funkcja beta
: 14 cze 2007, o 21:22
\(\displaystyle{ \int\limits_{0}^{1}x^{3}(1-x^{2})^{5}dx}\)
Podstawienie, żeby się nie zaliczyć:
\(\displaystyle{ t = 1- x^{2}\\
x^{2} = 1 - t\\
dt = -2xdx}\)
\(\displaystyle{ \int x^{3}(1 - x^{2})^{5}dx = -\frac{1}{2}\int (1 - t)t^{5}dt =\\
= -\frac{1}{2}\int (t^{5} - t^{6})dt = -\tfrac{1}{12}t^{6} + \tfrac{1}{14}t^{7} + C =\\
= -\tfrac{1}{12}(1 - x^{2})^{6} + \tfrac{1}{14}(1 - x^{2})^{7} + C}\)
Ze wzoru podstawowego mamy:
\(\displaystyle{ \int\limits_{0}^{1} x^{3}(1 - x^{2})^{5}dx =\\
= -\tfrac{1}{12}(1 - 1^{2})^{6} + \tfrac{1}{14}(1 - 1^{2})^{7} - \left(-\tfrac{1}{12}(1 - 0^{2})^{6} + \tfrac{1}{14}(1 - 0^{2})^{7}\right) = \\
= 0 - \left(-\tfrac{1}{12} + \tfrac{1}{14}\right) = \tfrac{1}{84}}\)