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Question

Let Im,n=01xm(1x)n dxI_{m,n} = \int_{0}^{1}x^{m}(1-x)^{n}\ dx for constant m,n.m,n. Show that Im,n=In,m.I_{m,n} = I_{n,m}.

Solution

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Using substitution, we have:

01xm(1x)ndx=[1x=tdx=dtx=t1x=0t=1x=1t=0]==10(1t)mtndt=01(1t)mtndt==01tn(1t)mdt=In,m\begin{align*} \int_{0}^{1} x^m(1-x)^n\,dx&= % matrica supstitucije \color{#4257b2} \left[ \small { \begin{array} {rlcrl} 1-x&=t & \Rightarrow & -dx&=dt \\ x&=t-1 \\ x &= 0 & \Rightarrow & t&=1 \\ x &=1 & \Rightarrow & t&=0 \end{array} } \right] \color {black} = \\[15pt] &=-\int_{1}^{0} (1-t)^m\,t^n\,dt=\int _{0}^{1} (1-t)^m \,t^n\,dt= \\[21pt] &=\int _{0}^{1} t^n\,(1-t)^m\,dt=I_{n,m} \end{align*}

\quad \quad \quad So,

Im,n=In,mI_{m,n}=I_{n,m}

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