Question-61042

Question Number 61042 by aliesam last updated on 28/May/19

Answered by ajfour last updated on 28/May/19

l e t N △ x = 5 - 2 = 3

x = 2 + n △ x = 2 + \frac{3 n}{N}

\int_{2}^{5} (x^{3} - 1) d x = \lim_{N \to \infty} \frac{3}{N} \underset{n = 1}{\sum^{N}} {{(2 + \frac{3 n}{N})}^{3} - 1}

= \lim_{N \to \infty} \frac{3}{N} {7 N + \frac{36}{N} Σ n + \frac{54}{N^{2}} Σ n^{2} + \frac{27}{N^{3}} Σ n^{3}}

= 21 + \lim_{N \to \infty} \frac{3}{N} {36 \times \frac{(N + 1)}{2} + 54 \times \frac{(N + 1) (2 N + 1)}{6 N} + 27 \times \frac{{(N + 1)}^{2}}{4 N}}

= 21 + 54 + 54 + \frac{81}{4}

= 149 \frac{1}{4} .

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