# (&int;)^(1)f '(x)f ''(x) &DifferentialD;x; 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.

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