Question

∫一-。,

1.∫1/x dx
2.∫4 1/（1+根号x）dx
0
3.∫5 1/(x-3)^2 dx
4
4求函数f（x,y）=x^2/2-xy+y^2+3x的极值
忽然发现没财富了，也不知道什么时候百度知道改了，赚点分数来再追加财富

Answer 1

1.ln (x)
2.设t = 根号x,变成∫2 -2t/(1+t)dt = 2ln(t+1) - 2t |2 = 2ln3 - 4
0 0
3.-1 / (x - 3) |5 = 1/2
4
4.
fx(x,y) = x - y + 3
fy(x,y) = -x + 2y
求出驻点为(-6,-3)
fxx(-6,-3) = 1 > 0
fyy(-6,-3) = 2
fxy(-6,-3) = -1
fxx * fyy - fxy^2 = 1 > 0
所以f(x,y)在(-6,-3)处取极小值-9