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How to show that the area $A$ of the region $$\{(x,y)\in\mathbb R^2 : |x|\le y\le e^{-x^2}\}$$ is such that $$0 < A = 2\int_{0}^{\beta}{(e^{-t^2}-t)dt} < 1$$ in which $\beta>0$ satisfies $e^{-\beta^2}=\beta$

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Consider the substitution $$ u = \mathrm{e}^{-x^2} \implies \mathrm{d}x = \mathrm{d}\sqrt{-\ln(u)} = -\frac{1}{2u\sqrt{-\ln(u)}}\mathrm{d}u$$ under which the bounds become $u(0) = 1, u(\beta) = \beta$: \begin{align} A & = 2\int_1^\beta \left(u - \sqrt{\ln(u)}\right)\frac{-1}{2u\sqrt{-\ln(u)}}\mathrm{d}u = \int_\beta^1 \frac{u-\sqrt{-\ln(u)}}{u\sqrt{-\ln(u)}}\mathrm{d}u \\ & = \int_\beta^1 \frac{u}{u\sqrt{-\ln(u)}}\mathrm{d}u - \int_\beta ^1 \frac{1}{u}\mathrm{d}u \leq \int_\beta^1 \frac{1}{u\sqrt{-\ln(u)}}\mathrm{d}u - \beta^2 = 2\beta - \beta^2 \end{align} Observe now that $\beta \leq 0.7$ because $\exp(-0.7^2) \approx 0.61 \leq 0.7$, so $A \leq 2\beta - \beta^2 \leq 0.91 < 1$.

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Hint: You can consider that: $$2\int_0^\beta e^{-t^2}dt\leq 2\max_{t\in[0,\beta]}[e^{-t^2}](\beta-0)\leq2\beta$$ and that: $$2\int_0^\beta-tdt=-\beta^2$$ so: $$A\leq 2\beta-\beta^2=\beta(2-\beta)$$ this function $\beta(2-\beta)$ have a only one point of max $\beta=1$, but we know that $\beta\neq1$ and that the function $e^{-t^2}$ is decreasing on $[0,\infty)$ and that $t$ is increasing so exist only one point for the intersection. If $\beta=1\implies A\leq1$.

So, $\forall \beta\in[0,\infty), \beta\neq1 \quad A<1$.

Otherwise we want show $A>0$. $$2\int_0^\beta e^{-t^2}dt\geq 2\min_{t\in[0,\beta]}[e^{-t^2}](\beta-0)\geq2e^{-\beta^2}\beta=2\beta^2$$ then we have that $$A\geq 2\beta^2-\beta^2=\beta^2$$