Całka do sprawdzenia

[Tom555]

\(\displaystyle{ \begin{array}{l}
\int\limits_{ - 1}^1 {\sqrt {1 - {x^2}} } \mathop = \limits^{\scriptstyle x = \sin t \hfill \atop
\scriptstyle dx = \cos tdt \hfill} \int\limits_{ - 1 = \sin t}^{1 = \sin t} {\sqrt {1 - {{\sin }^2}(t)} } \cos (t)dt = \int\limits_{{\textstyle{{ - \pi } \over 2}}}^{{\textstyle{\pi \over 2}}} {\sqrt {{{\cos }^2}(t)} } \cos (t)dt = \\
= \int\limits_{{\textstyle{{ - \pi } \over 2}}}^{{\textstyle{\pi \over 2}}} {\left| {\cos (t)} \right|} \cos (t)dt \\
\end{array}}\)


Ponieważ cos(t) w podanym przedziale jest nieujemny więc:

\(\displaystyle{ \int\limits_{{\textstyle{{ - \pi } \over 2}}}^{{\textstyle{\pi \over 2}}} {\left| {\cos (t)} \right|} \cos (t)dt = \int\limits_{{\textstyle{{ - \pi } \over 2}}}^{{\textstyle{\pi \over 2}}} {{{\cos }^2}(t)} dt}\)


\(\displaystyle{ {\cos ^2}(t) = \frac{{\cos (2t) + 1}}{2}}\)

Więc:

\(\displaystyle{ \begin{array}{l}
\int\limits_{{\textstyle{{ - \pi } \over 2}}}^{{\textstyle{\pi \over 2}}} {\frac{{\cos (2t) + 1}}{2}} dt = \frac{1}{2}\int\limits_{{\textstyle{{ - \pi } \over 2}}}^{{\textstyle{\pi \over 2}}} {(\cos (2t) + 1)} dt = \\
\frac{1}{2}\left( {\frac{1}{2}\sin (2t) + t} \right)_{{\textstyle{{ - \pi } \over 2}}}^{{\textstyle{\pi \over 2}}} = \frac{\pi }{2} \\
\end{array}}\)

Dobrze jest?

[loitzl9006]

Nie zauważyłem błędów.