# Why $\sum(x)$ is greater than $\int x \, dx$

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?

• You can interpret the sum as an area under the ‘curve’. Try to figure out what is this curve in your example. – lcv Dec 24 '18 at 10:06
• 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? – Sai Satwik Kuppili Dec 24 '18 at 10:15
• @SaiSatwikKuppili no the curve is a staricase – lcv Dec 24 '18 at 10:29

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.
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.