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I feel like this should be a no-brainer but this is my first set of variational problems so I'm probably just not used to how it should work.

Find extremal of $$F[u] = \int_0^1 xuu' dx$$ where $u(0) = 0$ and $u(1) = 1$.

So using the Euler-Lagrange formula with $f(x,u,u') = xuu'$ I get $$0 = f_u - \frac{d}{dx}f_{u'} = xu' - u - xu' = -u$$ so that $u(x) = 0$.

But clearly this does not satisfy the boundary condition, so now what?

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    What if we try changing variables: $w = {1 \over 2} u^2$. So, $$ u u' = w', $$ and the problem is now: $$ \min \rightarrow F[w] = \int_{0}^{1} x \; w' \; dx, \quad w(0)=0, w(1)=1. $$ – 2017-01-26

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Your analysis is right: there is no extremal function for this problem. Integration by parts makes this clear: $$ F[u]= \frac12 \int_0^1 x (u^2)'\,dx = \frac12 xu(x)^2\bigg|_0^1 - \frac12 \int_0^1 u(x)^2\,dx = \frac12 - \frac12 \int_0^1 u(x)^2\,dx $$ Under the given boundary conditions, the integral $\int_0^1 u^2$ can be arbitrarily large, or arbitrarily close to zero — but it cannot be equal to zero, since $u$ cannot be identically zero. Conclusion: $\inf_u F[u]=-\infty$, $\sup_u F[u] = 1/2$, and neither one is attained.

The Euler-Lagrange equation correctly suggested what $u$ would have to be to maximize $F$, but $u$ was prevented from being that by the boundary conditions.