作业帮一道关于大一微积分的问题.
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/05/05 04:14:40
一道关于大一微积分的问题.
lim[n趋向于无穷]π/n(cos^2 π/n+cos^2 2π/n+...+cos^2 (n-1)π/n)=?
求教……!
lim[n趋向于无穷]π/n(cos^2 π/n+cos^2 2π/n+...+cos^2 (n-1)π/n)=?
求教……!
解(定积分法)：
原式=lim(n->∞)[π/n(cos²(π/n)+cos²(2π/n)+...+cos²((n-1)π/n))]
=lim(n->∞)[π/n(cos²(π/n)+cos²(2π/n)+...+cos²((n-1)π/n)+cos²(nπ/n)-cos²(nπ/n))] (添加cos²(nπ/n)项)
=lim(n->∞)[π/n(cos²(π/n)+cos²(2π/n)+...+cos²((n-1)π/n)+cos²(nπ/n))]+lim(n->∞)[(π/n)cos²(nπ/n)]
=π∫(0,1)cos²(πx)dx+lim(n->∞)(π/n) (∫(0,1)表示从0到1积分)
=π/2∫(0,1)[1+cos(2πx)]dx
=π/2[x+sin(2πx)/(2π)]|(0,1)
=π/2(1+0-0-0)
=π/2.
原式=lim(n->∞)[π/n(cos²(π/n)+cos²(2π/n)+...+cos²((n-1)π/n))]
=lim(n->∞)[π/n(cos²(π/n)+cos²(2π/n)+...+cos²((n-1)π/n)+cos²(nπ/n)-cos²(nπ/n))] (添加cos²(nπ/n)项)
=lim(n->∞)[π/n(cos²(π/n)+cos²(2π/n)+...+cos²((n-1)π/n)+cos²(nπ/n))]+lim(n->∞)[(π/n)cos²(nπ/n)]
=π∫(0,1)cos²(πx)dx+lim(n->∞)(π/n) (∫(0,1)表示从0到1积分)
=π/2∫(0,1)[1+cos(2πx)]dx
=π/2[x+sin(2πx)/(2π)]|(0,1)
=π/2(1+0-0-0)
=π/2.