1∫f(x)dx=∫f(a-x)dx .... integrating between limits 0≤x≤a.
Now applying the above propr.in B .....
2B=0 ...... (B is an odd function) .....
B=0
Therefore (A/B)=∞
\hspace{-16}$If $\mathbf{A=\int_{0}^{1}\{x^{50}-(2-x)^{50}\}dx}$ and $\mathbf{B=\int_{0}^1\{x^{50}.(1-x)^{50}\}dx}$.\\\\\\ Then $\mathbf{\frac{A}{B}=}$
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7 Answers
71A and B can be evaluate easily separately , But then A/B would be very ugly.
A is simple while B is a Beta function which directly yields the result.
1
262A can be easily calculated
B can be calculated by parts
on applying the limits after each by part, one term will become zero .
\int_{0}^{1}{x^{50}(1-x)^{50}} = \left[- x^{50}(1-x)^{51} \right]^{1}_{0} + \int_{0}^{1}{\frac{50x^{49}(1-x)^{51}}{51}}
= \frac{49}{51}\int_{0}^{1}{{x^{49}(1-x)^{51}}}
this can be recursively done to calculate B
71That will turn out to be what like deriving the direct formula for the beta function ! :)
