Evaluate ^ 2 _ -1 (e^ 3x + 7x - 5) dx as a limits of sum. - Vidyadip

Question

Evaluate $\int\limits^{2}_{-1}(\text{e}^{3\text{x}} + 7\text{x} - 5) \text{dx}$ as a limits of sum.

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Answer

$\int\limits^{2}_{-1}(\text{e}^{3\text{x}} + 7\text{x} - 5) \text{dx}$ $\text{here h} = \frac{3}{\text{n}}$
$\lim\limits_{h \rightarrow 0}\text{h}[f (-1) + \text{f} ( -1 + \text{h} + \dots\dots\dots]$
$\lim\limits_{h \rightarrow 0} \text{h}[(\text{e}^{-3} - 12) + e^{-3 + 3\text{h}} + 7\text{h} - 12) + \dots\dots\text{+}(\text{e}^{-3 + \overline{\text{n} - 1}\text{h}} + 7 (\text{n - 1) h -12)}]$
$\lim\limits_{h \rightarrow 0}\text{h}[\text{e}^{-3}(1 + \text{e}^{\text{3h}} + \text{e}^{\text{6h}} + \dots\dots\text{+}\text{e}^{\text{3(n - 1)h}}) + \text{7h} ( 1 + 2 + 3 + \dots\dots\overline{\text{n - 1}}) -\text{12 nh}]$
$= \lim\limits_{h \rightarrow 0}\text{h} \bigg[\frac{\text{e}^{-3(\text{e}^{\text{3nh}} - 1) \text{h}}}{\text{e}^{\text{3 h}}-1}+ \frac{\text{ 7(nh) (nh - h)}}{2} - \text{12nh}\bigg]$
$ = \frac{\text{e}^{-3}(\text{e}^{9 - 1})}{3} + \frac{63}{2} - 36 = \frac{\text{e}^{9} - 1}{3\text{ e}^{3}} - \frac{9}{2}$

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