Question

How do you evaluate the definite integral `int(-sinx)dx` from `[pi/6,pi/2]`?

Answer 1

`int_(pi/6)^(pi/2)(-sinx)dx`

`=[cosx]_(pi/6)^(pi/2)`

`=cos(pi/2)-cos(pi/6)`

`=0-sqrt3/2`

`=-sqrt3/2`