Problem Analysis

We are given a thrice differentiable function $F(x)$, some values of the function, and two definite integrals involving its derivatives. We are asked to verify four statements related to the function $f(x) = xF(x)$. The solution will primarily rely on the technique of integration by parts.

First, let's list the given information:

1. $F(1) = 0$
2. $F(3) = -4$
3. $F(x) < 0$ for all $x \in (1/2, 3)$. Note: This condition contradicts $F(1)=0$, as $1 \in (1/2, 3)$. This suggests a typo in the problem statement, perhaps the interval should be $(1/2, 1) \cup (1, 3)$. However, the calculations for the options do not depend on this condition, so we can proceed using the other given data.
4. $f(x) = xF(x)$
5. $\int_1^3 x^2 F'(x) dx = -12$
6. $\int_1^3 x^3 F''(x) dx = 40$

We will evaluate the expressions in the options step-by-step.

Step 1: Evaluate $\int_1^3 f(x) dx$

This addresses options B and D. We need to compute the value of $\int_1^3 f(x) dx = \int_1^3 xF(x) dx$. We can use the given integral $\int_1^3 x^2 F'(x) dx = -12$ and apply integration by parts.

Let $I_1 = \int_1^3 x^2 F'(x) dx$. Using integration by parts formula $\int u \, dv = uv - \int v \, du$: Let $u = x^2$ and $dv = F'(x)dx$. Then $du = 2x \, dx$ and $v = F(x)$.

$I_1 = [x^2 F(x)]_1^3 - \int_1^3 F(x) (2x) dx$ $-12 = (3^2 F(3) - 1^2 F(1)) - 2 \int_1^3 x F(x) dx$

Substitute the given values $F(1) = 0$ and $F(3) = -4$: $-12 = (9(-4) - 1(0)) - 2 \int_1^3 x F(x) dx$ $-12 = -36 - 2 \int_1^3 x F(x) dx$

Now, we solve for the integral: $2 \int_1^3 x F(x) dx = -36 + 12$ $2 \int_1^3 x F(x) dx = -24$ $\int_1^3 x F(x) dx = -12$

Since $f(x) = xF(x)$, we have: $\int_1^3 f(x) dx = -12$

Comparing this result with the options:

• Option B: $\int_1^3 {f\left( x \right)dx = 12}$ is incorrect.
• Option D: $\int_1^3 {f\left( x \right)dx = -12}$ is correct.

Step 2: Evaluate the expressions involving $f'(x)$

This addresses options A and C. First, let's find the derivative of $f(x)$. $f(x) = xF(x)$ Using the product rule for differentiation: $f'(x) = \frac{d}{dx}(x F(x)) = 1 \cdot F(x) + x \cdot F'(x) = F(x) + xF'(x)$

Now, let's evaluate $f'(1)$ and $f'(3)$: $f'(1) = F(1) + 1 \cdot F'(1) = 0 + F'(1) = F'(1)$ $f'(3) = F(3) + 3 \cdot F'(3) = -4 + 3F'(3)$

To find a relationship between $F'(1)$ and $F'(3)$, we use the second given integral, $I_2 = \int_1^3 x^3 F''(x) dx = 40$, and apply integration by parts. Let $u = x^3$ and $dv = F''(x)dx$. Then $du = 3x^2 \, dx$ and $v = F'(x)$.

$I_2 = [x^3 F'(x)]_1^3 - \int_1^3 F'(x) (3x^2) dx$ $40 = (3^3 F'(3) - 1^3 F'(1)) - 3 \int_1^3 x^2 F'(x) dx$

We are given $\int_1^3 x^2 F'(x) dx = -12$. Substituting this value: $40 = (27F'(3) - F'(1)) - 3(-12)$ $40 = 27F'(3) - F'(1) + 36$ $27F'(3) - F'(1) = 40 - 36$ $27F'(3) - F'(1) = 4$

Now we can check options A and C. Let's check option C: $9f'(3) - f'(1) + 32 = 0$. Substitute the expressions for $f'(1)$ and $f'(3)$: $9f'(3) - f'(1) + 32 = 9(-4 + 3F'(3)) - (F'(1)) + 32$ $= -36 + 27F'(3) - F'(1) + 32$ $= (27F'(3) - F'(1)) - 4$

From our calculation above, we know $27F'(3) - F'(1) = 4$. Substituting this result: $= 4 - 4 = 0$ So, the expression in option C is correct.

Let's check option A for completeness: $9f'(3) + f'(1) - 32 = 0$. $9(-4 + 3F'(3)) + (F'(1)) - 32$ $= -36 + 27F'(3) + F'(1) - 32$ $= 27F'(3) + F'(1) - 68$ This would imply $27F'(3) + F'(1) = 68$, which is not what we derived. So, option A is incorrect.

Conclusion

Based on our calculations:

• Option C, $9f'(3) - f'(1) + 32 = 0$, is correct.
• Option D, $\int_1^3 {f\left( x \right)dx = -12}$, is correct.

Therefore, the correct expressions are C and D.