Evaluate $\int\limits_0^1 {{e^{2 - 3x}}} dx$ as a limit of a sum.

: Let $I = \int\limits_a^b {{e^{2 - 3x}}} dx$

Here $a = 0,b = 1$ $f\left( x \right) = {e^{2 - 3x}}$

$h = \cfrac{{1 - 0}}{n} = \cfrac{1}{n}$ $\Rightarrow$ $nh = 1$

$f\left( a \right) = f\left( 0 \right) = {e^2}$

$f\left( {a + h} \right) = f\left( {0 + h} \right) = {e^{2 - 3h}}$

$f\left( {a + 2h} \right) = f\left( {0 + 2h} \right) = {e^{2 - 6h}}$

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$f\left( {a + \left( {n - 1} \right)h} \right) = f\left( {a + \left( {n - 1h} \right)} \right) = {e^{3\left( {n - 1} \right)h}}$

$\therefore$ $\int\limits_0^1 {f\left( x \right)dx = \mathop {\lim }\limits_{h \to 0} } h\left[ {f\left( a \right) + f\left( {a + h} \right) + .... + f\left( {a + \left( {n - 1} \right)h} \right)} \right]$

$= \mathop {\lim }\limits_{h \to 0} \left[ {{e^2} + {e^{2 - 3h}} + {e^{2 - 6h}} + .... + {e^{2 - 3\left( {n - 1} \right)h}}} \right]$

$= \mathop {\lim }\limits_{h \to 0} \left[ {{e^2}\left( {1 + {e^{ - 3h}} + {e^{ - 6h}} + .... + {e^{ - 3\left( {n - 1} \right)h}}} \right)} \right]$

$= \mathop {\lim }\limits_{h \to 0} \left[ {{e^2}\left( {\cfrac{{1\left( {{e^{ - 3hn}} - 1} \right)}}{{{e^{ - 3h}} - 1}}} \right)} \right]$

$= \mathop {\lim }\limits_{h \to 0} \left[ {{e^2}\left( {\cfrac{{h\left( {{e^{ - 3}} - 1} \right)}}{{{e^{ - 3h}} - 1}}} \right)} \right]$

$= \mathop {\lim }\limits_{h \to 0} \left[ {{e^2}\left( {\cfrac{{ - 3h}}{{{e^{ - 3h}} - 1}} \times \cfrac{{{e^{ - 3}} - 1}}{{ - 3}}} \right)} \right] = \cfrac{{{e^2} \times \left( {{e^{ - 3}} - 1} \right)}}{{ - 3}} = \cfrac{1}{3}\left( {{e^2} - \cfrac{1}{e}} \right)$

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