Answer:
[tex]\mathbf{4 \lim \limits _{n \to \infty} \sum \limits ^n_{i=1} \Big ( \dfrac{n(n+4i)}{2n^3 +(n+4i)^3} \Big )}[/tex]
Step-by-step explanation:
Given integral:
[tex]\int ^5_1 \dfrac{x}{2+x^3} \ dx[/tex]
[tex]\mathbf{Using \ Riemann \ sums; \ we \ have: }[/tex]
[tex]\int ^b_a \ f(x) \ dx = \lim_{n \to \infty} \sum \limits ^n_{i =1} \ f( a + i \Delta x) \Delta x[/tex]
[tex]here; \ \Delta x = \dfrac{b-a}{n}[/tex]
∴
[tex]\int ^5_1 \dfrac{x}{2+x^3} \ dx = f(x) = \dfrac{x}{2+x^3}[/tex]
[tex]\implies \Delta x = \dfrac{5-1}{n} =\dfrac{4}{n}[/tex]
[tex]f(a + i \Delta x ) = f ( 1 + \dfrac{4i}{n})[/tex]
[tex]f( 1 + \dfrac{4i}{n}) = \dfrac{n^2 ( n+4i)}{2n^3 + (n + 4i)^3}[/tex]
[tex]\lim_{n \to \infty} \sum \limits ^n_{i=1} \ f(a + i \Delta x) \Delta x = \lim_{n \to \infty} \sum \limits ^n_{i=1} \Big ( \dfrac{n^2(n+4i)}{2n^3 +(n+4i)^3} \Big )\dfrac{4}{n}[/tex]
[tex]\mathbf{= 4 \lim \limits _{n \to \infty} \sum \limits ^n_{i=1} \Big ( \dfrac{n(n+4i)}{2n^3 +(n+4i)^3} \Big )}[/tex]
