Question

73-74 express the limit as a definite integral.

Answer 1

Pull out a factor of 1/n from the summand and the result follows.

$\displaystyle \lim_{n\to\infty} \sum_{i=1}^n \frac{i^4}{n^5} = \lim_{n\to\infty} \frac1n \sum_{i=1}^n \left(\frac in\right)^4 = \boxed{\int_0^1 x^4 \, dx}$