prove-that-the-function-f-x-x-2-x-1-4-is-Riemannian-integral-

Question Number 143193 by mohammad17 last updated on 11/Jun/21

p r o v e t h a t t h e f u n c t i o n f (x) = x^{2}, x ε [1, 4]

i s R i e m a n n i a n i n t e g r a l ?

Answered by mathmax by abdo last updated on 11/Jun/21

\int_{1}^{4} f (x) dx = \int_{1}^{4} x^{2} dx = {[\frac{x^{3}}{3}]}_{1}^{4} = \frac{1}{3} (4^{3} - 1) = \frac{1}{3} (63) = \frac{63}{3} = 21

\lim_{n \to + \infty} \frac{4 - 1}{n} \sum_{k = 1}^{n} f (1 + (4 - 1) \times \frac{k}{n})

= \lim_{n \to + \infty} \frac{3}{n} \sum_{k = 1}^{n} f (1 + \frac{3 k}{n})

= \lim_{n \to + \infty} \frac{3}{n} \sum_{k = 1}^{n} {(1 + \frac{3 k}{n})}^{2} = \lim_{n \to + \infty} \frac{3}{n} \sum_{k = 1}^{n} (1 + \frac{6 k}{n} + \frac{{9 k}^{2}}{n^{2}})

= 3 + \lim_{n \to + \infty} \frac{18}{n^{2}} \sum_{k = 1}^{n} k + \lim_{n \to + \infty} \frac{27}{n^{3}} \sum_{k = 1}^{n} k^{2}

= 3 + \lim_{n \to + \infty} \frac{18}{n^{2}} \frac{n (n + 1)}{2} + \lim_{n \to + \infty} \frac{27}{n^{3}} \frac{n (n + 1) (2 n + 1)}{6}

= 3 + 9 + 9 = 21 \Rightarrow f is Rieman integrable .