Let $g:\left[0.5,1\right]\rightarrow\mathbb{R}$ be a positive non
constant differentiable function such that
$g'(x)<2g(x)$ and $g(0.5)=1$. Then prove that
$\displaystyle \int^{1}_{0.5}g(x)dx\in\bigg(0,\frac{e-1}{2}\bigg)$
solution i try
$g'(x)-2g(x)<0$. multiply both side by $e^{-2x}$
so $g'(x)\cdot e^{-2x}-2g(x)\cdot e^{-2x}<0$
$\bigg(g(x)\cdot e^{-2x}\bigg)'<0$
i am not go ahead after that, help me
On
I think I have something that might help:
So, I treat this as a "differential inequality":
$g'(x) < 2g(x)$, so we rearrange or separate it as an ODE to get:
$\frac{g'(x)}{g(x)} < 2$, so "integrating" both sides w.r.t $x$, we get:
$\int \frac{g'(x)}{g(x)}dx < \int 2dx$, so we get:
$ log(g(x)) < 2x +c => g(x) < ce^{2x}$
Solving for the initial value problem $g(0.5) =1$, we can get that the constant $c =0.367879$, this is done same as you'd do for standard ODE IVP problem.
So, then we have that $g(x) < 0.367879e^{2x}$.
Since $g$ is assumed to be differentiable on the interval $[0.5,1]$, and more importantly, nonzero, then we have that it is integrable there.
So, we get that:
$\int_{0.5}^{1} g(x)dx < \int_{0.5}^{1}0.367879e^{2x}dx$ and after you calculate the integral on the right, it equals $0.85914$. That is, $\frac{e-1}{2}$, which is an upper bound for this integral. To obtain the lower bound, that is $0$, then of course the exponential function is never zero so hence the value of that integral on $[0.5,1]$, and hence $[0,1]$, is between $0$ and $\frac{e-1}{2}$.
I think your statement is incorrect. The correct conclusion should be:
Prove: $$0 < \displaystyle \int_{1/2}^1 g(x)dx < \dfrac{e-1}{2}$$.
Note that $g(t) > 0, \forall t \in [1/2, 1]$. Hence $\dfrac{g'(t)}{g(t)}< 2\implies \displaystyle \int_{1/2}^x \dfrac{g'(t)}{g(t)}dt< \displaystyle \int_{1/2}^x 2dt\implies \ln(g(x))=\ln\left(\dfrac{g(x)}{g(1/2)}\right)< 2\left(x-1/2\right)=2x-1\implies g(x) < e^{2x-1}\implies \displaystyle \int_{1/2}^1 g(x)dx < \displaystyle \int_{1/2}^1 e^{2x-1}dx= \displaystyle \int_{0}^1 \dfrac{e^x}{2}dx= \dfrac{e-1}{2}$ .