作业帮用 定积分的定义法求解这个 sinxdx (0
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用 定积分的定义法求解这个 sinxdx (0
把[0,π]平均分成n份,每小份取右端点作为函数的值
∑sin(kπ/n)*(π/n)(k从1到n)
=(π/n)/sin(π/n)∑sin(kπ/n)sin(π/n)
=(π/n)/sin(π/n)∑-1/2{[cos(k+1)π/n]-[cos(k-1)π/n]}
=(-1/2)(π/n)/sin(π/n){cosπ+cos[(n+1)π/n]-cos0-cos(π/n)}
lim(n→∞)∑sin(kπ/n)*(π/n)
=lim(n→∞)(-1/2)(π/n)/sin(π/n){cosπ+cos[(n+1)π/n]-cos0-cos(π/n)}
=(-1/2)*1(-1-1-1-1)=2
故∫[0,π]sinxdx=2
∑sin(kπ/n)*(π/n)(k从1到n)
=(π/n)/sin(π/n)∑sin(kπ/n)sin(π/n)
=(π/n)/sin(π/n)∑-1/2{[cos(k+1)π/n]-[cos(k-1)π/n]}
=(-1/2)(π/n)/sin(π/n){cosπ+cos[(n+1)π/n]-cos0-cos(π/n)}
lim(n→∞)∑sin(kπ/n)*(π/n)
=lim(n→∞)(-1/2)(π/n)/sin(π/n){cosπ+cos[(n+1)π/n]-cos0-cos(π/n)}
=(-1/2)*1(-1-1-1-1)=2
故∫[0,π]sinxdx=2