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प्रश्न
उत्तर
\[\int_a^b f\left( x \right) d x = \lim_{h \to 0} h\left[ f\left( a \right) + f\left( a + h \right) + f\left( a + 2h \right) . . . . . . . . . . . . . . . + f\left( a + \left( n - 1 \right)h \right) \right]\]
\[\text{where }h = \frac{b - a}{n}\]
\[\text{Here }a = 1, b = 3, f\left( x \right) = 3x - 2, h = \frac{3 - 1}{n} = \frac{2}{n}\]
Therefore,
\[I = \int_1^3 \left( 3x - 2 \right) d x\]
\[ = \lim_{h \to 0} h\left[ f\left( 1 \right) + f\left( 1 + h \right) + . . . . . . . . . . . . . . . . . . . . + f\left( 1 + \left( n - 1 \right)h \right) \right]\]
\[ = \lim_{h \to 0} h\left[ \left( 3 - 2 \right) + \left( 3 + 3h - 2 \right) + \left( 3 + 6h - 2 \right) . . . . . . . . . . . . . . . + \left( 3\left( n - 1 \right)h + 3 - 2 \right) \right]\]
\[ = \lim_{h \to 0} h\left[ n + 3h\left( 1 + 2 + 3 . . . . . . . . . + \left( n - 1 \right) \right) \right]\]
\[ = \lim_{h \to 0} h\left[ n + 3h\frac{n\left( n - 1 \right)}{2} \right]\]
\[ = \lim_{n \to \infty} \frac{2}{n}\left[ n + 3n - 3 \right]\]
\[ = \lim_{n \to \infty} 2\left( 4 - \frac{3}{n} \right)\]
\[ = 8\]
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