How I can evaluate $\int_{x_{i-1}}^{x_i}(x-x_{i-1})(x_i-x)dx $ with $ h=-x_{i-1}+x_i$?

Question

My friend sent me the following problem for help , let $ h=-x_{i-1}+x_i$ , I have tried to evaluate this $\int_{x_{i-1}}^{x_i}(x-x_{i-1})(x_i-x)dx $ using Darboux integral which uses Darboux sum upper sum and lower sum of it where $[x_{i-1}-x_i]$ is a subintervale of the partition of $[a,b]$ but I can't come up to $\dfrac{h^3}{6}$ which it is the result of computing $\int_{x_{i-1}}^{x_i}(x-x_{i-1})(x_i-x)dx $ which it were claimed by my friend , Any help ?

Answer 1

We have to compute (with $v=x_i,u=x_{i-1}, v-u=h$) $$\begin{aligned}\int_{u}^{v}(x-u)(v-x)dx =& \int_u^v ((v+u)x-uv-x^2)dx \\ =& \int_u^v((v+u)x-uv-x^2)dx \\ = & \left[\dfrac{(v+u)x^2}2-uvx-\dfrac{x^3}3\right]_u^v \\=& \dfrac{(v+u)(v^2-u^2)}2 - uv(v-u)-\dfrac{v^3-u^3}3 \\ \text{(taking $h=v-u$ common}) \ = & h\left(\dfrac{(v+u)(v+u)}2-uv-\dfrac{v^2+u^2+uv}3\right)\\ = & h\left(\dfrac{3v^2+3u^2+6uv-6uv-2v^2-2u^2-2uv}6\right)\\= & h\left(\dfrac{v^2+u^2-2vu}6\right)\\=& h\left(\dfrac{(v-u)^2}6\right)=h\cdot\dfrac{h^2}6=\dfrac{h^3}6\end{aligned}$$

Answer 2

Or shorter, set $x-x_{i-1}=sh$, $x_i-x=h-sh$, $s\in[0,1]$, then the integral becomes $$ \int_{0}^{1}(sh)(h-sh)d(sh)=h^3\int_0^1 s(1-s)ds=h^3(\tfrac12-\tfrac13)=\frac{h^3}{6}. $$