Question

\[\int\limits_1^2 \left( x^2 - 1 \right) dx\]

Answer 1

\[\int_a^b f\left( x \right) d x = \lim_{h \to 0} h\left[ f\left( a \right) + f\left( a + h \right) + f\left( a + 2h \right) . . . . . . . . . . . . . . . + f\left( a + \left( n - 1 \right)h \right) \right]\]
\[\text{where }h = \frac{b - a}{n}\]

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