Bounding Definite Integral of Function Satisfying Differential Inequality

To find the interval in which the definite integral $\int_{1/2}^1 f(x) \, dx$ lies, we analyze the given differential inequality:

Given:

1. $f(x) > 0$ for all $x \in \left[\frac{1}{2}, 1\right]$
2. $f\left(\frac{1}{2}\right) = 1$
3. $f'(x) < 2f(x)$ for all $x \in \left[\frac{1}{2}, 1\right]$

Rearranging the differential inequality: $f'(x) - 2f(x) < 0$

Multiplying both sides by the positive integrating factor $e^{-2x}$: $e^{-2x}f'(x) - 2e^{-2x}f(x) < 0$ $\frac{d}{dx}\left(e^{-2x}f(x)\right) < 0$

Let us define a helper function $g(x) = e^{-2x}f(x)$. Since its derivative $g'(x) < 0$, $g(x)$ is a strictly decreasing function on $\left[\frac{1}{2}, 1\right]$.

Therefore, for any $x \in \left(\frac{1}{2}, 1\right]$: $g(x) < g\left(\frac{1}{2}\right)$ $e^{-2x}f(x) < e^{-2(1/2)}f\left(\frac{1}{2}\right)$

Substitute $f\left(\frac{1}{2}\right) = 1$: $e^{-2x}f(x) < e^{-1} \cdot 1$ $f(x) < e^{2x - 1}$

Since $f(x)$ is given to be a positive function, we have: $0 < f(x) < e^{2x - 1} \quad \text{for all } x \in \left(\frac{1}{2}, 1\right]$

Integrating all parts of the inequality from $x = \frac{1}{2}$ to $x = 1$: $0 < \int_{1/2}^1 f(x) \, dx < \int_{1/2}^1 e^{2x - 1} \, dx$

Evaluating the upper bound integral: $\int_{1/2}^1 e^{2x - 1} \, dx = \left[ \frac{e^{2x - 1}}{2} \right]_{1/2}^1 = \frac{e^{2(1)-1} - e^{2(1/2)-1}}{2} = \frac{e - 1}{2}$

Thus, we obtain: $0 < \int_{1/2}^1 f(x) \, dx < \frac{e - 1}{2}$

Hence, the value of the definite integral lies in the interval $\left(0, \frac{e - 1}{2}\right)$.

Therefore, the correct option is (D).