Cini_u2_a1
Enviado por jmcisnerog • 20 de Octubre de 2014 • 290 Palabras (2 Páginas) • 246 Visitas
a) y=x^2,y=4
x=0,x=4 donde se cruzan
2^2=4<4(2)=8
a). A=∫_0^4▒〖[4x-x^2 ]dx=├ (4x^2)/2-x^3/3]_0^4 〗
2x^2-├ x^3/3]_0^4=32-64/3=(96-64)/3=32/3≈10.67μ^2
b) y=x+1,y=9-x^2,x=-1,x=2
0+1=1<9-0^2=9
A=∫_(-1)^2▒[9-x^2-(x+1)]dx
=∫_(-1)^2▒〖[-x^2-1+8]dx=├ (-x^3)/3-x^2/2+8x]_(-1)^2=〗 ├ (-2x^3-3x^3+48x)/6]_(-1)^2
(-2(2)^3-3(2)^2+48(2))/6-((2-3-48)/6)=68/6+49/6=117/6=7/2=19.5μ^2
c) y=x, y=3x,x+y=4
y=4-x
x=0,〖 y〗_1=0,〖 y〗_2=0,y_3=4
x=1,〖 y〗_1=1,〖 y〗_2=3,y_3=3
x=2,〖 y〗_1=2,〖 y〗_2=9,y_3=2
1/1<3/2 3/2<4-3/2=9/2
[0,1] y_1<y_2 [1,2] y_1<y_3
A=∫_0^1▒〖[3x-x]dx+∫_1^2▒[4-x-x]dx=∫_0^1▒〖2xdx+∫_1^2▒〖4-2xdx〗〗〗
=├ x^2 ]_0^1+4x-├ x^2 ]_1^2=1+4-3=2μ^2
d)y=1/x, y=1/x^2 , x=2
x=1 y_1=1 y_2=1
1/((3/2) )=2/3>1/(3/2)^2 =1/((9/4) )=9/4
y_1>y_2
A=∫_1^2▒〖1/x-1/x^2 dx=ln|x|-├ (〖-x〗^(-1) )]_1^2=ln|x|+├ 1/x]_1^2≈1.19-1≈.19μ^2 〗
e) y=4x^2, y=x^2+3
4x^2=x^2+3 3x^2=3 x^2=1 x=±1
4(0)^2=0 < 0^2+3 =3
y_1< y_2
A=∫_(-1)^1▒〖[x^2+3-4x^2 ]dx=∫_(-1)^1▒[3-3x^2 ]dx〗=3 x-├ x^3 ]_(-1)^1=2-(-2)=4μ^2
f) y=x√(x^2-9), y=0, x=5
y=x√(x^2-9)=0 x^2=9 x= ±3
x√(4^2-9)>0
A=∫_3^5▒〖x√(x^2-9) dx=1/2 ∫_0^16▒〖√udu=├ 1/2∙u^(3/2)/(3/2)]_0^16 〗= √(u^3 )/3〗=√4096/3=64/3≈21.33μ^2
g) y=x^3-x, y=3x
x^3-x=3x x^3-4x=0
x(x^2-4)=0 x=0 x=±2
1^3-1=0 < 3(1)=3
y_1=y_2
A=∫_0^2▒〖[3x-(x^3-x)]dx=∫_0^2▒〖4x-x^3 dx=2x^2 ├ -x^4/4]_0^2=8-16/4=16/4≈4μ^2 〗〗
h) y=cosx, y=〖sec〗^2 x, x=π/4, x=π/4
cos〖0=0 < 〖sec〗^2 (0)=1〗
A=∫_(π/4)^4▒├ 〖sec〗^2 x-cosxdx=tanx-sinx]_(-π/4)^(π/4)
=tan(π/4)-sin(π/4)-(tan(-π/4)-sin(-π/4))
≈1-0.7071-(-1+0.7071)=≈0.29-(-.29)=0.58μ^2
i) y=sinx, y=sin(2x), x=0, x=π/2
sin(1)≈0.8414< sin(2(1))≈0.9092
y_1< y_2
A=∫_0^(π/2)▒〖sin(2x)-sinxdx=├ -cos(2x)/2+cosx]_0^(π/2) 〗
=-cos(2(π/2))/2+cos(π/2)-(-cos(0)/2+cos(0) )
=1/2+0-(-1/2+1)=1/2-1/2=0μ^2
j) y=e^x, y=x, x=0, x=1
e^(1/2)≈1.65
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