Question

\[\int\limits_3^5 \left( 2 - x \right) dx\]

Answer 1

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

\[\text{Here }a = 3, b = 5, f\left( x \right) = 2 - x, h = \frac{5 - 3}{n} = \frac{2}{n}\]

Therefore,

\[I = \int_3^5 \left( 2 - x \right) d x\]
\[ = \lim_{h \to 0} h\left[ f\left( 2 \right) + f\left( 2 + h \right) + . . . + f\left( 2 + \left( n - 1 \right)h \right) \right]\]
\[ = \lim_{h \to 0} h\left[ \left( 2 - 2 \right) + \left( 2 - h - 2 \right) + . . . + \left( 2 - \left( n - 1 \right)h - 2 \right) \right]\]
\[ = \lim_{h \to 0} h\left[ - h\left( 1 + 2 + 3 + . . . + \left( n - 1 \right) \right) \right]\]
\[ = \lim_{h \to 0} h\left[ - 2h\frac{n\left( n - 1 \right)}{2} \right]\]
\[ = \lim_{n \to \infty} \frac{2}{n}\left[ - 2n + 2 \right]\]
\[ = \lim_{n \to \infty} 2\left( - 2 + \frac{2}{n} \right)\]
\[ = - 4\]