Question
Evaluate $\int\limits_1^3(\text{2x}^{2}+\text{5x})$ dx as a limit of a sum. 

Answer

$\int\limits_1^3(\text{2x}^{2}+\text{5x})\text{dx}=\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\text{h}[\text{f(1) + f(1 + h) + f(1 + 2 h)+..........+ f(1 }+\overline{\text{n - 1}}\text{ h)}]$
where $f(x) = 2x^2 + 5x$ and $h = \frac{2}{\text{n}}$ or nh $2$
$f(1) = 7$
$f(1 + h) = 2 (1 + h)^2 + 5 (1 + h) = 7 + 9h + 2h^2$
$f(1 + 2h) = 2 (1 + 2h)^2 + 5 (1 + 2h) = 7 + 18h + 2.2^2h^2$
$f(1 + 3h) = 2 (1 + 3h)^2 + 5 (1 + 3h) = 7 + 27h + 2.3^2h^2$
$f(1 + (n – 1) h) = 7 + 9 (n – 1) h + 2.(n – 1)^2 h^{2.}$
$\text{I}=\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\Bigg[\text{h}[\text{7n + 9h}\frac{\text{n(n-1)}}{{2}}+\text{2h}^{2}\cdot\frac{\text{n(n-1)(2n-1)}}{6}\Bigg]$
$=\DeclareMathOperator*{\median}{\text{lim}} \median_{\text{h}\rightarrow0}\Bigg[\text{7nh}+\frac{9}{2}\text{nh (nh - h)}+\frac{1}{3}\text{nh (nh - h)(2nh - h)}\Bigg]$
$= 14+18+\frac{16}{3}=\frac{112}{3}$.

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