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I just noticed that $$\sum_{i=1}^n (a_i - a_{i-1})=a_n-a_0$$ and $$\int_a^b f'(x)\mathrm{d}x=f(b)-f(a)$$ look really similar. We can consider $a_i-a_{i-1}$ a discrete analog to the derivative of continuous functions.

Is there anything deep between those equations?

Gabriel
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1 Answers1

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Consider

$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}h$$

So we have

$$\int_a^bf'(x)dx=\lim_{h\to0}\int_a^b\frac{f(x+h)-f(x)}hdx$$

$$=\lim_{h\to0}\frac1h\left(\int_a^bf(x+h)-f(x)dx\right)$$

$$=\lim_{h\to0}\frac1h\left(\int_a^bf(x+h)dx-\int_a^bf(x)dx\right)$$

$$=\lim_{h\to0}\frac1h\left(\int_{a+h}^{b+h}f(x)dx-\int_a^bf(x)dx\right)$$

$$=\lim_{h\to0}\frac1h\left(\int_b^{b+h}f(x)dx-\int_a^{a+h}f(x)dx\right)\tag1$$

$$=f(b)-f(a)\tag2$$

You may graph or look at Riemann sums to see the last step above $(2)$.

Consider your telescoping sum for $(1)$.

Simply Beautiful Art
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  • Why can you "comute" the limit with the integral in your second equation? And I didn't understood why (1) is true or why it is related to a telescopic sum. – Gabriel Aug 10 '16 at 15:17
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    @GabrielRibeiro The limit is with respect to $h$, which is a constant with respect to $x$, and any constant can be comuted or factored out through an integral. $(1)$ is related to telescoping sums if you graph it (the gray area is where the integrals cancel, in a "telescoping" like manner): https://www.desmos.com/calculator/a4qbpc8ckb – Simply Beautiful Art Aug 11 '16 at 00:41
  • The graph was really helpful! Thanks – Gabriel Aug 11 '16 at 01:07