Let $f:[a,b] \to \mathbb{R}^2$ be a continuous curve on the plane.



> 
> **Question:** Are there numbers $a \leq x \leq c \leq y \le b$ such that $$(c-a)f(x)+(b-c)f(y) = \int\_a^b f(t) \, dt \ ?$$
> 
> 
> 


In other words, is there a Riemann sum with two terms that hits the bull's-eye?


**EDIT:** Prompted by a down-vote, maybe I should give some **motivation:** 


It's not difficult to convince oneself (but not completely trivial to prove (\*)) that the barycenter $\frac{1}{b-a}\int\_a^b f(t) \, dt$ is a convex combination of **two** values of $f$, say $f(x)$ and $f(y)$ with $x<y$. But this doesn't answer my question, because depending on the weights we may be unable to find the partition point $c$. 


For curves in $\mathbb{R}^n$, I could ask if there is a Riemann sum with $n$ terms that hits the bull's-eye, but dimension $n=2$ already seems tricky enough.


(\*) *Footnote:* A theorem of Korobkov [1], improving a theorem of McLeod [2], says that if $F: [a,b] \to \mathbb{R}^n$ is continuous on $[a,b]$ and differentiable on $(a,b)$ then $\frac{F(b)-F(a)}{b-a}$ is a convex combination of $n$ values of the derivative $F'$ (which is not necessarily continuous). 


*References:*


[1] Korobkov, M.V. -- A generalization of Lagrange's mean value theorem to the case of vector-valued mappings. Sibirsk. Mat. Zh. 42 (2001), no. 2, 349--353, ii; translation in Siberian Math. J. 42 (2001), no. 2, 297--300, [doi: 10.1023/A:1004889013835](http://dx.doi.org/10.1023/A:1004889013835). 


[2] McLeod, R.M. -- Mean value theorems for vector valued functions. Proc. Edinburgh Math. Soc. (2) 14 1964/1965, 197--209, [doi: 10.1017/S0013091500008786](http://dx.doi.org/10.1017/S0013091500008786). 



Bull's-eye Riemann sum
Not necessarily. Take small $\delta>0$. Spend time $\frac 23(1-\delta)$ at $(0,1)$. Then, in time $\delta/4$ travel the route $(0,1)\to(0,2)\to(-1,2)\to(-1,-2)$ so that the integral of the vertical coordinate is $0$. Then stay at $(-1,-2)$ for the time $\frac 16(1-\delta)$ and return to $(0,1)$ running the same movement backwards. Make a symmetric (with respect to the $y$ axis) round trip to the right. The resulting function has integral $(0,0)$ over the interval $[0,1]$. On the other hand, any two points on the associated curve that lie on the same line through the origin on the opposite sides of the origin can cancel each other only with coefficients $1/2$. However, the corresponding partition of $[0,1]$ puts them in the same interval.


This will certainly make an excellent extra credit question for multivariate calculus students. To make the life even more interesting, let us ask if any $n$ is sufficient on the plane. 


