Step-by-Step Solution:

1. Analyze the given differential inequality. We are given a function $f(x)$ on the interval $[1/2, 1]$ with the properties:

   • $f(x) > 0$ (positive)
   • $f(x)$ is non-constant
   • $f(x)$ is differentiable
   • $f'(x) < 2f(x)$
   • $f(1/2) = 1$

   The inequality is $f'(x) < 2f(x)$. We can rewrite it as: $f'(x) - 2f(x) < 0$

2. Introduce an integrating factor. To solve this type of differential inequality, we use an integrating factor, similar to solving linear differential equations. The integrating factor is $e^{\int -2 dx} = e^{-2x}$. Since $e^{-2x}$ is always positive for real $x$, multiplying the inequality by $e^{-2x}$ does not change its direction: $e^{-2x}f'(x) - 2e^{-2x}f(x) < 0$

3. Recognize the derivative of a product. The left side of the inequality is the derivative of the product $e^{-2x}f(x)$ with respect to $x$, according to the product rule $(uv)' = u'v + uv'$: rac{d}{dx} ig( e^{-2x} f(x) ig) < 0

4. Analyze the behavior of the new function. Let's define a new function $g(x) = e^{-2x}f(x)$. The inequality from the previous step becomes: $g'(x) < 0$ A function with a negative derivative on an interval is strictly decreasing on that interval. Thus, $g(x)$ is a strictly decreasing function on $[1/2, 1]$.

5. Use the property of the decreasing function to find a bound for f(x). Since $g(x)$ is strictly decreasing, for any $x$ in the interval $(1/2, 1]$, we have $x > 1/2$, which implies: $g(x) < g(1/2)$ Now, we substitute the definition of $g(x)$ back into the inequality: $e^{-2x} f(x) < e^{-2(1/2)} f(1/2)$ We are given that $f(1/2) = 1$. So: $e^{-2x} f(x) < e^{-1} imes 1$ $e^{-2x} f(x) < e^{-1}$ To find an upper bound for $f(x)$, we multiply by $e^{2x}$ (which is positive): $f(x) < e^{2x} e^{-1}$ $f(x) < e^{2x-1}$ This gives us an upper bound for the function $f(x)$ for all $x ext{ in } (1/2, 1]$.

6. Integrate the bounds to find the interval for the integral. We want to find the value of $I = \int_{1/2}^1 f(x) dx$. We can integrate the inequality we found for $f(x)$.

   Upper bound for the integral: $I = \int_{1/2}^1 f(x) dx < \int_{1/2}^1 e^{2x-1} dx$ Let's evaluate the integral on the right: $\int_{1/2}^1 e^{2x-1} dx = \left[ \frac{e^{2x-1}}{2} \right]_{1/2}^1$ $= \left( \frac{e^{2(1)-1}}{2} \right) - \left( \frac{e^{2(1/2)-1}}{2} \right)$ $= \frac{e^1}{2} - \frac{e^0}{2} = \frac{e}{2} - \frac{1}{2} = \frac{e-1}{2}$ So, we have the upper bound: $I < \frac{e-1}{2}$.

   Lower bound for the integral: We are given that $f(x)$ is a positive function, i.e., $f(x) > 0$. Since $f(x)$ is continuous and positive on the interval $[1/2, 1]$, its definite integral over this interval must be strictly positive. $I = \int_{1/2}^1 f(x) dx > 0$

7. Combine the bounds and determine the correct interval. Combining the lower and upper bounds, we get: $0 < \int_{1/2}^1 f(x) dx < \frac{e-1}{2}$ Therefore, the value of the integral lies in the interval $\left( 0, \frac{e-1}{2} \right)$.

8. Compare with the given options. The options are: A: $\left( {2e - 1,2e} \right)$ B: $\left( {e - 1,\,2e - 1} \right)$ C: $\left( {{{e - 1} \over 2},e - 1} \right)$ D: $\left( {0,{{e - 1} \over 2}} \right)$

   Our derived interval matches option D exactly.