Sa facem notatia: [tex]f(x)=xe^{x^2}[/tex].
Acum, limita ta se scrie:
[tex]\lim_{n\to\infty}\dfrac{1}{n^2}\left[e^{\frac{1^2}{n^2}}+...+ne^{\frac{n^2}{n^2}}\right]=\\ \\ =\lim\limits_{n\to\infty}\sum\limits_{i=1}^{n}\dfrac{i}{n^2}e^{\left(\frac{i}{n}\right)^2}=\\ \\ \\ =\lim\limits_{n\to\infty}\sum\limits_{i=1}^{n}\left[\left(\dfrac{i}{n}\right)e^{\left(\frac{i}{n}\right)^2\right]\cdot\dfrac{1}{n}= \\ \\ \\ =\lim\limits_{n\to\infty}\sum\limits_{i=1}^{n}f\left(\frac{i}{n}\right)\cdot\dfrac{1-0}{n}=[/tex]
Am scris suma sub forma aceasta, ca sa fie evident ca avem de-a face cu o suma infinita Riemann, care de fapt defineste o integrala de la 0 la 1. Asa ca suma este egala cu:
[tex]=\int\limits_{0}^{1}xe^{x^2}dx.[/tex]
Deci totul se reduce la a calcula integrala data.
Facem schimbarea de variabila [tex]t=x^2[/tex]
Rezulta:
[tex]dt=2xdx\\ \\ t(0)=0\\t(1)=1[/tex]
Integrala va deveni:
[tex]I=\dfrac{1}{2}\int\limits_0^1 e^tdt=\\ \\ =\dfrac{1}{2}(e-1).[/tex]
Si e gata!
