Question

利用定积分的性质，估计积分∫【-(√3)/3到√3】xarctanxdx

Answer 1

∫[-(√3)/3,√3]xarctanxdx

=1/2∫[-(√3)/3,√3]arctanxdx^2
=1/2x^2arctanx[-(√3)/3,√3]-1/2∫[-(√3)/3,√3]x^2darctanx
=1/2x^2arctanx[-(√3)/3,√3]-1/2∫[-(√3)/3,√3]x^2/(1+x^2)dx
=1/2x^2arctanx[-(√3)/3,√3]-1/2∫[-(√3)/3,√3][1-1/(1+x^2)]dx
=1/2(x^2arctanx-x+arctanx)[-(√3)/3,√3]
=1/2(3*π/3-√3+π/3+1/3*π/6-√3/3+π/6［
=11π/18-2√3/3

Answer 2

精确值是7π/9-2/√3~1.288760，用以对比的
现在来估算。
梯形法：
取n=4，Δx=[√3-(-√3/3)]/4=1/√3
∫(-√3/3到√3) xarctanx dx
~(Δx/2)(y0+2y1+2y2+2y3+y4)
=[1/(2√3)][f(a)+2f(a+Δx)+2f(a+2Δx)+2f(a+3Δx)+f(b)]
=[1/(2√3)][0.302+0+0.6046+1.978+1.81

辛普森法则：
取n=4，Δx=[√3-(-√3/3)]/4=1/√3
∫(-√3/3到√3) xarctanx dx
~(Δx/3)[y0+4y1+2y2+4y3+y4]
=[1/(3√3)][f(a)+4f(a+Δx)+2f(a+2Δx)+4f(a+3Δx)+f(b)]
=[1/(3√3)][

误差就不算了。