if $f'(x)<2f(x)$ and $f(\frac{1}{2})=1$ and $f$ is a function from $[\frac{1}{2},1]$ to $\mathbb R$ find the interval in which $\int_{\frac{1}{2}}^1 f(x)dx$ lies, given $f(x)$ is non-negative .
My attempt:-
from $f'(x)<2f(x)$ we have,
$$f(x)<e^{2x} +C$$
substituting the value of $f(\frac{1}{2})$ we get
$$1-e <C$$
integrating $f(x)<e^{2x} +C$ between $\frac{1}{2}$ and $1$
we get
$$\int_{\frac{1}{2}}^1 f(x) < 2e(e-1)+\frac{C}{2}$$ to find the upper limit one can substitute C to obtain
$$\int_{\frac{1}{2}}^1 f(x)< 2e(e-1)+\frac{1-e}{2}$$
As $f(x)$ is non-negative, the minimum value of $C$ is $e^2$ is $-e^2$ so
$$e^2-2e<\int_{\frac{1}{2}}^1 f(x)dx< 2e(e-1)+\frac{1-e}{2}$$
however, my book says that the answer is $(0, \frac{e-1}{2})$
I'm not sure I understand where I'm going wrong. What am I doing wrong?