Trebuie demonstrat că:
[tex]\displaystyle\int_{0}^{1}e^{-x^2}dx<\dfrac{\pi}{4}[/tex]
Voi folosi expansiunea Taylor:
[tex]e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+...[/tex]
Substituim x cu x²:
[tex]e^{x^2}=1+x^2+\dfrac{x^4}{2!}+\dfrac{x^6}{3!}+...[/tex]
Se observă clar că:
[tex]e^{x^2}>1+x^2 \\ \\\Rightarrow \dfrac{1}{e^{x^2}}<\dfrac{1}{1+x^2}\\ \\\Rightarrow e^{-x^2}<\dfrac{1}{1+x^2}\\ \\\Rightarrow \displaystyle\int_{0}^{1}e^{-x^2}dx<\displaystyle\int_{0}^{1}\dfrac{1}{1+x^2}\,dx\\\\ \Rightarrow \displaystyle\int_{0}^{1}e^{-x^2}dx<\arctan(x)\bigg|_{0}^{1}\\ \\\Rightarrow \displaystyle\int_{0}^{1}e^{-x^2}dx<\dfrac{\pi}{4}-0 \\ \\\Rightarrow \displaystyle\int_{0}^{1}e^{-x^2}dx<\dfrac{\pi}{4}[/tex]
