Question

\[\int\limits_{- 1}^1 e^{2x} dx\]

Answer 1

\[\text{Here }a = - 1, b = 1, f\left( x \right) = e^{2x} , h = \frac{1 + 1}{n} = \frac{2}{n}\]

Therefore,

\[ \int_{- 1}^1 e^{2x} d x = \lim_{h \to 0} h\left[ f\left( a \right) + f\left( a + h \right) + f\left( a + 2h \right) + . . . . . . . . . . . . + f\left( a + \left( n - 1 \right)h \right) \right]\]

\[ = \lim_{h \to 0} h\left[ f\left( - 1 \right) + f\left( - 1 + h \right) + . . . . . . . . . . + f\left( - 1 + \left( n - 1 \right)h \right) \right]\]

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

\[ = \lim_{h \to 0} h e^{- 2} \left[ \frac{\left( e^{2h} \right)^n - 1}{e^{2h} - 1} \right]\]

\[ = \lim_{h \to 0} e^{- 2} \left[ \frac{e^4 - 1}{\frac{e^{2h} - 1}{2h}} \right] \times \frac{1}{2} .......................\left(\text{Since, nh = 2 }\right)\]

\[ = \frac{1}{2}\left( e^2 - e^{- 2} \right)\]