Let \begin{equation} N:=\inf\{n\ge2\colon X_{n-1}>X_n\}, \end{equation} where $X_1,X_2,\dots$ are independent random variables uniformly distributed on $[0,1]$. We want to find \begin{equation} EX_N=\sum_{n=2}^\infty EX_n\,1(X_1\le\cdots\le X_{n-1}>X_n). \end{equation}
We have \begin{equation} \begin{aligned} &EX_n\,1(X_1\le\cdots\le X_{n-1}>X_n) \\ &=EX_n\,1(X_1\le\cdots\le X_{n-1}) \\ &-EX_n\,1(X_1\le\cdots\le X_{n-1}\le X_n), \end{aligned} \end{equation} \begin{equation} \begin{aligned} &EX_n\,1(X_1\le\cdots\le X_{n-1}) \\ &=EX_n\,P(X_1\le\cdots\le X_{n-1})=\frac12\,\frac1{(n-1)!}, \end{aligned} \end{equation} \begin{equation} \begin{aligned} &E(1-X_n)\,1(X_1\le\cdots\le X_{n-1}\le X_n) \\ &=E\int_0^1 dx\,1(X_1\le\cdots\le X_{n-1}\le X_n\le x) \\ &=\int_0^1 dx\,P(X_1\le\cdots\le X_{n-1}\le X_n\le x) \\ &=\int_0^1 dx\,x^n\frac1{n!}= \frac1{(n+1)!}, \end{aligned} \end{equation} \begin{equation} \begin{aligned} &EX_n\,1(X_1\le\cdots\le X_{n-1}\le X_n) \\ &=E1(X_1\le\cdots\le X_{n-1}\le X_n) \\ &-E(1-X_n)\,1(X_1\le\cdots\le X_{n-1}\le X_n) \\ &=P(X_1\le\cdots\le X_{n-1}\le X_n)-\frac1{(n+1)!} \\ &=\frac1{n!}-\frac1{(n+1)!}, \end{aligned} \end{equation} \begin{equation} \begin{aligned} EX_N&=\sum_{n=2}^\infty \Big(\frac12\,\frac1{(n-1)!}-\frac1{n!}+\frac1{(n+1)!}\Big) \\ &=\frac e2-1\approx0.359. \end{aligned} \end{equation}
One may also note that \begin{equation} \begin{aligned} &EN=E\sum_{n=0}^\infty1(N>n)=\sum_{n=0}^\infty P(N>n) \\ &=\sum_{n=0}^\infty P(X_1\le\cdots\le X_n) =\sum_{n=0}^\infty \frac1{n!}=e\approx2.72. \end{aligned} \end{equation}
Simulation with Mathematica appears to confirm these results (click on the image below to enlarge it):
