Question 438696
Note that if we were to approximate the integral {{{int(x^(1/3), dx, 1, 1000)}}}, we can use 1000 rectangles of width 1, and sum from the right-hand endpoints. This approximation is actually equal to {{{sum(i^(1/3), i=1,1000)}}}, since the base of each rectangle is 1 and the height is x^(1/3).


According to Wolfram Alpha,

{{{int(x^(1/3), dx, 1, 1000) = 7499.25}}}


{{{sum(i^(1/3), i=1,1000) = 7504.72}}} (rounded to two decimal places)