$\int\limits_{ - 1}^1 {{e^x}dx}$

Let $f\left( x \right) = {e^x}$ $\Rightarrow$ $\int\limits_{ - 1}^1 {f\left( x \right)dx} = \int\limits_{ - 1}^1 {{e^x}dx}$

Where $a = - 1,b = 1,nh = b - a = 1 + 1 = 2$

Now, $f\left( a \right) = f\left( { - 1} \right) = {e^{ - 1}}$

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

$f\left( {a + 2h} \right) = f\left( { - 1 + 2h} \right) = {e^{ - 1 + 2h}}$
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$f\left( {a + \left( {n - 1} \right)h} \right) = f\left( { - 1 + \left( {n - 1} \right)h} \right) = {e^{ - 1 + \left( {n - 1} \right)h}}$

By definition $\int\limits_a^b {f\left( x \right)dx}$

$= \mathop {\lim }\limits_{h \to 0} h\left[ {f\left( a \right) + f\left( {a + h} \right) + f\left( {a + 2h} \right) + ... + f\left( {a + \left( {n - 1} \right)h} \right)} \right]$

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

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

$= \mathop {\lim }\limits_{h \to 0} h \cdot \cfrac{1}{e} \cdot \cfrac{{1 \cdot \left( {{e^{nh}} - 1} \right)}}{{{e^h} - 1}} = \mathop {\lim }\limits_{h \to 0} \left( {\cfrac{1}{e}} \right) \cdot \cfrac{{1 \cdot \left( {{e^2} - 1} \right)}}{{\cfrac{{\left( {{e^h} - 1} \right)}}{h}}} = \cfrac{1}{e} \cdot \cfrac{{{e^2} - 1}}{{\log e}} = e - \cfrac{1}{e}$