MathDB

Today's calculation of Integral 34

Source: created by kunny

June 4, 2005

calculusintegrationprobabilitycalculus computations

Problem Statement

Let

p

be a constant number such that

0<p<1

. Evaluate

\sum_{k=0}^{2004} \frac{p^k (1-p)^{2004-k}}{\displaystyle \int_0^1 x^k (1-x)^{2004-k} dx}