Odpowiedź:
Szczegółowe wyjaśnienie:
[tex]f_{sr}[/tex] na przedziale <0, π>
[tex]f_{sr}=\frac{1}{\pi -0} \int\limits^\pi _0 {sinx} \, dx =\frac{1}{\pi } [-cosx]^\pi _0=-\frac{1}{\pi } (cos\pi -cos0)=-\frac{1}{\pi } (-1-1)=\frac{2}{\pi }[/tex]
[tex]f_{sr}=f(c)[/tex]
[tex]\frac{2}{\pi } =sin c\\c=arcsin\frac{2}{\pi }[/tex]
b.
[tex]f_{sr}[/tex] na przedziale <-1, 1>
[tex]f_{sr}=\frac{1}{1-(-1)} \int\limits^1_{-1} {\frac{1}{x^2+1} } \, dx =\frac{1}{2} [arctgx]^1_{-1}=\frac{1}{2} [arctg (1)-arctg (-1)] =\\=\frac{1}{2} (\frac{\pi }{4} -(-\frac{\pi }{4} ))=\frac{\pi }{4}[/tex]
[tex]f_{sr}=f(c)[/tex]
[tex]\frac{\pi }{4} =\frac{1}{c^2+1} \\\pi (c^2+1)=4\\c^2+1=\frac{4}{\pi } \\c^2=\frac{4}{\pi }-1\\c=\sqrt{\frac{4}{\pi }-1}[/tex]
