How do you evaluate #int sin(x^2)dx# from -1 to 2?

1 Answer
Jul 31, 2017

I = 1.084

Explanation:

If you wrote the problem correctly, the answer
involves Fresnel Integrals and that is Graduate Level
Math for a Math Major.

I suspect you meant to write

#I = int (sin x)^2 dx , x = -1, 2#
#I = 1/2 [ x - sin x cos x ] + C#

For x = 2 radians
#I = 1/2 [ 2 - sin 2 cos 2 ] = 1/2 [ 2 - ( 0.909 ) ( - 0.416 ) ]#
#I = .811#

For x = - 1 radian
#I = 1/2 [ -1 - sin (-1) cos (-1) ] = 1/2 [ -1 - ( -0.84 ) ( 0.54 ) ]#
#I = - 0.273#
#I = 0.811 - ( - 0.273 ) = 1.084#