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