Evaluate: \int_0^1 c^{k+1} (1-c)^{n-k}dc

Question:

Evaluate: {eq}\int_0^1 c^{k+1} (1-c)^{n-k}dc {/eq}

Beta Function:


The beta function is two parameter composition of gamma function.

The beta function is defined by

{eq}\displaystyle {\beta (m,n) = \int\limits_0^1 {{x^{m - 1}}{{(1 - x)}^{n - 1}}dx} ,\,\,m > 0,n > 0} {/eq}


Properties of the beta function:

{eq}\displaystyle \eqalign{ & 1.\,\,\beta (m,n) = \frac{{\Gamma (m)\Gamma (n)}}{{\Gamma (m + n)}} \cr & 2.\,\Gamma (n + 1) = n\Gamma (n) \cr} {/eq}


Answer and Explanation:


Let {eq}\displaystyle I = \int\limits_0^1 {{c^{k + 1}}{{(1 - c)}^{n - k}}dc} {/eq}

We will evaluate this integral using beta function.

{eq}\displaystyle \eqalign{ & \left[ {\therefore \beta (m,n) = \int\limits_0^1 {{x^{m - 1}}{{(1 - x)}^{n - 1}}dx} } \right] \cr & I = \int\limits_0^1 {{c^{k + 1 + 1 - 1}}{{(1 - c)}^{n - k + 1 - 1}}dc} \cr & I = \int\limits_0^1 {{c^{(k + 2) - 1}}{{(1 - c)}^{(n - k + 1) - 1}}dc} \cr & I = \beta (k + 2,\,n - k + 1) \cr & \left[ {\therefore \beta (m,n) = \frac{{\Gamma (m)\Gamma (n)}}{{\Gamma (m + n)}}} \right] \cr & I = \frac{{\Gamma (k + 2)\Gamma (n - k + 1)}}{{\Gamma (k + 2 + n - k + 1)}} \cr & I = \frac{{\Gamma (k + 2)\Gamma (n - k + 1)}}{{\Gamma (n + 3)}} \cr} {/eq}

Therefore, {eq}\boxed { \int\limits_0^1 {{c^{k + 1}}{{(1 - c)}^{n - k}}dc} = \frac{{\Gamma (k + 2)\Gamma (n - k + 1)}}{{\Gamma (n + 3)}}. } {/eq}




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