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Please help me find the value of the following integral:
$$\frac{(5050)\int^1_0(1-x^{50})^{100} dx}{\int^1_0(1-x^{50})^{101} dx}$$ I tried solving both numerator and denominator via by-parts but it isn't giving me a conclusive solution. Any other suggestions?

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marked as duplicate by Ian Miller, Michael Hoppe, Nosrati, Claude Leibovici calculus Apr 16 '17 at 5:55

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Let's try integrating by parts and see what happens. $$ \int_0^1 (1-x^n)^m \, dx = \left[ x(1-x^n)^m \right]_0^1 - \int_0^1 -nmx^n(1-x^n)^{m-1} \, dx = 0+ nm\int_0^1 x^n(1-x^n)^{m-1} \, dx. $$ We can fiddle with the right-hand side to get it into a more familiar form: $$ \int_0^1 x^n(1-x^n)^{m-1} \, dx = \int_0^1 \left( 1 -(1-x^n) \right) (1-x^n)^{m-1} \, dx = \int_0^1 (1-x^n)^{m-1} \, dx - \int_0^1 (1-x^n)^{m} \, dx. $$ Collecting the copies of the integrals together, we find $$ (1+nm)\int_0^1 (1-x^n)^m \, dx = nm \int_0^1 (1-x^n)^{m-1} \, dx, $$ or $$ \frac{\int_0^1 (1-x^n)^{m-1} \, dx}{\int_0^1 (1-x^n)^m \, dx} = \frac{1+nm}{nm}. $$ Putting $n=50$, $m=101$, the right-hand side becomes $ 5051/5050 $.

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  • $\begingroup$ What happened to $[ x(1-x^n)^m ]$ ? $\endgroup$ – A---B Apr 15 '17 at 1:09
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    $\begingroup$ Yep, that should be evaluated at both endpoints and disappear. Fixed. Thanks. $\endgroup$ – Chappers Apr 15 '17 at 1:10
  • $\begingroup$ very nice answer..@ Uddeshya Singh you can accept this answer because it answers fully your question $\endgroup$ – Marios Gretsas Apr 15 '17 at 1:13
  • $\begingroup$ This answer is very clever.. $\endgroup$ – A---B Apr 15 '17 at 1:35
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one way to solve it besides the change of variable,is by using the binomial theorem in both denominator and enumerator.

$\sum_{n=0}^{100}\binom{100}{n}(x^{50})^{100-n}(-1)^n=(1-x^{50})^{100}$

$\binom{n}{k}=\frac{n!}{k!(n-k)!}$

Use the linearity of the integral and integrate each term of the series.

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  • $\begingroup$ I'm sorry but m not really well versed in usage of binomial theorem, Can u please guide further $\endgroup$ – The Dead Legend Apr 15 '17 at 0:51
  • $\begingroup$ i edited ...if you want something else let me know $\endgroup$ – Marios Gretsas Apr 15 '17 at 0:55
  • $\begingroup$ Got it. Done. answer is 5051? $\endgroup$ – The Dead Legend Apr 15 '17 at 0:56
  • $\begingroup$ i dont think so.. $\endgroup$ – Marios Gretsas Apr 15 '17 at 1:01

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