MaxWong

avatar
BenutzernameMaxWong
Punkte7023
Stats
Fragen 159
Antworten 2617

 #6
avatar+7023 
0

Well. I got quick ways for all of them.

 

Formula: \(\displaystyle\int^{1}_{0}x^p(1-x^r)^s\; dx = \dfrac{s}{p + rs + 1}B(\dfrac{p+1}{r},s)\)

Where \(B(x,y) = \dfrac{\Gamma(x) \cdot \Gamma(y)}{\Gamma(x+y)}\)

 

1)

 \(\displaystyle \int^{1}_{0}\sqrt{1-\sqrt{x}}dx\\ =\displaystyle \int^{1}_{0}x^{0}(1-x^{1/2})^{1/2}\; dx \\ =\dfrac{1/2}{0+1/2\cdot1/2 + 1}B(\dfrac{0+1}{1/2},1/2)\\ =\dfrac{2}{5}B(2,1/2)\\ =\dfrac{2}{5}\dfrac{\Gamma(2)\cdot\Gamma(1/2)}{\Gamma(2+1/2)}\\ =\dfrac{8}{15}\)

 

2)

\(\displaystyle\int^{1}_{0}\sqrt{\sqrt[3]{x}-\sqrt{x}}\; dx\\ =\displaystyle\int^{1}_{0}\sqrt{\sqrt[3]{x}(1-\sqrt[6]{x})}\;dx\\ =\displaystyle\int^{1}_{0}x^{1/6}(1-x^{1/6})^{1/2}\;dx\\ =\dfrac{1/2}{1/6+1+1/6\cdot1/2}B(\dfrac{1/6+1}{1/6},1/2)\;dx\\ =\dfrac{2}{5}B(7,1/2)\\ =\dfrac{4096}{15015}\)

3)

\(\displaystyle\int^{1}_{0}(x^{2/5}-\sqrt[3]{x})^{4/7}dx\\ =\displaystyle\int^{1}_{0}(x^{2/5})^{4/7}(1-x^{-1/15})^{4/7}dx\\ =\displaystyle\int^{1}_{0}x^{8/35}(1-x^{-1/15})^{4/7}dx\\ =\dfrac{\frac{4}{7}}{\frac{8}{35}+1+\frac{-1}{15}\cdot\frac{4}{7}}B(\dfrac{\frac{8}{35}+1}{\frac{-1}{15}},\dfrac{4}{7})\\ =\dfrac{12}{25}B(-\dfrac{129}{7},\dfrac{4}{7})\\ \approx -0.0626\)

4)

\(\displaystyle\int^{\pi/2}_{0}\cos^{15}xdx\\ =\displaystyle\int^{\pi/2}_{0}\cos x (1 - \sin^2 x)^7dx\\ u = \sin x\\ =\displaystyle\int^{1}_{0}(1-u^2)^7du\\ =\dfrac{7}{0+1+2\times7}B(\dfrac{0+1}{2},7)\\ =\dfrac{7}{15}B(1/2,7)\\ =\dfrac{2048}{6435}\)

5)

\(\displaystyle\int^{\pi/2}_{0}\sin^{13}xdx\\ =\displaystyle\int^{\pi/2}_{0}(-\sin x)(1-\cos^2x)^6\;dx\\ u = \cos x\\ =\displaystyle\int^{1}_{0}(1-u^2)^6du\\ =\dfrac{6}{0+1+2\times 6}B(\dfrac{0+1}{2},6)\\ =\dfrac{6}{13}B(\dfrac{1}{2},6)\\ =\dfrac{1024}{3003}\)

MaxWong 08.11.2017