Integrais - Exercícios Resolvidos

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Exercícios 
 
1.º) Resolva as Integrais 
 
 
a) ∫( 10√x + x 4/5 + x – ½ ) dx 
= ∫10√xdx + ∫ x 4/5 dx+ ∫x – ½ dx 
= 10 ∫x1/2dx + ∫ x 4/5 dx+ ∫x – ½ dx 
 
= 10. x3/2 + x9/5 + x1/2 
 3/2 9/5 1/2 
 
= 10. 2 x3/2 + 5 x9/5 + 2x1/2 
 3 9 
 
= 20x3/2 + 5 √x9/5 + 2√x + c 
 3 9 
 
b) ∫( 3√x + x 1/5 + x–3 ) dx 
 4 
= ∫x1/3dx +∫ x 1/5dx + ∫ x–3dx 
 4 
 
= 1 ∫x1/3dx +∫ x 1/5dx + ∫ x–3dx 
 4 
 
= 1 . x4/3 + x6/5 + x-2 
 4 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 + c 
 16 2x2 
 
 
 
 
 
 
 
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 
½ 1/5 5 -2 
 
= 2. x0,5 + 5 . x0,2 + 1 . x5 + 1 
 1 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 – 3 
 3 2 3 2 
 = 240 + 2.000 + 900 – 24 – 2 – 9 
 6 
 = 3.105 
 6 
 
= 517,5 
 
 
 
 
 
 
 
 
 
 
 
 
e) 
 8 
∫1 1 dx 
 x3 
 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) dx 
 0 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 = 2,08 
 3 4 12 12 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
g) 
 1 
∫0 (1 x + 1 x2 + 5x + 1) dx 
 2 4 
 1 1 1 1 
 = 1 ∫0 xdx + 1 ∫0 x2 dx + 5 ∫0 xdx + 1 ∫0 dx 
 2 4 
 
= 1 . 1 x2 + 1 . 1 x3 + 5 . 1 x2 + x 
 2 2 4 3 2 
 1 
= 1 x2 + 1 x3 + 5 x2 + x = ( 1 . 12 + 1 . 13 + 5 . 12 + 1 ) - 0 
 4 12 2 0 4 12 2 
 
 = 1 + 1 + 5 + 1 = 3 + 1 + 30 + 12 = 46 = 3,83 
 4 12 2 12 12 
 
 
h) 
 3 
∫1 (5x – 2x2 + 3) dx 
 
 3 3 3 
∫1 5x - ∫1 2x2 dx + ∫1 3 dx 
 
 3 
= 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 – 3 
 2 3 2 3 
 = 135 – 108 + 54 – 15 + 4 – 18 
 6 
 = 52 = 8,67 
 6 
 
 
 
 
 
 
 
 
 
 
i) 
 3a 
∫a ( a + x) dx 
 3a 3a 
∫a a dx + ∫a x dx 
 3a 3a 
a ∫a dx + ∫a x dx 
 3a 
a.x + x2 = a.3a + 3a2 – (a . a + a2) 
 2 a 2 2 
 = 3a2 + 3a2 – a2 – a2 = 6a2 + 3a2 – 2a2 – a2 = 6a2 = 3a2 
 2 2 2 2 
 
 
j) 
 4 
∫1 (√x – x) dx 
 4 4 
∫1 x1/2 - ∫1 x dx 
 4 
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 2 
 2 = ( 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 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
k) 
 1 
∫-3 ( 1 - 1 ) dx 
 x3 x2 
 1 
∫-3 ( 1 - 1 ) dx = x-3 – x-2 
 1 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 ) = 1–1 - ( 1 + 1 ) = - 1 - 1 = -1 – 3 = - 4 
 x2 x -3 12 1 -32 -3 9 3 9 3 9 9 
 
 
l) 
 1 
∫-1 ( 3x2 – 2x3) dx 
 1 1 
∫-1 3x2 dx - ∫-1 2x3 dx 
 1 
= 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 = 2 
 2 2 2 2 
 
 
 
 
 
 
 
 
 
 
 
 
 
m) 
 3 
∫2 ( 2x3 – 1x + 4) dx 
 3 
 3 3 3 
∫2 2x3dx - ∫2 1x dx + ∫2 4 dx 
 3 
 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 – 8 
 2 6 6 
 = 243 – 9 + 72 – 48 + 4 – 48 
 6 
 = 214 = 35,67 
 6 
 
 
 
n) 
 1 
∫1 ( x3 - x2 + 1) dx 
 1 1 1 
∫1 x3 dx - ∫1 x2 dx + ∫1 dx 
 1 
= x4 - x3 + x = ( 14 – 13 + 1) – (14 – 13 + 1) = 0 
 4 3 1 4 3 4 3 
 
 
 
 n 
Conclui-se que ∫n f(x) dx = 0 
 
 
 
 
 
 
 
 
 
o) 
 2 1 
∫1 x2 dx e ∫2 (x2 dx 
 2 2 
∫1 x2 dx = x3 = 23 – 13 = 8 - 1 = 7 
 3 1 3 3 3 3 3 
 1 1 
∫2 x2 dx = x3 = 13 – 23 = 1 - 8 = - 7 
 3 2 3 3 3 3 3 
 
 m n 
Conclui-se que ∫n f(x) dx = - ∫m f(x) dx 
 
 
 
 
2.º) Dada a função de custo marginal e custo fixo, determine as funções de 
CT e CM. 
 
a) Cmg = 5 + 2√x, CF = 2 
b) Cmg = x3 – 6x2 + 7x, CF = 3 
c) Cmg = 900 – 360x + 9x2, CF = 30 
 
a) 
CT = ∫Cmg 
CT = ∫(5 + 2√x) dx 
CT =∫5 dx + ∫2 x1/2 dx 
 
CT = 5x + 2.x3/2 
 3/2 
CT = 5x + 2 . 2 x3/2 
 3 
CT = 5x + 4 x3/2 + c 
 3 
 
 
 
 
 
CT = CF p/ x = 0 
 
2 = 5. 0 + 4 03/2 + c 
 3 
2 = c 
 
CT = 5x + 4 x3/2 + 2 
 3 
 
CM = CT 
 x 
 
CM = 5x + 4 x3/2 + 2 
 3 
 x 
 
CM = 5 + 4 x1/2 + 2 
 3 x 
 
 
 
 
 
b) 
CT = ∫Cmg 
CT = ∫ (x3 – 6x2 + 7x) dx 
CT =∫ x3 dx - ∫6x2 dx + ∫7x dx 
 
CT= x4 – 6x3 + 7x2 + c 
 4 3 2 
 
CT = 1x4 – 2x3 + 7x2 + c 
 4 2 
 
 
 
 
CT = CF p/ x = 0 
 
3 = c 
 
CT = 1x4 – 2x3 + 7x2 + 3 
 4 2 
 
CT = 0,25x4 – 2x3 + 3,5x2 + 3 
 
CM = CT 
 x 
 
CM = 0,25x4 – 2x3 + 3,5x2 + 3 
 x 
 
CM = 0,25x3 – 2x2 + 3,5x + 3 
 x 
 
 
 
 
 
c) 
 
CT = ∫Cmg 
CT = ∫ (900 – 360x + 9x2) dx 
CT =∫ 900 dx - ∫360x dx + ∫9x2 dx 
 
CT = 900x – 360x2 + 9x3 
 2 3 
CT = 900x – 180x2 + 3x3 + c 
 
 
CT = CF p/ x = 0 
 
30 = c 
 
CT = 900x – 180x2 + 3x3 + 30 
 
CM = CT 
 x 
 
CM = 900x – 180x2 + 3x3 + 30 
 x 
CM = 900 – 180x + 3x2 + 30 
 x 
 
 
 
 
 
 
 
 
 
 
3.º) Um monopolista tem a seguinte curva de rendimento marginal 
Rmg = 1.000 – 10x. Determine RT e RM p/ x = 10, sabendo-se que a RT = 95 
para x = 1. 
 
RT = ∫Rmg 
 
RT = ∫(1.000 – 10x) dx 
 
RT = ∫1.000 dx – ∫10x dx 
 
RT = 1.000x – 10x2 
 2 
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 = RT 
 x 
 
RM = 1.000x – 5x2 – 900 
 x 
 
RM = 1.000 – 5x - 900 
 x 
 
RM x=10 =1.000 – 5.10 – 900 
 10 
RM x=10 = 1.000 – 50 – 90 
 
RM x=10 = $ 860,00 
4.º) Se a curva de custo marginal é dada por Cmg = 3x2 – 5x + 10 determinar 
o custo total e o custo médio sabendo-se que o custo fixo associado é 20. 
 
CT = ∫Cmg 
CT = ∫ (3x2 – 5x + 10) dx 
CT =∫ 3x2 dx - ∫5x dx + ∫10 dx 
 
CT = 3x3 – 5x2 + 10x + c 
 3 2 
 
CT = CF p/ x = 0 
 
20 = c 
 
 
CT = 3x3 – 5x2 + 10x + 20 
 3 2 
 
CT = x3 – 2,5x2 + 10x + 20 
 
 
CM = CT 
 x 
 
CM = x3 – 2,5x2 + 10x + 20 
 x 
 
CM = x2 – 2,5x + 10 + 20 
 x