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

Question

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

Aurora Alvarado · Accepted Answer

I = 1.084

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#

Ezra Adams · Answer

undefined To evaluate ∫sin(x^2)dx from -1 to 2, you can use numerical methods such as the trapezoidal rule, Simpson's rule, or Gaussian quadrature. These methods approximate the integral by dividing the interval [-1, 2] into smaller subintervals and approximating the function within each subinterval. Then, the integral is approximated by summing up the contributions from all the subintervals. The choice of method depends on the desired level of accuracy and the computational resources available.