: 1. 1 : ∫((1-x^7)^(1/3)-(1-x^3)^(1/7))dx : 0 1 1 (1-x^7)^(1/3) 1 (1-y^3)^(1/7) 1 ∫(1-x^7)^(1/3) dx = ∫∫ dydx = ∫∫ dxdy = ∫(1-y^3)^(1/7) dy 0 0 0 0 0 0 所以说 原式 = 0 : 2. lim{∫([bx+a(1-x)]^n)dx)}^(1/n),b>a>0. : n→0 [(b-a)x + a]^(n+1) ∫[(b-a)x + a]^n dx = -------------------- + c (n+1)(b-a) 这题是不是有缺条件啊...? : 3.一曲面x^2/4+y^2+z^2/9=3,求其在(-2,1,3)这点的tangent plane和normal line. f(x) = x^2/4 + y^2 + z^2/9 - 3, del[f(x)] = (x/2, 2y, 2z/9) del[f(-2,1,3)] = (-1,2,2/3) = (-3,6,2) tangent plane: -3(x+2) + 6(y-1) + 2(z-3) = 0, 3x-6y-2z+18=0 x+2 y-1 z-3 normal line: ----- = ----- = ----- -3 6 2 : 4.一个碗里面有水,水蒸发的速率和水面面积(就是水和空气接触的面积)成正比,请证明水 : 面下降的速率(水深减少的速率)为定值(和碗的形状无关). -dV -Adz -dz ----- = kA, ------ = kA, ----- = v = k, k为一常数 dt dt dt : 5. ∞ : Σ (1/(1+n^2)) 是否收敛? : n=1 1 ------- 1+n^2 1 lim --------- = lim ----------- = 1, 原函数收敛 n->∞ 1 n->∞ 1 ----- ----- + 1 n^2 n^2 : 6.一曲线为x^2+y^2=1及x-y+z=1的交线,而f(x,y,z)=x+2y+3z,求在曲线上f的最大值. del(x+2y+3z) = a[del(x^2+y^2-1)] + b[del(x-y+z-1)] 1 = 2xa + b, 2 = 2ya - b, 3 = b -1 = xa, 5/2 = ya, x = (-2/5)y y = +-5/(29)^(1/2), x = -+2/(29)^(1/2), z = 1 +- 7/(29)^(1/2) f(-2/(29)^(1/2), 5/(29)^(1/2), 1 + 7/(29)^(1/2)) = 3 + (29)^(1/2) = max : 7.R为y^2=4-4x和y^2=4+4x所围成的区域,其中y≧0,求∫∫ydA : R 0 (4+4x)^(1/2) 1 (4-4x)^(1/2) ∫∫ydA = ∫ ∫ ydydx + ∫ ∫ ydydx = 1 + 1 = 2 R -1 0 0 0 -- ※ 编辑: hsnuyi (118.168.236.205 台湾), 09/12/2019 01:07:58