Question

\[\int\limits_0^3 \left( 2 x^2 + 3x + 5 \right) dx\]

Answer 1

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

\[\text{Here }a = 0, b = 3, f\left( x \right) = 2 x^2 + 3x + 5, h = \frac{3 - 0}{n} = \frac{3}{n}\]
Therefore,
\[I = \int_0^3 \left( 2 x^2 + 3x + 5 \right) d x\]
\[ = \lim_{h \to 0} h\left[ f\left( 0 \right) + f\left( 0 + h \right) + . . . . . . . . . . . . . . . . . . . . + f\left\{ 0 + \left( n - 1 \right)h \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ \left( 0 + 0 + 5 \right) + \left( 2 h^2 + 3h + 5 \right) + . . . . . . . . . . . . . . . + \left\{ 2 \left( n - 1 \right)^2 h^2 + 3\left( n - 1 \right)h + 5 \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ 5n + 2 h^2 \left( 1^2 + 2^2 + 3^2 . . . . . . . . . + \left( n - 1 \right)^2 \right) + 3h\left\{ 1 + 2 + . . . . . . . + \left( n - 1 \right)h \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ 5n + 2 h^2 \frac{n\left( n - 1 \right)\left( 2n - 1 \right)}{6} + 3h\frac{n\left( n - 1 \right)}{2} \right]\]
\[ = \lim_{n \to \infty} \frac{3}{n}\left[ 5n + \frac{3\left( n - 1 \right)\left( 2n - 1 \right)}{n} + \frac{9\left( n - 1 \right)}{2} \right]\]
\[ = \lim_{n \to \infty} 3\left[ 5 + 3\left( 1 - \frac{1}{n} \right)\left( 2 - \frac{1}{n} \right) + \frac{9}{2}\left( 1 - \frac{1}{n} \right) \right]\]
\[ = 15 + 18 + \frac{27}{2}\]
\[ = \frac{93}{3}\]