Question

Express \(\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to {\rm{\currency}}} \sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{sin}}} {{\rm{x}}_{\rm{i}}}{\rm{\Delta x}}\)as a definite integral on the interval \({\rm{(0,\pi )}}\) and then evaluate the integral.

Short answer

Value of integral

Step-by-step solution

Definition of definite integral

Definite integral is defined as the difference between the values of an integral for an upper value b and an of the independent variable x for a particular function f(x).

According to definition of definite integral, \(\int_a^b f (x)dx = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n f \left( {{x_i}} \right)\Delta x\)

Calculating definite integral for interval \({\rm{(0,}}\pi {\rm{)}}\)

\(f(x)\)is given as \(\sin x\).

\(\begin{aligned}{c}\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\sin } \left( {{x_i}} \right)\Delta x &= \int_0^\pi {\sin } xdx\\ &= ( - \cos x)_0^\pi \\ &= - \cos (\pi ) - ( - \cos (0))\\ &= - ( - 1) - ( - 1)\\ &= 1 + 1\\ &= 2\end{aligned}\)

Therefore, \(\mathop {\lim }\limits_{n \to \currency} \sum\limits_{i = 1}^n {\sin } {x_i}\Delta x\)expressed as \(\int_0^\pi {\sin } xdx\)and this integral evaluated as 2.