Using limit of sum evaluate the integral \[\int_0^2 {{e^x}} dx\] ~~~~~[NCERT Exemp. Q. 28,Page 165 ]

Let \[ I = \int_0^2 {{e^x}} dx\]


Here, \[ a = 0\] and \[ b = 2\]


therefore,\[ h = \frac{{b – a}}{n}\]


\[ \Rightarrow \] \[ nh = 2\] and \[ f(x) = {e^x}\]


Now, \[ \int_0^2 {{e^x}} dx = \mathop {\lim }\limits_{h \to 0} h[f(0) + f(0 + h) + f(0 + 2h) + \ldots + f\{ 0 + (n – 1)h\} ]\]

therefore,\[ I = \mathop {\lim }\limits_{h \to 0} h\left[ {1 + {e^h} + {e^{2h}} + \ldots + {e^{(n – 1)h}}} \right]\]


\[ = \mathop {\lim }\limits_{h \to 0} h\left[ {\frac{{1 \cdot {{\left( {{e^h}} \right)}^n} – 1}}{{{e^h} – 1}}} \right] = \mathop {\lim }\limits_{h \to 0} h\left( {\frac{{{e^{nh}} – 1}}{{{e^h} – 1}}} \right)\]


therefore,\[ I = \mathop {\lim }\limits_{h \to 0} h\left[ {1 + {e^h} + {e^{2h}} + \ldots + {e^{(n – 1)h}}} \right]\]

\[ = \mathop {\lim }\limits_{h \to 0} h\left[ {\frac{{1 \cdot {{\left( {{e^h}} \right)}^n} – 1}}{{{e^h} – 1}}} \right] = \mathop {\lim }\limits_{h \to 0} h\left( {\frac{{{e^{nh}} – 1}}{{{e^h} – 1}}} \right)\]


\[ = \mathop {\lim }\limits_{h \to 0} h\left( {\frac{{{e^2} – 1}}{{{e^h} – 1}}} \right)\]


\[ = {e^2} – 1 = {e^2} – 1\]

 

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