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

Question

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?

Answer 1

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.

Answer 2

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.