How to use the limit definition (Riemann sum) to evaluate the following integral ?

Aug 5, 2018

$I = {\int}_{0}^{2} \left({x}^{3} + x\right) \mathrm{dx} = 6$

Explanation:

We know that ,

$\left(1\right) {\int}_{a}^{b} f \left(x\right) \mathrm{dx} = {\lim}_{n \to \infty} h {\sum}_{i = 1}^{n} f \left(a + i h\right) , w h e r e , h = \frac{b - a}{n}$

We have ,

$I = {\int}_{0}^{2} \left({x}^{3} + x\right) \mathrm{dx} \implies f \left(x\right) = {x}^{3} + x , \mathmr{and} a = 0 , b = 2$

$f \left(a + i h\right) = f \left(0 + i h\right) = f \left(i h\right) = {\left(i h\right)}^{3} + \left(i h\right) = {i}^{3} {h}^{3} + i h$

$\therefore h = \frac{b - a}{n} = \frac{2 - 0}{n} = \frac{2}{n}$

$I = {\lim}_{n \to \infty} h {\sum}_{i = 1}^{n} \left\{{i}^{3} {h}^{3} + i h\right\}$

$= {\lim}_{n \to \infty} h \left\{{h}^{3} {\sum}_{i = 1}^{n} {i}^{3} + h {\sum}_{i = 1}^{n} i\right\}$

$= {\lim}_{n \to \infty} \frac{2}{n} \left\{\frac{8}{n} ^ 3 {\sum}_{i = 1}^{n} {i}^{3} + \frac{2}{n} {\sum}_{i = 1}^{n} i\right\} \ldots . \to \left[\because h = \frac{2}{n}\right]$

$= {\lim}_{n \to \infty} \frac{16}{n} ^ 4 {\sum}_{i = 1}^{n} {i}^{3} + {\lim}_{n \to \infty} \frac{4}{n} ^ 2 {\sum}_{i = 1}^{n} i$

$= {\lim}_{n \to \infty} \frac{16}{n} ^ 4 \cdot {n}^{2} / 4 {\left(n + 1\right)}^{2} + {\lim}_{n \to \infty} \frac{4}{n} ^ 2 \cdot \frac{n}{2} \left(n + 1\right)$

$= {\lim}_{n \to \infty} \frac{4}{n} ^ 2 {\left(n + 1\right)}^{2} + {\lim}_{n \to \infty} \frac{2}{n} \left(n + 1\right)$

$= 4 {\lim}_{n \to \infty} {\left(\frac{n + 1}{n}\right)}^{2} + 2 {\lim}_{n \to \infty} \left(\frac{n + 1}{n}\right)$

$= 4 {\lim}_{n \to \infty} {\left(1 + \frac{1}{n}\right)}^{2} + 2 {\lim}_{n \to \infty} \left(1 + \frac{1}{n}\right)$

$= 4 {\left(1 + 0\right)}^{2} + 2 \left(1 + 0\right) \ldots \to \left[\because {\lim}_{n \to \infty} \frac{1}{n} = 0\right]$

$= 4 + 2$

$= 6$

Note :
$\diamond {\sum}_{i = 1}^{n} i = \frac{n}{2} \left(n + 1\right)$

$\diamond {\sum}_{i = 1}^{n} {i}^{3} = {n}^{2} / 4 {\left(n + 1\right)}^{2}$