Question

\[\int\limits_0^2 e^x dx\]

Answer 1

\[\int_a^b f\left( x \right) d x = \lim_{h \to 0} h\left[ f\left( a \right) + f\left( a + h \right) + f\left( a + 2h \right) . . . . . . . . . . . . . . . + f\left( a + \left( n - 1 \right)h \right) \right]\]
\[\text{where }h = \frac{b - a}{n}\]

\[\text{Here }a = 0, b = 2, f\left( x \right) = e^x , h = \frac{2 - 0}{n} = \frac{2}{n}\]
Therefore,
\[I = \int_0^2 e^x d x\]
\[ = \lim_{h \to 0} h\left[ f\left( 0 \right) + f\left( 0 + h \right) + . . . . . . . . . . . . . . . . . . . . + f\left\{ 0 + \left( n - 1 \right)h \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ e^0 + e^h + e^{2h} + . . . . . . . + e^\left( n - 1 \right)h \right]\]
\[ = \lim_{h \to 0} h\left[ \frac{\left( e^h \right)^n - 1}{e^h - 1} \right]\]
\[ = \lim_{h \to 0} \left[ \frac{e^2 - 1}{\frac{e^h - 1}{h}} \right]\]
\[ = \frac{e^2 - 1}{1}\]
\[ = e^2 - 1\]