高等数学2不会做，有没有学霸，求答案，非常谢谢

Question

如题所述

Answer 1

(1) x = 1-1/(1+t), x' = 1/(1+t)^2;
y = 1+1/t, y' = -1/t^2; z' =2t
当 t = 1 时， 曲线上的点为 (1/2, 2, 1),
切线向量是 {1/4, -1, 2},
切线方程是 (x-1/2)/(1/4) = (y-2)/(-1) = (z-1)/2
即 2(2x-1)/1 = (y-2)/(-1) = (z-1)/2
法平面方程是 (1/4)(x-1/2)-(y-2)+2(z-1) = 0
即 (2x-1)-8(y-2)+16(z-1) = 0，
2x-8y+16z-1 = 0
(2) I = 4∫<0,1>dx ∫<0,1>(x^2+y^2)dy
= 4∫<0,1>dx[x^2y+y^3/3]<y=0, y=1>
= 4∫<0,1>(x^2+1/3)dx = 4[x^3/3+x/3]<0, 1> = 8/3
(3) 原式 = lim<x→π>3cos3x/[5(sec5x)^2] = -3/(-5) = 3/5
(4) y = e^x/x, dy = [(xe^x-e^x)/x^2]dx = [(x-1)e^x/(x^2)]dx