(∫)[0]^(1)f '(x)f ''(x) ⅆx; if f '(0)=3 and f '(1)=2

You need to use the formula of integration by parts, such that:

`int u dv = uv - int vdu`

Considering `u = f'(x)`   and `dv = f''(x)` yields:

`u = f'(x) => du = f''(x) dx `

`dv = f''(x) dx => v = f'(x)`

Using the formula yields:

`int_0^1 f'(x) f''(x) dx = f'(x)f'(x)|_0^1 - int_0^1 f'(x) f''(x) dx`

You should come up with the following notation for `int_0^1 f'(x) f''(x) dx` such that:

`int_0^1 f'(x) f''(x) dx = I`

`I = f'(x)f'(x)|_0^1 - I`

You need to move the terms that contain `I` to the left side, such that:

`I + I = f'(x)f'(x)|_0^1 => 2I = f'(x)f'(x)|_0^1 => I = (f'(x)f'(x)|_0^1)/2`

You need to use the fundamental theorem of calculus, such that:

`I = (1/2)(f'(1)f'(1) - f'(0)f'(0))`

Since the problem provides the values `f'(1) = 2, f'(0) = 3` , yields:

`I = (1/2)(2*2 - 3*3) => I = -5/2`

Hence, evaluating the given definite integral, using the formula of integration by parts and the fundamental theorem of calculus, yields `I = -5/2.`