summation to integral? calculus

[tex]\displaystyle\sum_{k=1}^n\left(\dfrac{3k}n-1\right)^4\frac3n[/tex]

You should recognize the [tex]\dfrac3n=\dfrac{b-a}n[/tex] term on the right to represent the width of each equally-spaced subinterval of the integration interval [tex][a,b][/tex]. We don't know the endpoints' exact values just yet.

[tex]n[/tex] represents the total number of subintervals. If we fix it to some manageably small number, we can get a better picture of what the [tex]\left(\dfrac{3k}n-1\right)^4[/tex] term represents.

If [tex]n=3[/tex], the partial sum reduces to

[tex]\displaystyle\sum_{k=1}^3\left(\dfrac{3k}3-1\right)^4\dfrac33=\sum_{k=1}^3(k-1)^4=0^4+1^4+2^4[/tex]

This suggests the sum represents the left-endpoint Riemann summation to find the integral of [tex]x^4[/tex], and in particular, the integral seems to be taken over the interval [0, 3]. Taking the limit, we get

[tex]\displaystyle\lim_{n\to\infty}\sum_{k=1}^n\left(\dfrac{3k}n-1\right)^4\frac3n=\int_0^3x^4\,\mathrm dx[/tex]