大学数学 高等数学 微积分 求解!急! 50
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二. 1. 由洛必达法则得 原式 = lim<x→0>[1-e^(-x^2)]/(3x^2) (0/0)
= im<x→0>2xe^(-x^2)/(6x) = 1/3
2. 分子分母同乘以 [√(x^2+y^2+1) + 1], 得
原式 = lim<x→0, y→0>[√(x^2+y^2+1) + 1]/1 = 2
三。 D = ∫<0, 3>(6-y-√y)dy = [6y-y^2/2-2y^(3/2)/3]<0, 3> = 27/2-2√3
一. 1. I = ∫<-1, 1>xdx/(1+x^2) + ∫<-1, 1>|x|dx/(1+x^2)
= 0 + 2∫<0, 1>xdx/(1+x^2) = ∫<0, 1>d(1+x^2)/(1+x^2)
= [ln(1+x^2)]<0, 1> = ln2
2. I = ∫<1, 2>e^(1/x)dx/x^2 = -∫<1, 2>e^(1/x)d(1/x)
= -[e^(1/x)]<1, 2> = e - √e
3. I = ∫<0, 1>xe^(2x)dx = (1/2)∫<0, 1>xde^(2x)
= (1/2)[xe^(2x)]<0, 1> - (1/2)∫<0, 1>e^(2x)dx
= e^2/2 - (1/4)[e^(2x)]<0, 1> = (1/4)(1+e^2)
4. 联立解 xy = 2, y= 1+x^2 得交点 (1, 2). 则
I = ∫<1, 2>x^2dx ∫<2/x, 1+x^2>dy/y^2
= ∫<1, 2>x^2dx[-1/y]<2/x, 1+x^2>
= ∫<1, 2>x^2[x/2-1/(1+x^2)]dx
= ∫<1, 2>[x^3/2-x^2/(1+x^2)]dx
= [x^4/8 - x + arctanx]<1, 2>
= 7/8 - π/4 +arctan2
= im<x→0>2xe^(-x^2)/(6x) = 1/3
2. 分子分母同乘以 [√(x^2+y^2+1) + 1], 得
原式 = lim<x→0, y→0>[√(x^2+y^2+1) + 1]/1 = 2
三。 D = ∫<0, 3>(6-y-√y)dy = [6y-y^2/2-2y^(3/2)/3]<0, 3> = 27/2-2√3
一. 1. I = ∫<-1, 1>xdx/(1+x^2) + ∫<-1, 1>|x|dx/(1+x^2)
= 0 + 2∫<0, 1>xdx/(1+x^2) = ∫<0, 1>d(1+x^2)/(1+x^2)
= [ln(1+x^2)]<0, 1> = ln2
2. I = ∫<1, 2>e^(1/x)dx/x^2 = -∫<1, 2>e^(1/x)d(1/x)
= -[e^(1/x)]<1, 2> = e - √e
3. I = ∫<0, 1>xe^(2x)dx = (1/2)∫<0, 1>xde^(2x)
= (1/2)[xe^(2x)]<0, 1> - (1/2)∫<0, 1>e^(2x)dx
= e^2/2 - (1/4)[e^(2x)]<0, 1> = (1/4)(1+e^2)
4. 联立解 xy = 2, y= 1+x^2 得交点 (1, 2). 则
I = ∫<1, 2>x^2dx ∫<2/x, 1+x^2>dy/y^2
= ∫<1, 2>x^2dx[-1/y]<2/x, 1+x^2>
= ∫<1, 2>x^2[x/2-1/(1+x^2)]dx
= ∫<1, 2>[x^3/2-x^2/(1+x^2)]dx
= [x^4/8 - x + arctanx]<1, 2>
= 7/8 - π/4 +arctan2
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