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प्रश्न
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
उत्तर
\[\text{Here }a = 0, b = 2, f\left( x \right) = x^2 + 2, h = \frac{2 - 0}{n} = \frac{2}{n}\]
Therefore,
\[ \int_0^2 \left( x^2 + 2 \right) d x = \lim_{h \to 0} h\left[ f\left( a \right) + f\left( a + h \right) + f\left( a + 2h \right) + . . . . . . . . . . . . + f\left( a + \left( n - 1 \right)h \right) \right]\]
\[ = \lim_{h \to 0} h\left[ f\left( 0 \right) + f\left( 0 + h \right) + . . . . . . . . . . + f\left( 0 + \left( n - 1 \right)h \right) \right]\]
\[ = \lim_{h \to 0} h\left[ 0 + 2 + \left( 0 + h \right)^2 + 2 + \left( 0 + 2h \right)^2 + 2 + . . . . . . . . . + \left( \left( n - 1 \right)h \right)^2 + 2 \right]\]
\[ = \lim_{h \to 0} h\left[ 2n + h^2 \left( 1^2 + 2^2 + . . . . . . . . . . . . . . \left( n - 1 \right)^2 \right) \right]\]
\[ = \lim_{h \to 0} h\left[ 2n + h^2 \frac{n\left( n - 1 \right)\left( 2n - 1 \right)}{6} \right]\]
\[ = \lim_{n \to 0 } \left[ 4 + \frac{4}{3}\left( 1 - \frac{1}{n} \right)\left( 2 - \frac{1}{n} \right) \right]\]
\[ = 4 + \frac{8}{3}\]
\[ = \frac{20}{3}\]
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