Question

\[\int\limits_0^3 \left( x + 4 \right) dx\]

Answer 1

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

\[Here, a = 0, b = 3, f\left( x \right) = x + 4, h = \frac{3 - 0}{n} = \frac{3}{n}\]
\[\text{Therefore, }I = \int_0^3 \left( x + 4 \right) d x\]
\[ = \lim_{h \to 0} h\left[ f\left( 0 \right) + f\left( 0 + h \right) + . . . . . . . + f\left( 0 + \left( n - 1 \right)h \right) \right]\]
\[ = \lim_{h \to 0} h\left[ \left( 0 + 4 \right) + \left( h + 4 \right) + . . . . . . . + \left( \left( n - 1 \right)h + 4 \right) \right]\]
\[ = \lim_{h \to 0} h\left[ 4n + h\left( 1 + 2 + . . . . . . . + \left( n - 1 \right) \right) \right]\]
\[ = \lim_{h \to 0} h\left[ 4n + h\frac{n\left( n - 1 \right)}{2} \right]\]
\[ = \lim_{n \to \infty} \frac{3}{n}\left[ 4n + \frac{3}{n} \times \frac{n\left( n - 1 \right)}{2} \right]\]
\[ = \lim_{n \to \infty} \left[ 12 + \frac{9}{2}\left( 1 - \frac{1}{n} \right) \right]\]
\[ = 12 + \frac{9}{2} = \frac{33}{2}\]