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Beniamin Bogosel wrote:Calculati \( \int_{0}^{\pi}x\cos^2 x dx \).

Cât despre o dovadă geometrice?

\( \ln \left( {1 + e^{ - x} } \right) = \int_0^{\exp ( - x)} {\frac{1}
{{1 + u}}\,du} . \)

By reversing integration order we get a single integral which can be solved using the geometric series, the rest follows by using the Basel problem.

Cezar Lupu wrote:Sa se calculeze

\( \lim_{n\to\infty}\int_0^n ln(1+e^{-x})dx \).

A double integration trick yields the answer too.

Evaluate

\int_{-\infty}^{\infty}e^{-x^2}dx . I don't speak romanian, but I think you may understand me. \begin{eqnarray*} I &=& \int_{ - \infty }^\infty {e^{ - x^2 } \,dx}\\ I^2 &=& \int_{ - \infty }^\infty {e^{ - x^2 } \,dx} \, \cdot \int_{ - \infty }^\infty {e^{ - y^2 } \,dy}\\ &=&...

Okay, that's clear.

Thanks for answering.

I know that, but I mean for example



Here

\( $\displaystyle\int_{-\infty}^\infty e^{-x^2}\,dx=\sqrt{\pi}$ \)

You see the difference?, that's what I'm talkin' about, MimeTeX.

Hi

Sorry, but I dunno romanian language.

Just wonderin', isn't LaTeX installed yet?, 'cause I see you have MimeTeX.