Question

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

Answer 1

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

\[Here\ a = 0, b = 2, f\left( x \right) = x + 3, h = \frac{2 - 0}{n} = \frac{2}{n}\]
Therefore,
\[I = \int_0^2 \left( x + 3 \right) d x\]
\[ = \lim_{h \to 0} h\left[ f\left( 0 \right) + f\left( 0 + h \right) + . . . . . . . . . . . . . . . . . . . . + f\left( 0 + \left( n - 1 \right)h \right) \right]\]
\[ = \lim_{h \to 0} h\left[ \left( 0 + 3 \right) + \left( 0 + h + 3 \right) + . . . . . . . . . . . . . . . + \left( 0 + \left( n - 1 \right)h + 3 \right) \right]\]
\[ = \lim_{h \to 0} h\left[ 3n + h\left\{ 1 + 2 + 3 . . . . . . . . . + \left( n - 1 \right) \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ 3n + h\frac{n\left( n - 1 \right)}{2} \right]\]
\[ = \lim_{n \to \infty} \frac{2}{n}\left[ 3n + n - 1 \right]\]
\[ = \lim_{n \to \infty} 2\left( 4 - \frac{1}{n} \right)\]
\[ = 8\]