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प्रश्न
\[\int\limits_2^3 e^{- x} dx\]
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उत्तर
\[\text{Here }a = 2, b = 3, f\left( x \right) = e^{- x} , h = \frac{3 - 2}{n} = \frac{1}{n}\]
Therefore,
\[ \int_2^3 e^{- x} d x = \lim_{h \to 0} h\left[ f\left( a \right) + f\left( a + h \right) + f\left( a + 2h \right) + . . . . . . . . . . . . + f\left( a + \left( n - 1 \right)h \right) \right]\]
\[ = \lim_{h \to 0} h\left[ f\left( 2 \right) + f\left( 2 + h \right) + . . . . . . . . . . + f\left( 2 + \left( n - 1 \right)h \right) \right]\]
\[ = \lim_{h \to 0} h\left[ e^{- 2} + e^{- \left( 2 + h \right)} + e^{- \left( 2 + 2h \right)} + . . . . . . . + e^{- \left( 2 + \left( n - 1 \right)h \right)} \right]\]
\[ = \lim_{h \to 0} h e^{- 2} \left[ \frac{\left( e^{- h} \right)^n - 1}{e^{- h} - 1} \right]\]
\[ = \lim_{h \to 0} e^{- 2} \left[ \frac{e^{- 1} - 1}{\frac{e^{- h} - 1}{- h}} \right] \times - 1 ....................\left(\text{Since nh = 1 }\right)\]
\[ = \left( e^{- 2} - e^{- 3} \right)\]
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