Ejercícios resueltos Cálculo Integral

Jaime Andrés Silva Velosa - 1010209749
Cálculo integral en una variable
Profesor Gustavo Rubiano
Tarea 1, entregada el miércoles 21 de Agosto

1

Problemas 5.1

En los ejercicios 1 a 4, utilice aproximaciones nitas para estimar el área debajo de la gráca de la función; para ello
emplee
a. una suma inferior con dos rectángulos del mismo ancho.
b. una suma inferior con cuatrorectángulos del mismo ancho.
c. una suma superior con dos rectángulos del mismo ancho.
d. una suma superior con cuatro rectángulos del mismo ancho.

1. f (x) = x2 entre x = 0 y x = 1
a) ∆x = 1 , x0 = 0, x1 = 0.5, x2 = 1
2
a = f (x0 )∆x + f (x1 )∆x
a=

0
2

+

c) ∆x = 1 , x0 = 0, x1 = 0.5, x2 = 1
2
a = f (x1 )∆x + f (x2 )∆x

0.25
2

a=

a = 0.125

0.25
2

1
2

+

a = 0.625b) ∆x = 1 , x0 = 0, x1 = 0.25, x2 = 0.5, x3 = 0.75,
4

d) ∆x = 1 , x0 = 0, x1 = 0.25, x2 = 0.5, x3 = 0.75,
4

x4 = 1

x4 = 1

a = f (x0 )∆x + f (x1 )∆x + f (x2 )∆x + f (x3 )∆x

a = f (x1 )∆x + f (x2 )∆x + f (x3 )∆x + f (x4 )∆x

a=

0
4

+

0.0625
4

+

0.25
4

+

0.5625
4

a = 0.21875

a=

0.0625
4

0.25
4

+

+

0.5625
4

+

1
4

a =0.46875

2. f (x) = x3 entre x = 0 y x = 1
a) ∆x = 1 , x0 = 0, x1 = 0.5, x2 = 1
2
a = f (x0 )∆x + f (x1 )∆x
a=

0
2

+

a = f (x1 )∆x + f (x2 )∆x

0.125
2

a=

a = 0.0625

x4 = 1

+

+

1
2

d) ∆x = 1 , x0 = 0, x1 = 0.25, x2 = 0.5, x3 = 0.75,
4
x4 = 1

a = f (x0 )∆x + f (x1 )∆x + f (x2 )∆x + f (x3 )∆x
a=

0.125
2

a = 0.5625

1
b) ∆x = 4 , x0 = 0, x1 = 0.25,x2 = 0.5, x3 = 0.75,

0
4

c) ∆x = 1 , x0 = 0, x1 = 0.5, x2 = 1
2

0,015625
4

a = 0.140625

+

0.125
4

+

0,421875
4

a = f (x1 )∆x + f (x2 )∆x + f (x3 )∆x + f (x4 )∆x
a=

0,015625
4

+

a = 0.390625

0.125
4

+

0,421875
4

+

1
4

3. f (x) = 1/x entre x = 1 y x = 5
a) ∆x = 2, x0 = 1, x1 = 3, x2 = 5
a = f (x1 )∆x + f (x2 )∆x
a=

2
3

a=16
15

+

2
5

c) ∆x = 2, x0 = 1, x1 = 3, x2 = 5
a = f (x0 )∆x + f (x1 )∆
2
3

a=2+
a=

= 1.06

b) ∆x = 1, x0 = 1, x1 = 2, x2 = 3, x3 = 4,

8
3

= 2.6

d) ∆x = 1, x0 = 1, x1 = 2, x2 = 3, x3 = 4,

x4 = 5

x4 = 5

a = f (x1 )∆x + f (x2 )∆x + f (x3 )∆x + f (x4 )∆x

a = f (x0 )∆x + f (x1 )∆x + f (x2 )∆x + f (x3 )∆x

a=

1
2

a=

77
60

+

1
3

+

14

+

1
5

1
2

a=1+

= 1.283

a=

25
12

+

1
3

+

1
4

= 2.083

4. f (x) = 4 − x2 entre x = −2 y x = 2
a) ∆x = 2, x0 = −2, x1 = 0, x2 = 2

c) ∆x = 2, x0 = −2, x1 = 0, x2 = 2

a = f (x0 )∆x + f (x2 )∆x

a = f (x1 )∆(x) + f (x2 )∆x

a=0+0

a=8+8

a=0

a = 16

b) ∆x = 1, x2 = −2, x1 = −1, x2 = 0, x3 = 1,

d) ∆x = 1, x0 = −2, x1 = −1, x2 = 0, x3 =1,

x4 = 2

x4 = 2

a = f (x0 )∆x + f (x1 )∆x + f (x3 )∆x + f (x4 )∆x

a = f (x1 )∆x + f (x2 )∆x + f (x2 )∆x + f (x3 )∆x

a=0+3+3+0

a=3+4+4+3

a=6

a = 14

Utilice rectángulos cuyas alturas estén dadas por el valor de la función en el punto medio de la base del rectángulo (la
regla del punto medio); además, estime el área debajo de las grácas de las siguientes funciones usandoprimero dos
y luego cuatro rectángulos.
5. f (x) = x2 entre x = 0 y x = 1
1
a) ∆x = 2 , x1 = 1 , x2 =
4

3
4

a = f (x1 )∆x + f (x2 )∆x
a=

1
32

+

9
32

a=

10
32

= 0.3125

1
1
3
5
b) ∆x = 4 , x1 = 8 , x2 = 8 , x3 = 8 , x4 =

7
8

a = f (x1 )∆x + f (x2 )∆x + f (x3 )∆x + f (x4 )∆x
a=

1
256

a=

21
64

+

9
256

+

25
256

+

49
256= 0.328125

6. f (x) = x3 entre x = 0 y x = 1
a) ∆x = 1 , x1 = 1 , x2 =
2
4

3
4

a = f (x1 )∆x + f (x2 )∆x
a=

1
128

a=

7
32

1
1
3
5
b) ∆x = 4 , x1 = 8 , x2 = 8 , x3 = 8 , x4 =

a = f (x1 )∆x + f (x2 )∆x + f (x3 )∆x + f (x4 )∆x

27
128

a=

1
2048

= 0.21875

a=

496
512

+

7
8

+

27
2048

+

125
2048

= 0.2421875

+

343...