How do you use the trapezoid rule for #int 2 sin x^2 dx# from x = 0 to x = 1/2 with n = 4?

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Answer 1 · 2015-06-19T23:27:12

#int_0^(1/2) 2 sin x^2 dx~~T_4~~0.114489#

First of all let's calculate 4 sub-intervals of [0,1/2]:
#[0,1/8],[1/8,1/4],[1/4,3/8],[3/8,1/2]#

Now, let's apply the trapezoidal rule to these interval:
#T_4=[2sin0^2+2*2sin(1/8)^2+2*2sin(1/4)^2+2*2sin(3/8)^2+2sin(1/2)^2]*((1/2-0)/8)#

#T_4=[2*0+4sin(1/64)+4sin(1/16)+4sin(9/64)+2sin(1/4)]*1/16#

#T_4~~[0.062497+0.249837+0.560647+0.958851]*1/16#

#T_4~~[1.831832]*1/16=0.114489#