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.

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].