Show-that-the-function-x-x-3-is-of-Riemann-within-the-interval-1-2-then-calculate-1-2-x-2-dx-

Question Number 92799 by Ar Brandon last updated on 09/May/20

Show that the function x \to x^{3} is

of Riemann within the interval [- 1, 2]

then calculate \int_{- 1}^{2} x^{2} dx

Commented by mathmax by abdo last updated on 09/May/20

\int_{- 1}^{2} x^{2} d x = {[\frac{1}{3} x^{3}]}_{- 1}^{2} = \frac{1}{3} (2^{3} - {(- 1)}^{3}) = \frac{1}{3} (8 + 1) = 3

w e k n o w

\int_{a}^{b} f (x) d x = {l i m}_{n \to \infty} \frac{b - a}{n} \sum_{k = 1}^{n} f (a + \frac{k (b - a)}{n}) \Rightarrow

\int_{- 1}^{2} x^{2} d x =_{x = t - 1} \int_{0}^{3} {(t - 1)}^{2} d t = \int_{0}^{3} (t^{2} - 2 t + 1) d t

= \int_{0}^{3} t^{2} d t - 2 \int_{0}^{3} t d t + 3

\int_{0}^{3} t^{2} d t = {l i m}_{n \to + \infty} \frac{3}{n} \sum_{k = 1}^{n} {(\frac{3 k}{n})}^{2} = {l i m}_{n \to + \infty} \frac{27}{n^{3}} \times \frac{n (n + 1) (2 n + 1)}{6}

= {l i m}_{n \to + \infty} \frac{27}{6} \times \frac{2 n^{3}}{n^{3}} = 9

\int_{0}^{3} t d t = {l i m}_{n \to + \infty} \frac{3}{n} \sum_{k = 1}^{n} \frac{3 k}{n} = {l i m}_{n \to + \infty} \frac{9}{n^{2}} \times \frac{n (n + 1)}{2}

= \frac{9}{2} \Rightarrow \int_{- 1}^{2} x^{2} d x = 9 - 2 \times \frac{9}{2} + 3 = 3

Commented by Ar Brandon last updated on 10/May/20

thanks bro