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

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

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

\[ = {e^2} – 1 = {e^2} – 1\]

Buy Best Mathematics E-Books Visit : https://mathstudy.in/

Buy Mathematics Formula Book for Class XI,XII,JEE and other Engineering Competition Exam https://mathstudy.in/product/mathematics-formula-book/

Buy Mathematics Workbook for Class XII ( Fully Solved ) : https://mathstudy.in/product/work-book-class-xii-c-b-s-e-fully-solved/

Buy Mathematics Chapter Tests for Class XII ( Fully Solved) : https://mathstudy.in/product/mathematics-chapter-tests-class-xii-c-b-s-e/

Buy Objective Type Question Bank Class XII (Fully Solved ) : https://mathstudy.in/product/objective-type-question-bank-for-mathematics-class-xii-c-b-s-e/