Question

\[\int\limits_a^b e^x dx\]

Answer 1

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

\[\text{Here }a = a, b = b, f\left( x \right) = e^x , h = \frac{b - a}{n}\]

Therefore,

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