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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS4 Marks
Question
$\text{Evaluate} \int\limits^{3}_{1} (e^{2} - 3x + x^{2} + 1)\text{dx as a limit of a sum.} $
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Answer

$\int\limits^{3}_{1}(e^{2 - 3x} + \text{x}^{2} + 1) \text{dx here h} = \frac{2}{\text{n}} $
$\lim\limits_{h\rightarrow{0}} \text{h}[\text{f(1) + f (1 + h) +f (1 + 2h) + }\dots\dots\dots\text{+ f (n - 1) h)] }$
$\lim\limits_{h\rightarrow{0}} \text{h}[(\text{e}^{-1} + 2) +\text{(e}^{-1 - 3\text{h}} + 2 + 2\text{h} +\text{h}^{2}) + (\text{e}^{-1-6\text{h}} + 2 + 4\text{h + 4h)}^{2}+ \dots\dots\dots$
$+ (\text{e}^{-1 - 3\text{(n - 1)h}} + 2 + 2 \text{(n - 1) h + (n - 1)}^{2}\text{h)}^{2}]$
$= \lim\limits_{h\rightarrow{0}} \text{h} \bigg[e^{-1}\bigg(1 + e^{-3\text{h}} + \text{e}^{-6\text{h}} + \dots\dots\text{+e}^{-3\text{(n- 1)h}}\bigg) + 2\text{n + 2h}\bigg(1 + 2 +\dots\dots\text{+ (n- 1)}\bigg) + \text{h}^{2}\bigg(1^{2} + 2^{2} +\dots\dots\text{+ (n - 1)}^{2}\bigg)\bigg]$
$= \lim\limits_{h\rightarrow{0}} \text{h} \bigg(e^{-1}.\frac{e^{\text-{3nh}} -1}{e^{-3} -1}. \text{h + 2nh + 2} \frac{\text{nh(nh -h)}}{2} + \frac{\text{nh (nh - h) (2nh -h)}}{6}\bigg)$
$= \text{e}^{-1}. \frac{(\text{e}^{-6} - 1)}{-3} + 4 + 4 + \frac{8}{3} = - \text{e}^{-1} \frac{(\text{e}^{-6} - 1)}{3} + \frac{32}{3}$

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