Question

\[\int\limits_0^5 \left( x + 1 \right) dx\]

Answer 1

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

\[\text{Here, }a = 0, b = 5, f\left( x \right) = x + 1, h = \frac{5 - 0}{n} = \frac{5}{n}\]
Therefore,
\[I = \int_0^5 \left( x + 1 \right) d x\]
\[ = \lim_{h \to 0} h\left[ f\left( 0 \right) + f\left( 0 + h \right) + . . . . . . . . . . . . . . . . . . . . + f\left\{ 0 + \left( n - 1 \right)h \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ \left( 0 + 1 \right) + \left( h + 1 \right) + . . . . . . . . . . . . . . . + \left\{ \left( n - 1 \right)h + 1 \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ n + h\left\{ 1 + 2 + 3 . . . . . . . . . + \left( n - 1 \right) \right\} \right]\]
\[ = \lim_{h \to 0} h\left[ n + h\frac{n\left( n - 1 \right)}{2} \right]\]
\[ = \lim_{n \to \infty} \frac{5}{n}\left[ n + \frac{5n - 5}{2} \right]\]
\[ = \lim_{n \to \infty} 5\left( \frac{7}{2} - \frac{5}{n} \right)\]
\[ = \frac{35}{2}\]