Question

\[\int\limits_1^4 \left( 3 x^2 + 2x \right) dx\]

Answer 1

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

\[\text{Here }a = 1, b = 4, f\left( x \right) = 3 x^2 + 2x, h = \frac{4 - 1}{n} = \frac{3}{n}\]
Therefore,
\[I = \int_1^4 \left( 3 x^2 + 2x \right) d x\]
\[ = \lim_{h \to 0} h\left[ f\left( 1 \right) + f\left( 1 + h \right) + . . . . . . . . . . . . . . . . . . . . + f\left( 1 + \left( n - 1 \right)h \right) \right]\]
\[ = \lim_{h \to 0} h\left[ \left( 3 . 1^2 + 2 \times 1 \right) + \left( 3 \left( 1 + h \right)^2 + 2\left( 1 + h \right) \right) + . . . . . . . . . . . . . . . + \left\{ 3 \left( 1 + \left( n - 1 \right)h \right)^2 + 2\left( 1 + \left( n - 1 \right)h \right) \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ 3\left\{ 1^2 + \left( 1 + h \right)^2 + \left( 1 + 2h \right)^2 + . . . . . . . . . . . + \left( 1 + \left( n - 1 \right)h \right)^2 \right\} + 2\left\{ 1 + \left( 1 + h \right) + . . . . . . . . . . + \left( 1 + \left( n - 1 \right)h \right) \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ 3n + 3 h^2 \left\{ 1^2 + 2^2 + 3^2 . . . . . . . . . + \left( n - 1 \right)^2 \right\} + 6h\left\{ 1 + 2 + . . . . . . . . . \left( n - 1 \right)h \right\} + 2n + 2h\left\{ 1 + 2 + . . . . . . . . . . + \left( n - 1 \right)h \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ 5n + 3 h^2 \frac{n\left( n - 1 \right)\left( 2n - 1 \right)}{6} + 8h\frac{n\left( n - 1 \right)}{2} \right]\]
\[ = \lim_{n \to \infty} \frac{3}{n}\left[ 5n + \frac{9\left( n - 1 \right)\left( 2n - 1 \right)}{2n} + 12n - 12 \right]\]
\[ = \lim_{n \to \infty} 3\left[ 17 - \frac{12}{n} + \frac{9}{2}\left( 1 - \frac{1}{n} \right)\left( 2 - \frac{1}{n} \right) \right]\]
\[ = 51 + 27\]
\[ = 78\]