Given, $$f(x)$$ be a positive, non-constant and differentiable function and $$f\left(\frac{1}{2}\right)=1$$.

Also given $$f^{\prime}(x)<2 f(x)$$

$$\Rightarrow f^{\prime}(x)-2 f(x)<0$$

Multiply both side by $$e^{-2 x}$$

$$\begin{aligned} & \Rightarrow e^{-2 x} f^{\prime}(x)-2 e^{-2 x} f(x)<0 \\ & \Rightarrow d\left(e^{-2 x} f(x)\right)<0 \end{aligned}$$

Let $$g(x)=e^{-2 x} f(x) \forall x \in\left[\frac{1}{2}, 1\right]$$

$$\Rightarrow d(g(x))<0$$

Hence, $$g(x)$$ is a decreasing function in the Given, domain $$x \in\left[\frac{1}{2}, 1\right]$$

$$\begin{aligned} & \because \quad g(x)< g\left(\frac{1}{2}\right) \\ & \Rightarrow e^{-2 x} f(x) < e^{-1} \cdot f\left(\frac{1}{2}\right) \\ & \Rightarrow f(x) < e^{2 x-1} \cdot 1 \\ & \therefore \quad f(x) \text { is always positive } \\ & \therefore \quad 0 < f(x) < e^{2 x-1} \\ & \Rightarrow 0 < \int_{\frac{1}{2}}^1 f(x) d x < \int_{\frac{1}{2}}^1 e^{2 x-1} d x \\ & \Rightarrow 0 < \int_{\frac{1}{2}}^1 f(x) d x < \left[\frac{e^{2 x-1}}{2}\right]_{\frac{1}{2}}^1 \\ & \Rightarrow 0 < \int_{\frac{1}{2}}^1 f(x) d x < \frac{e-1}{2} \\ & \Rightarrow \int_{\frac{1}{2}}^1 f(x) d x \in\left(0, \frac{e-1}{2}\right) \end{aligned}$$

Hints:

$$\Rightarrow$$ (i) if $$f(x)$$ is a decreasing function for $$x \in[a, b]$$, then $$f(b) < f(x) < f(a)$$

(ii) Recall that $$e^{-2 x} f^{\prime}(x)-2 e^{-2 x} f(x)=d\left(e^{-2 x} f(x)\right)$$