Integrais_-_Exercícios_Resolvidos
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Transcript of Integrais_-_Exercícios_Resolvidos
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7/31/2019 Integrais_-_Exerccios_Resolvidos
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Exerccios
1.) Resolva as Integrais
a) ( 10x + x 4/5 + x ) dx= 10xdx + x 4/5 dx+ x dx= 10 x1/2dx + x 4/5 dx+ x dx
= 10. x3/2 + x9/5 + x1/23/2 9/5 1/2
= 10. 2 x3/2
+ 5 x9/5
+ 2x1/2
3 9
= 20x3/2 + 5 x9/5 + 2x + c
3 9
b) ( 3x + x 1/5 + x3 ) dx4
= x1/3dx + x 1/5dx + x3dx4
= 1 x1/3dx + x 1/5dx + x3dx4
= 1 . x4/3 + x6/5 + x-24 4/3 6/5 -2
= 1 . 3 . x4/3 + 5 x6/5 - 1 . 1
4 4 6 2 x2
= 3 x4/3 + 5 x6/5 1 + c16 2x2
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c) (x0,2 + x0,5 + x4 x-3) dx
= x0,2 dx + x0,5 dx + x4 dx x-3 dx
= x1/2 + x1/5 + x5 x-2 1/5 5 -2
= 2. x0,5 + 5 . x0,2 + 1 . x5 + 11 1 5 x2
2
= 2x0,5 + 5x0,2 + 0,2x5 + 0,5x2 + c
d)
10
1 (4 + x2 + 3x) dx
10 10 10
1 4 dx +1 x2 dx + 1 3x dx
10= 4x + x3 + 3x2 = (4 . 10 + 103 + 3.102) - (4 . 1 + 13 + 3.12)3 2 1 3 2 3 2
= 40 + 1000 + 300 4 1 33 2 3 2
= 240 + 2.000 + 900 24 2 96
= 3.1056
= 517,5
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e)
8
1 1 dxx3
8
1 x-3 dx
8= x -2 = 1 = 1 . 1 = - 1 = ( - 1 ) - ( - 1 )
-2 x2 x2 -2 2x2 1 2.82 2.12-2 = - 1 + 1 = - 1 + 64 = 63 = 0,49
128 2 128 128
f)
0
-1 (x2 + x3 + 2) dx0 0 0
-1 x2dx + -1 x3dx + -1 2dx
0
x3
+ x4
+ 2x = (03
+ 04
+ 2.0) (-13
+ -14
+ 2.-1)3 4 -1 3 4 3 4
= 1 1 + 2 = 4 3 + 24 = 25 = 2,083 4 12 12
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g)
1
0 (1 x + 1 x2 + 5x + 1) dx2 4
1 1 1 1
= 10 xdx + 1 0 x2dx + 5 0 xdx + 1 0 dx2 4
= 1 . 1 x2 + 1 . 1 x3 + 5 . 1 x2 + x2 2 4 3 2
1
= 1 x2 + 1 x3 + 5 x2 + x = ( 1 . 12 + 1 . 13 + 5 . 12 + 1 ) - 04 12 2 0 4 12 2
= 1 + 1 + 5 + 1 = 3 + 1 + 30 + 12 = 46 = 3,834 12 2 12 12
h)
3
1 (5x 2x2 + 3) dx
3 3 3
1 5x - 1 2x2 dx + 1 3 dx3
= 5 x2 2x3 + 3x = ( 5 . 32 2 . 33 + 3 . 3 ) (5 . 12 2 . 13 + 3 . 1)2 3 1 2 3 2 3
= 45 54 + 9 5 + 2 32 3 2 3
= 135 108 + 54 15 + 4 186
= 52 = 8,676
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i)
3a
a ( a + x) dx3a 3a
a a dx + a x dx3a 3a
a a dx + a x dx3a
a.x + x2 = a.3a + 3a2 (a . a + a2)2 a 2 2
= 3a2 + 3a2 a2 a2 = 6a2 + 3a2 2a2 a2 = 6a2 = 3a22 2 2 2
j)
4
1 (x x) dx4 4
1 x1/2 - 1 x dx4
x3/2
x2
= 2 x3/2
x2
= ( 2 (4)3/2
42
) ( 2 (1)3/2
12
)3 2 3 2 1 3 2 3 22 = ( 2 (22)3/2 16 ) 2 + 1 =
3 2 3 2= 2 . 23 8 - 2 + 1
3 3 2= 16 8 2 + 1 = 32 48 4 + 3 = - 17 = - 2,83
3 3 2 6 6
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k)
1
-3 ( 1 - 1 ) dxx3 x2
1
-3 ( 1 - 1 ) dx = x-3 x-21 1
x-3 x-2
1
-3 (x-3 x-2) dx
1 1
-3 x-3 dx - -3 x
-2 dx
1
= x-2 x-1 = 1 - 1 = ( 1 - 1 ) ( 1 - 1 ) = 11 - ( 1 + 1 ) = - 1 - 1 = -1 3 = - 4x2 x -3 12 1 -32 -3 9 3 9 3 9 9
l)
1
-1 ( 3x2 2x3) dx1 1
-1 3x2 dx - -1 2x3 dx1
= 3x3 2x4 = x3 1x4 = ( 13 1.14 ) ( -13 1. -14) = 1 1 ( - 1 1 )3 4 2 -1 2 2 2 2
= 1 1 + 1 + 1 = 2 1 + 2 + 1 = 4 = 22 2 2 2
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m)
3
2 ( 2x3 1x + 4) dx3
3 3 3
2 2x3dx - 2 1x dx + 2 4 dx3
3
= 2x4 1 . x2 + 4x = 1x4 x2 + 4x = ( 1 . 34 32 + 4.3) (1 . 24 22 + 4.2)4 3 2 2 6 2 2 6 2 6
= 81 9 + 12 ( 8 4 + 8)2 6 6
= 81 9 + 12 8 + 4 82 6 6
= 243 9 + 72 48 + 4 486
= 214 = 35,676
n)
1
1 ( x3 - x2 + 1) dx1 1 1
1 x3 dx - 1 x2 dx + 1 dx1
= x4 - x3 + x = ( 14 13 + 1) (14 13 + 1) = 04 3 1 4 3 4 3
n
Conclui-se que n f(x) dx = 0
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o)
2 1
1 x2 dx e 2 (x2 dx2 2
1 x2 dx = x3 = 23 13 = 8 - 1 = 73 1 3 3 3 3 3
1 1
2 x2 dx = x3 = 13 23 = 1 - 8 = - 73 2 3 3 3 3 3
m n
Conclui-se que
n f(x) dx = -
m f(x) dx
2.) Dada a funo de custo marginal e custo fixo, determine as funes deCT e CM.
a)Cmg = 5 + 2x, CF = 2b)Cmg = x3 6x2 + 7x, CF = 3c)Cmg = 900 360x + 9x2, CF = 30a)
CT = CmgCT = (5 + 2x) dxCT =5 dx + 2 x1/2 dx
CT = 5x + 2.x3/23/2
CT = 5x + 2 . 2 x3/23
CT = 5x + 4 x3/2 + c3
CT = CF p/ x = 0
2 = 5. 0 + 4 03/2 + c3
2 = c
CT = 5x + 4 x3/2 + 23
CM = CTx
CM = 5x + 4 x3/2 + 23x
CM = 5 + 4 x1/2 + 23 x
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b)
CT = CmgCT = (x3 6x2 + 7x) dxCT = x3 dx - 6x2 dx + 7x dx
CT = x4 6x3 + 7x2 + c4 3 2
CT =1x4 2x3 + 7x2 + c4 2
CT = CF p/ x = 0
3 = c
CT =1x4 2x3 + 7x2 + 34 2
CT = 0,25x4 2x3 + 3,5x2 + 3
CM = CTx
CM = 0,25x4 2x3 + 3,5x2 + 3
x
CM = 0,25x3 2x2 + 3,5x + 3x
c)
CT =
Cmg
CT = (900 360x + 9x2) dxCT = 900 dx - 360x dx + 9x2 dx
CT = 900x 360x2 + 9x32 3
CT = 900x 180x2 + 3x3 + c
CT = CF p/ x = 0
30 = c
CT = 900x 180x2 + 3x3 + 30
CM = CTx
CM = 900x 180x2 + 3x3 + 30x
CM = 900 180x + 3x2 + 30x
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3.) Um monopolista tem a seguinte curva de rendimento marginalRmg = 1.000 10x. Determine RT e RM p/ x = 10, sabendo-se que a RT = 95para x = 1.
RT =
Rmg
RT = (1.000 10x) dx
RT = 1.000 dx 10x dx
RT = 1.000x 10x22
RT = 1.000x 5x2 + c
95 = 1.000 . 1 5 . 12 + c
95 = 1.000 5 + c
95 1.000 + 5 = c
- 900 = c
RT = 1.000x 5x2 900
RT x=10 = 1.000 . 10 5 . 102 - 900
RT x=10 = 10.000 500 900
RT x=10 = $ 8.600,00
RM = RTx
RM = 1.000x 5x2 900x
RM = 1.000 5x - 900x
RM x=10 =1.000 5.10 90010
RM x=10 = 1.000 50 90
RM x=10 = $ 860,00
4.) Se a curva de custo marginal dada por Cmg = 3x2 5x + 10 determinaro custo total e o custo mdio sabendo-se que o custo fixo associado 20.
CT = CmgCT = (3x2 5x + 10) dxCT = 3x2 dx - 5x dx + 10 dx
CT = 3x3 5x2 + 10x + c3 2
CT = CF p/ x = 0
20 = c
CT = 3x3 5x2 + 10x + 203 2
CT = x3 2,5x2 + 10x + 20
CM = CTx
CM = x3 2,5x2 + 10x + 20x
CM = x2 2,5x + 10 + 20x
