Kako koristite trapezoidno pravilo s n = 4 za procjenu integralnog int_0 ^ (pi / 2) cos (x ^ 2) dx?

Odgovor:

# Int_0 ^ (pi / 2) cos (x ^ 2) dx ~~ 0.83 #

Obrazloženje:

Trapezno pravilo nam govori da:

# Int_b ^ af (x) dx ~~ h / 2 f (x_0) + f (x_n) +2 f (x_1) + f (x_2) + cdotsf (x_ (n-1)) # gdje # H = (b-a) / n #

# H = (pi / 2-0) / 4 = pi / 8 #

Dakle, imamo:

# Int_0 ^ (pi / 2) cos (x ^ 2) dx ~~ pi / 16 f (0) + f (pi / 2) + 2 f (pi / 8) + f (pi / 4) + f ((3pi) / 8) #

# = Pi / 16 cos ((0) ^ 2) + cos ((pi / 2) ^ 2), +2 cos ((pi / 8) ^ 2) + cos ((pi / 4) ^ 2) + cos (((3pi) / 8) ^ 2) #

# ~~ pi / 16 1-0.78 + 1,97 + 1,63 + 0,36 #

# ~~ pi / 16 4.23 #

#~~0.83#