Integrating sum 
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*$\displaystyle \sum_{k=1}^n k=\dfrac{n(n+1)}{2}$


Which is equivalent of saying, $$1+2+3+4+5+\cdots+n=\dfrac{n(n+1)}{2}$$
Now how do I integrate the left side of the equality and how that would look like?
Integrating the right side gives $\int \dfrac{n(n+1)}{2} \, \mathrm dn =\dfrac{1}{2}(\dfrac{n^3}{3}+\dfrac{n^2}{2})$
Which is quite similar to the sum $\dfrac{1}{2}\displaystyle \sum_{k=1}^n k^2=\dfrac{1}{2}(\dfrac{n^3}{3}+\dfrac{n^2}{2}+\dfrac{n}{6})$
Can someone explain these?

I'm choosing to interpret "taking the integral" as finding an anti-derivative, in the sense of indefinite integrals. So, we're looking for a function $F$ such that
$$F'(n) = 1 + 2 + \ldots + n.$$
In that sense, $\frac{n^3}{6} + \frac{n^2}{4}$ fits the bill; it's a function whose derivative agrees with the sum at integer points.
It is worth noting that we'll never get a function that agrees with the sum at non-integer points, as the sum doesn't make sense for non-integer points. Note that
$$1 + 2 + \ldots + 7.5$$
is a poorly-formed expression. The derivative of a function must be defined on an interval, including non-integer points (except if it's defined at a single point!). So, $\frac{n^3}{6} + \frac{n^2}{4} + C$ is about the best you're going to do.
Note also that, with the sum $\frac{1}{2} \sum_{k=1}^n k^2$, you are not integrating with respect to $n$. You are integrating the dummy variable $k$, which doesn't seem to have any conceptual connection with the original sum.
