4/4 - 2^4/4) - 3(3^3 - 2^3) + (23/2)(3^2 - 2^2) - 15*(3-2)

Int f(x)dx = (81-16)/4 - 3(27-8) + 23*5/2 - 15

Int f(x)dx = 65/4 - 57 + 115/2 - 15

Int f(x)dx = (65+230)/4 - 72

Int f(x) dx = 295/4 - 72

Int f(x)dx = 73.75 - 72

Int f(x)dx = 1.75 > 0

**So, the definite integral of f(x) is strictly positive over the interval [2,3].**