Integrales

1.- f(x)={█(6x(1-x),&0≤x ≤1@0, en otro caso)┤

ρ(0≤x ≤1)= ∫_0^1▒〖6x(1-x)〗 dx=∫_0^1▒〖(6x-〖6x〗^2)〗 dx=∫_0^1▒6x dx-∫_0^1▒〖6x〗^2 dx

=6∫_0^1▒x dx-6∫_0^1▒x^2 dx=3x^2-2├ x^3 ┤| ■(1@0)

〖3(1)〗^2-〖2(1)〗^3=1

μx=∫_0^1▒x f(x)dx=∫_0^1▒x [6x(1-x)]dx=∫_0^1▒x (6x-〖6x〗^2 )dx=∫_0^1▒〖〖(6x〗^2-〖6x〗^3)〗 dx

=∫_0^1▒〖6x〗^2 dx-∫_0^1▒〖6x〗^3 dx=6∫_0^1▒x^2 dx-6∫_0^1▒x^3 dx=〖2x〗^3-3/2 ├ x^4 ┤| ■(1@0)

=〖2(1)〗^3-3/2 (1)^4=1/2

σ^2 x=∫_0^1▒x^2 f(x)dx-μ^2 x=∫_0^1▒x^2 [6x(1-x)]=∫_0^1▒x^2 (6x-〖6x〗^2 )dx=∫_0^1▒〖(6x〗^3 -〖6x〗^4)dx

=∫_0^1▒〖6x〗^3 dx-∫_0^1▒〖6x〗^4 dx=6∫_0^1▒x^3 dx-6∫_0^1▒x^4 dx=3/2 x^4-6/5 ├ x^5 ┤| ■(1@0)

=3/2 〖(1)〗^4-6/5 (1)^5=3/10-(1/2)^2=1/20

σ√(1/20)=√1/√20

2.- f(x)={█(1/8,&0≤x ≤8@0, en otro caso)┤

ρ(0≤x ≤8)= ∫_0^8▒1/8 dx=1/8 ∫_0^8▒dx=├ 1/8 x┤| ■(8@0)=1/8(8)=1

μx=∫_0^8▒x f(x)dx=∫_0^8▒x 1/8 dx=1/8 ∫_0^8▒x dx=├ x^2/16┤| ■(8@0)=〖(8)〗^2/16=4

σ^2 x=∫_0^8▒x^2 f(x)dx-μ^2 x= ∫_0^8▒x^2 1/8 dx=1/8= ∫_0^8▒x^2 dx=├ x^3/24┤| ■(8@0)=(8)^3/24=64/3

64/3-〖(4)〗^2=16/3

σ√(16/3)=4√(1/3)

3.- f(x)={█(1/3 x^2,&-1≤x ≤2@0, en otro caso)┤

ρ(-1≤x ≤2)= ∫_(-1)^2▒〖1/3 x^2 〗 dx=1/3 ∫_(-1)^2▒x^2 dx=├ x^3/9┤| ■(2@-1)=〖(2)〗^3/9-((〖-1)〗^3)/9=1

μx=∫_(-1)^2▒x f(x)dx=∫_(-1)^2▒x 1/3 x^2 dx=1/3 ∫_(-1)^2▒x^3 dx=├ x^4/12┤| ■(2@-1)

〖(2)〗^4/12-((〖-1)〗^4)/12=5/3

σ^2 x=∫_(-1)^2▒x^2 f(x)dx-μ^2 x=∫_(-1)^2▒x^2 1/3 x^2 dx=1/3 ∫_(-1)^2▒x^4 dx=├ x^5/15┤| ■(2@-1)

〖(2)〗^5/15-((〖-1)〗^5)/15=31/15-(5/3)^2=-32/45

4.- f(x)={█((2x+2)/3,&0≤x ≤1@0, en otro caso)┤

ρ(0≤x ≤1)= ∫_0^1▒(2x+2)/3 dx=∫_0^1▒2/3 xdx+∫_0^1▒2/3 dx=2/3 ∫_0^1▒xdx+2/3 ∫_0^1▒dx=1/3 x^2+├ 2/3 x┤| ■(1@0)

1/3 〖(1)〗^2-2/3 (1)=1

μx=∫_0^1▒x f(x)dx=∫_0^1▒x ((2x+2)/3)dx=∫_0^1▒2/3 x^2 dx+∫_0^1▒2/3 xdx=2/3 ∫_0^1▒x^2 dx+2/3 ∫_0^1▒x dx

=2/9 x^3+2/6 ├ x^2 ┤| ■(1@0)=2/9 〖(1)〗^3+2/6 〖(1)〗^2=5/9

σ^2 x=∫_0^1▒x^2 f(x)dx-μ^2 x=∫_0^1▒x^2 ((2x+2)/3)dx=∫_0^1▒2/3 x^3 dx+∫_0^1▒2/3 x^2 dx=2/3 ∫_0^1▒x^3 dx+2/3 ∫_0^1▒x^2 dx

1/6 x^4+├ 2/9 x^3 ┤| ■(1@0)=1/6 〖(1)〗^4+2/9 〖(1)〗^3=7/18-(5/9)^2=13/162

σ√(13/162)=√13/√162

5.- f(x)={█(〖3x〗^2-5x+4,1≤x ≤3@0, en otro caso)┤

ρ(1≤x ≤3)=∫_1^3▒〖(〖3x〗^2-5x+4)〗 dx=∫_1^3▒〖3x〗^2 dx-∫_1^3▒5x dx+∫_1^3▒4 dx=3∫_1^3▒x^2 dx-5∫_1^3▒x dx+4∫_1^3▒dx

x^3-5/2 x^2+├ 4x┤| ■(3@1)

[〖(3)〗^3-5/2 (3)^2+4(3)]-[(1)^3-5/2 (1)^2+4(1)]=14

μx=∫_0^1▒x f(x)dx=∫_0^1▒x (〖3x〗^2-5x+4)dx=∫_1^3▒〖3x〗^3 dx-∫_1^3▒〖5x〗^2 dx+∫_1^3▒4x dx=3∫_1^3▒x^3 dx-5∫_1^3▒x^2 dx+4∫_1^3▒xdx= 3/4 x^4-5/3 x^3+├ 〖2x〗^2 ┤| ■(3@1)

= [3/4 〖(3)〗^4-5/3 (3)^3+〖2(3)〗^2 ]-[3/4 (1)^4-5/3 (1)^3+〖2(1)〗^2 ]=41

σ^2 x=∫_1^3▒x^2 f(x)dx-μ^2 x=∫_1^3▒x^2 (〖3x〗^2-5x+4)dx= ∫_1^3▒〖3x〗^4 dx-∫_1^3▒〖5x^3 〗 dx+∫_1^3▒〖4x〗^2 dx=3∫_1^3▒x^4 dx-5∫_1^3▒x^3 dx+4∫_1^3▒x^2 dx= 3/5 x^5-5/4 x^4+├ 4/3 x^3 ┤| ■(3@1)

= [3/5 〖(3)〗^5-5/4 (3)^4+4/3 〖(3)〗^3 ]-[3/5 (1)^5-5/4 (1)^4+4/3 (1)^3 ]=153/4-(41)^2=-1642 3/4

6.- f(x)={█(x^2-4/3 x,0≤x ≤1@0, en otro caso)┤

ρ(0≤x ≤1)=∫_0^1▒〖(x^2+4/3 x)〗 dx=∫_0^1▒x^2 dx+∫_0^1▒〖4/3 x〗 dx=∫_0^1▒x^2 dx+4/3 ∫_0^1▒x dx=x^3/3+├ 4/6 x^2 ┤| ■(1@0)

〖(1)〗^3/3+4/6 〖(1)〗^2=1

μx=∫_0^1▒x f(x)dx=∫_0^1▒〖x〖(x〗^2+4/3 x)〗 dx=∫_0^1▒x^3 dx+∫_0^1▒〖4/3 x^2 〗 dx=∫_0^1▒x^3 dx+4/3 ∫_0^1▒x^2 dx=x^4/4+├ 4/9 x^3 ┤| ■(1@0)= 〖(1)〗^4/4+4/9 〖(1)〗^3=5/6

σ^2 x=∫_0^1▒x^2 f(x)dx-μ^2 x=∫_0^1▒x^2 (x^2+4/3 x)dx=∫_0^1▒x^4 dx+∫_0^1▒〖4/3 x^3 〗 dx=∫_0^1▒x^4 dx+4/3 ∫_0^1▒x^3 dx=x^5/5+├ 4/12 x^4 ┤| ■(1@0)=(1)^5/5+4/12 (1)^4=1/2-(5/6)^2=-7/36

σ=-√(7/36)=1/6 √7

7.- f(x)={█(〖3/5 x〗^2+12/5 x-2/5,0≤x ≤1@0, en otro caso)┤

ρ(0≤x ≤1)=∫_0^1▒〖(〖3/5 x〗^2+12/5 x-2/5)〗 dx=∫_0^1▒3/5 x^2 dx+∫_0^1▒12/5 xdx-∫_0^1▒2/5 dx= 3/5 ∫_0^1▒x^2 dx+12/5 ∫_0^1▒xdx-2/5 ∫_0^1▒dx= 1/5 x^3+6/5 x^2-2/5 ├ x┤| ■(1@0)

1/5 〖(1)〗^3+6/5 〖(1)〗^2-2/5 (1)=1

μx=∫_0^1▒x f(x)dx=∫_0^1▒〖x〖(3/5

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