Ensino Superior ⇒ Integral Impropria

[iceman]

Alguém poderia me ajudar no calculo dessa integral? agradeço!

[tex3]\int\limits_{-\infty }^{+\infty }xdx[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\int\limits_{-\infty }^{+\infty }xdx[/tex3]

Resposta

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[Rafa2604]

[tex3]\;\; \rightarrow \;\int\limits_{-\infty }^{+\infty }xdx = \lim_{c \to \infty}\int\limits_{-c }^{+c }xdx = \lim_{c \to \infty}\left[\frac{x^{2}}{2}\right]_{-c}^{c} = \lim_{c \to \infty}\left[\frac{c^{2}}{2} - \frac{(-c)^{2}}{2}\right] = \lim_{c \to \infty}\left[\frac{c^{2}}{2} - \frac{c^{2}}{2}\right] = 0[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\;\; \rightarrow \;\int\limits_{-\infty }^{+\infty }xdx = \lim_{c \to \infty}\int\limits_{-c }^{+c }xdx = \lim_{c \to \infty}\left[\frac{x^{2}}{2}\right]_{-c}^{c} = \lim_{c \to \infty}\left[\frac{c^{2}}{2} - \frac{(-c)^{2}}{2}\right] = \lim_{c \to \infty}\left[\frac{c^{2}}{2} - \frac{c^{2}}{2}\right] = 0[/tex3]