Question

\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]

Answer 1

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

 

\[\text{Here, }a = 1, b = 3, f\left( x \right) = 2 x^2 + 5x, h = \frac{3 - 1}{n} = \frac{2}{n}\]
Therefore,
\[I = \int_1^3 \left( 2 x^2 + 5x \right) d x\]
\[ = \lim_{h \to 0} h\left[ f\left( 1 \right) + f\left( 1 + h \right) + . . . . . . . . . . . . . . . . . . . . + f\left\{ 1 + \left( n - 1 \right)h \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ \left( 2 + 5 \right) + \left\{ 2 \left( 1 + h \right)^2 + 5\left( 1 + h \right) \right\} + . . . . . . . . . . . . . . . + \left\{ 2 \left( 1 + \left( n - 1 \right)h \right)^2 + 5\left( 1 + \left( n - 1 \right)h \right) \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ 2\left\{ 1^2 + \left( 1 + h \right)^2 + . . . . . . . . . . . . + \left\{ 1 + \left( n - 1 \right)h \right\}^2 \right\} + 5\left\{ 1 + \left( 1 + h \right) + \left( 1 + 2h + . . . . . . . . + \left( 1 + \left( n - 1 \right)h \right) \right) \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ 2n + 2 h^2 \left( 1^2 + 2^2 + 3^2 . . . . . . . . . + \left( n - 1 \right)^2 \right) + 4h\left\{ 1 + 2 + . . . . . . + \left( n - 1 \right) \right\} + 5n + 5h\left\{ 1 + 2 + . . . . . . + \left( n - 1 \right) \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ 7n + 2 h^2 \frac{n\left( n - 1 \right)\left( 2n - 1 \right)}{6} + 9h\frac{n\left( n - 1 \right)}{2} \right]\]
\[ = \lim_{n \to \infty} \frac{2}{n}\left[ 7n + \frac{4\left( n - 1 \right)\left( 2n - 1 \right)}{3n} + 9n - 9 \right]\]
\[ = \lim_{n \to \infty} 2\left[ 16 + \frac{4}{3}\left( 1 - \frac{1}{n} \right)\left( 2 - \frac{1}{n} \right) - \frac{9}{n} \right]\]
\[ = 32 + \frac{16}{3}\]
\[ = \frac{112}{3}\]