Question

\[\int\limits_2^3 \left( 2 x^2 + 1 \right) dx\]

Answer 1

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

\[\text{Here }a = 2, b = 3, f\left( x \right) = 2 x^2 + 1, h = \frac{3 - 2}{n} = \frac{1}{n}\]
Therefore,
\[I = \int_2^3 \left( 2 x^2 + 1 \right) d x\]
\[ = \lim_{h \to 0} h\left[ f\left( 2 \right) + f\left( 2 + h \right) + . . . . . . . . . . . . . . . . . . . . + f\left\{ 2 + \left( n - 1 \right)h \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ 2\left( 2 . 2^2 \right) + 1 + \left\{ 2 \left( 2 + h \right)^2 + 1 \right\} + . . . . . . . . . . . . . . . + \left\{ 2 \left( \left( 2 + n - 1 \right)h \right)^2 + 1 \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ n + 2\left\{ 2^2 + \left( 2 + h \right)^2 + . . . . . . . . . . . . . \left( \left( 2 + n - 1 \right)h \right)^2 \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ n + 8n + 2 h^2 \left\{ 1^2 + 2^2 + 3^2 . . . . . . . . . + \left( n - 1 \right)^2 \right\} + 8h\left\{ 1 + 2 + . . . . . . . + \left( n - 1 \right) \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ 9n + h^2 \frac{2n\left( n - 1 \right)\left( 2n - 1 \right)}{6} + 8h\frac{n\left( n - 1 \right)}{2} \right]\]
\[ = \lim_{n \to \infty} \frac{1}{n}\left[ 9n + \frac{\left( n - 1 \right)\left( 2n - 1 \right)}{3n} + 4n - 4 \right]\]
\[ = \lim_{n \to \infty} \left\{ 13 + \frac{1}{3}\left( 1 - \frac{1}{n} \right)\left( 2 - \frac{1}{n} \right) - \frac{4}{n} \right\}\]
\[ = 13 + \frac{2}{3} = \frac{41}{3}\]