another definite Integral

\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 $\mathbf{\frac{A}{B}=}$

7 Answers

71
Vivek @ Born this Way ·

A 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
ARKA(REEK) ·

∫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)=∞

71
Vivek @ Born this Way ·

What is this? I don't agree.

262
Aditya Bhutra ·

completely miscued .

262
Aditya Bhutra ·

A 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

71
Vivek @ Born this Way ·

That will turn out to be what like deriving the direct formula for the beta function ! :)

262
Aditya Bhutra ·

well i dont know anything about "beta" function !

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