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Evaluate $\int_2^54 \,dx$ by the direct (Riemann) definition of the integral.

Let $2=x_0<x_1<\cdots<x_{n-1}<x_n=5$ be any partition of $[2,5]$, and let $\Delta_ix=x_i-x_{i-1}$. Then an approximating sum for $\int_5^24 \,dx$ is

$$\begin{align} \sum_{i=1}^nf(x_i^*)\Delta_ix &=\sum_{i=1}^n4\Delta_ix=4\left[\left(x_1-x_0\right)+\left(x_2-x_1\right)+\cdots+\left(x_n-x_{n-1}\right)\right]\\ &=4\left(x_n-x_0\right)=4(5-2)=4\cdot3=12 \end{align}$$

Could anyone explain how the $4\left[\left(x_1-x_0\right)+\left(x_2-x_1\right)+\cdots+\left(x_n-x_{n-1}\right)\right]$ becomes $4\left(x_n-x_0\right)$?

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    $\begingroup$ Look at the first two terms $x_{1}-x_{0}+x_{2}-x_{1}...$ this is the same as $x_{2}-x_{0}+...$ which means by extension, the only terms that will appear after the final sum will be $x_{n}-x_{0}$. $\endgroup$ – Autolatry Feb 19 '15 at 14:37
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If you look at the sum, everything else cancels out except for $x_0$ and $x_n$.

i.e.

$(x_1 - x_0) + (x_2 - x_1) = (x_1 - x_1) + (x_2 - x_0) = x_2 - x_1$

$(x_1 - x_0) + (x_2 - x_1) + (x_3 - x_2) = (x_1 - x_1) + (x_2 - x_2) + (x_3 - x_0) = x_3 - x_0$

......

$(x_1 - x_0) + (x_2 - x_1) + \ ...\ + (x_n - x_{n - 1}) = (x_1 - x_1) + (x_2 - x_2) +\ ...\ + (x_{n - 1} + x_{n - 1}) + (x_n - x_0) = x_n - x_0$.

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The terms cancel out except for $x_0$ and $x_n$ because it is telescoping.

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