# (a) Determine Δx and xi. Δx= xi= (b) Using the definition mentioned above, evaluate the integral. Value of integral: ?

Aug 30, 2015

#### Explanation:

$\Delta x = \frac{b - a}{n} = \frac{2 - 1}{n} = \frac{1}{n}$

${x}_{i} = a + i \Delta x$ for $i = 0 , 1 , 2 , 3 , . . . n$

${x}_{i} = 1 + \frac{i}{n}$

$f \left({x}_{i}\right) = 2 {\left(1 + \frac{i}{n}\right)}^{3} = 2 \left(1 + 3 \frac{i}{n} + 3 {i}^{2} / {n}^{2} + {i}^{3} / {n}^{3}\right)$

${\sum}_{i = 1}^{n} f \left({x}_{i}\right) \Delta x$

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

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

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

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

$= 2 {\sum}_{i = 1}^{n} \frac{1}{n} + 2 {\sum}_{i = 1}^{n} 3 \frac{i}{n} ^ 2 + 2 {\sum}_{i = 1}^{n} 3 {i}^{2} / {n}^{3} + 2 {\sum}_{i = 1}^{n} {i}^{3} / {n}^{4}$

$= \frac{2}{n} {\sum}_{i = 1}^{n} 1 + \frac{6}{n} ^ 2 {\sum}_{i = 1}^{n} i + \frac{6}{n} ^ 3 {\sum}_{i = 1}^{n} {i}^{2} + \frac{2}{n} ^ 4 {\sum}_{i = 1}^{n} {i}^{3}$

$= \frac{2}{n} \left(n\right) + \frac{6}{n} ^ 2 \left(\frac{n \left(n + 1\right)}{2}\right) + \frac{6}{n} ^ 3 \left(\frac{n \left(n + 1\right) \left(2 n + 1\right)}{6}\right) + \frac{2}{n} ^ 4 \left(\frac{{n}^{2} {\left(n + 1\right)}^{2}}{4}\right)$

$= 2 + 3 \left(\frac{n \left(n + 1\right)}{n} ^ 2\right) + \left(\frac{n \left(n + 1\right) \left(2 n + 1\right)}{n} ^ 3\right) + \frac{1}{2} \left(\frac{{n}^{2} {\left(n + 1\right)}^{2}}{n} ^ 4\right)$

So

${\int}_{0}^{1} 2 {x}^{3} \mathrm{dx} = {\lim}_{n \rightarrow \infty} {\sum}_{i = 1}^{n} f \left({x}_{i}\right) \Delta x$

 = lim_(nrarroo) ( 2+3((n(n+1))/n^2) + ((n(n+1)(2n+1))/n^3)+ 1/2 ((n^2(n+1)^2)/n^4)

$= 2 + \left(3\right) + \left(2\right) + \frac{1}{2} \left(1\right) = 7 \frac{1}{2}$