How do you use the trapezoidal rule with n = 4 to estimate the integral #int_0^(pi/2)cos(x^2)dx#?

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Answer 1 · 2018-05-18T10:20:06

#int_0^(pi/2)cos(x^2)dx~~0.83#

The trapezoidal rule tells us that:
#int_b^af(x)dx~~h/2[f(x_0)+f(x_n)+2[f(x_1)+f(x_2)+cdotsf(x_(n-1))]]# where #h=(b-a)/n#

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

So we have:
#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#