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I studied that difference between integration and summation is Summation is evaluated on discrete values while integration is evaluated on continuous values. If we consider a strictly increasing function like $f(x)=x$, $\sum_{i=1}^2 x=3$ is greater than $\int_{i=1}^2 x=1.5$ but according to above statement Integration value should be more since in integration every small value is adding up to the sum. But it is not happening. Am I going wrong with basic definitions of Integration and summations?

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    $\begingroup$ You can interpret the sum as an area under the ‘curve’. Try to figure out what is this curve in your example. $\endgroup$ – lcv Dec 24 '18 at 10:06
  • $\begingroup$ Here, curve is y=x and area made by y=x with x-axis between the interval 1 to 2 is 1.5 which is equal to integral of x from 1 to 2. But my doubt is, does integral of x from 1 to 2 mean summation of x for 1,1.000001,1.000002,....2 is same? $\endgroup$ – Sai Satwik Kuppili Dec 24 '18 at 10:15
  • $\begingroup$ @SaiSatwikKuppili no the curve is a staricase $\endgroup$ – lcv Dec 24 '18 at 10:29
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Actually you are treating unfair and $\sum_{x=1}^2x$ (a sum of $2$ terms) should be compared with $\int_0^2xdx$ (an integral over $2$ intervals of length $1$).

But also then: $$\sum_{x=1}^2x=3>2=\int_0^2xdx$$ More generally: $$\sum_{x=1}^nx=\int_0^n\lceil x\rceil\;dx>\int_0^nx\;dx$$ because $\lceil x\rceil> x$ almost everywhere.

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You want an explanation of $\sum_{k=1}^n k>\int_0^n x dx=\sum_{k=1}^n\int_{k=1}^k xdx$. It suffices to show $k>\int_{k=1}^k xdx$, i.e. the height-$k$ width-$1$ rectangle is larger than the trapezium under $x$ from $x=k-1$ to $x=k$. From a diagram, that's trivial. More generally, integrating an increasing function gives such a rectangle-exceeds-area result.

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