Find (x^ 2 + e^ 2x + 1) dx as the limit of a sum. - Vidyadip

Question

Find $\int\limits(\text{x}^{2} + e^{\text{2x}} + 1) \text{dx}$ as the limit of a sum.

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Answer

$\int\limits^{2}_{0}(\text{x}^{2} + \text{e}^{2\text{x} + 1})\text{dx}$
$\text{h} = \frac{2}{\text{h}}$
$\int\limits^{2}_{0}(\text{x}^{2} + \text{e}^{2\text{x} + 1})\text{dx} = \lim\limits_{x \rightarrow 0}\text{[h(0) + f (0 + h) + f( 0 + 2h)}+ \dots\dots\dots\dots$
$+\dots\dots\dots\dots\text{f(0 + n - 1) h ]}$
$= \lim\limits_{x \rightarrow 0}\text{h} [\text{h}^{2}( 1^{2} + 2^{2} + \dots\dots\dots \text{+} (\text{n} - 1)^{2}$
$+ \text{e}(1 + \text{e}^{2\text{h}} + \text{e}^{4\text{h}} + \dots\dots\dots\text{e}^{2(\text{n -1)h}})\bigg] $
$= \lim\limits_{\text{h} \rightarrow 0} \frac{\text{(nh)(nh - h)(2nh - h)}}{6} + \lim\limits_{\text{h} \rightarrow 0}\text{e.h}\bigg(\frac{\text{e}^{\text{2nh}} - 1}{\text{e}^{\text{2h}} - 1}\bigg)$
$= \frac{8}{3} + \frac{(\text{e}^{4} - 1)\text{e}}{2} = \frac{8}{3} + \frac{\text{e}^{5} - \text{e}}{2}$

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