Write the limit as n goes to infinity of the summation from k equals 1 of the product of the 4th power of the quantity negative 1 plus 3 times k over n and 3 over n as a definite integral.

Answer 1

so, it appears you have the sum
n
∑ (-1+3/n k)(3/n)
k=1

It appears we have divided the interval of width 3 into n parts. That would make the limit
∫[-1,2] x dx