MathDB

34

Part of 2005 Today's Calculation Of Integral

Problems(1)

Today's calculation of Integral 34

Source: created by kunny

6/4/2005
Let pp be a constant number such that 0<p<10<p<1. Evaluate ∑k=02004pk(1−p)2004−k∫01xk(1−x)2004−kdx\sum_{k=0}^{2004} \frac{p^k (1-p)^{2004-k}}{\displaystyle \int_0^1 x^k (1-x)^{2004-k} dx}
calculusintegrationprobabilitycalculus computations