Question

\[\int\limits_2^3 x^2 dx\]

Answer 1

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

\[\text{Here }a = 2, b = 3, f\left( x \right) = x^2 , h = \frac{3 - 2}{n} = \frac{1}{n}\]
Therefore,
\[I = \int_2^3 x^2 d x\]
\[ = \lim_{h \to 0} h\left[ f\left( 2 \right) + f\left( 2 + h \right) + . . . . . . . . . . . . . . . . . . . . + f\left\{ 2 + \left( n - 1 \right)h \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ 2^2 + \left( 2 + h \right)^2 + . . . . . . . . . . . + \left\{ 2 + \left( n - 1 \right)h \right\}^2 \right]\]
\[ = \lim_{h \to 0} h\left[ 4n + h^2 \left\{ 1^2 + 2^2 + 3^2 . . . . . . . . . + \left( n - 1 \right)^2 \right\} + 4h\left\{ 1 + 2 + . . . . . . + \left( n - 1 \right)h \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ 4n + h^2 \frac{n\left( n - 1 \right)\left( 2n - 1 \right)}{6} + 4h\frac{n\left( n - 1 \right)}{2} \right]\]
\[ = \lim_{n \to \infty} \frac{1}{n}\left[ 4n + \frac{\left( n - 1 \right)\left( 2n - 1 \right)}{6n} + 2n - 2 \right]\]
\[ = \lim_{n \to \infty} \left[ 6 + \frac{1}{6}\left( 1 - \frac{1}{n} \right)\left( 2 - \frac{1}{n} \right) - \frac{2}{n} \right]\]
\[ = 6 + \frac{1}{3}\]
\[ = \frac{19}{3}\]