GUIA 1

1. Calcular: ∑ 𝟓 𝒊𝟐 − 𝟐𝒏

𝒊=𝟑 en función de n.

Solución:

∑ 5 𝑖2 − 2𝑛

𝑖=3 = ∑ (5 𝑖2 − 2)

2

𝑖=1 + ∑ (5 𝑖2 − 2)

𝑛

𝑖=3

∑ (5 𝑖2 − 2)𝑛

𝑖=3 = ∑ (5 𝑖2 − 2)

𝑛

𝑖=1 + ∑ (5 𝑖2 − 2)

2

𝑖=1

5 . (𝑛(𝑛+1) (2𝑛+1)

6 - 2n - 5 .

2(3) (5)

6 + 4

5𝑛(𝑛+1) (2𝑛+1)

6− 2𝑛 − 10 + 4

5𝑛(𝑛+1) (2𝑛+1)

6 - 2n – 6

5𝑛(𝑛 + 1) (2𝑛 + 1) − 12𝑛 − 36

6

𝑛(5(𝑛 + 1) (2𝑛 + 1) − 12) − 36

6

𝑛(5(2𝑛2 + 3𝑛 + 1) − 12) − 36

6

𝑛(10𝑛2 + 15𝑛 + 5 − 12) − 36

6

R// ∑ (5 𝑖2 − 2)𝑛

𝑖=3 =

𝑛(10𝑛2+15𝑛+5−7)−36

6

2. ∑ (𝒊𝟐 − 𝟐𝒊)𝒏−𝟐

𝒊=𝟏 en función de n.

Solución:

∑ (𝑖2 − 2𝑖)𝑛−2

𝑖=1 =

(𝑛−2)(𝑛−2+1) (2𝑛−4+1)

6 -

2(𝑛−2) (𝑛−2+1)

2

= (𝑛−2)(𝑛−1) (2𝑛−3)

6− (𝑛 − 2) (𝑛 − 1)

= (𝑛−2)(𝑛−1)

6 (2𝑛 − 3 − 1)

= (𝑛−2)(𝑛−1)

6 (2𝑛 − 4)

= (𝑛−2)(𝑛−1)(2𝑛−4)

6

3. ∑ (𝒊 + 𝟏)𝟑𝒏

𝒊=𝟏

∑ (𝑖 + 1)3𝑛

𝑖=1 = ∑ (𝑖3 + 3𝑖2 + 3𝑖 + 1)𝑛

𝑖=1

= 𝑛2(𝑛+1)2

4 +

3𝑛(𝑛+1)(2𝑛+1)

6 +

3(𝑛+1)𝑛

2 + 𝑛

= 𝑛 (𝑛(𝑛+1)2

4+

3(𝑛+1)(2𝑛+1)

6+

3(𝑛+1)

2+ 1)

= 𝑛 ((𝑛 + 1) (𝑛(𝑛+1)

4+

(2𝑛+1)

2+

3(𝑛+1)

2) + 1)

= 𝑛 (𝑛 + 1) (𝑛(𝑛+1)+4𝑛+2+6𝑛+2

4) + 𝑛

= 𝑛 (𝑛 + 1) (𝑛2+𝑛+4𝑛+2+6𝑛+2

4) + 𝑛

= 𝑛 (𝑛 + 1) (𝑛2+11𝑛+4

4) + 𝑛

4. Simplificar ∑ (𝒊 + 𝟏)𝟑𝒏

𝒊=𝟐

∑ (𝑖 + 1)3𝑛

𝑖=2 = ∑ (𝑖 + 1)3𝑛

𝑖=1- ∑ (𝑖 + 1)3𝑛

𝑖=1

= ∑ (𝑖3 − 3𝑖2 + 3𝑖 − 1)𝑛𝑖=1

= (𝑛2(𝑛+1)2)

4− 3

(𝑛(𝑛+1)(2𝑛+1))

6+

3(𝑛(𝑛+1)

2− 𝑛

= 𝑛2(𝑛+1)2

4−

3(𝑛(𝑛+1)(2𝑛+1))

6+

3𝑛(𝑛+1)

2− 𝑛

= 𝑛(𝑛 + 1) (𝑛(𝑛+1)

4−

(2𝑛+1)

2+

3

2) − 𝑛

= 𝑛(𝑛 + 1) (𝑛(𝑛+1)−4𝑛−2+6

4) − 𝑛

= 𝑛(𝑛 + 1) (𝑛2+𝑛−4𝑛−2+6

4) − 𝑛

= 𝑛(𝑛 + 1) (𝑛2−3𝑛+4

4) − 𝑛

5. 𝐥𝐢𝐦𝒏→∞

𝟏𝟑+𝟐𝟑+𝟑𝟑+⋯+𝒏𝟑

𝒏𝟒 . 𝐥𝐢𝐦

𝒏→∞

𝟏𝟐+𝟐𝟐+𝟑𝟐+⋯+(𝒏−𝟏)𝟐

𝒏𝟑

lim𝑛→∞

(13

𝑛4+

23

𝑛4+

33

𝑛4+ ⋯ +

𝑛3

𝑛4) ( lim

𝑛→∞

12

𝑛3+

22

𝑛3+ ⋯ +

𝑛2 − 2𝑛 + 1

𝑛3)

( lim𝑛→∞

13

𝑛4+ lim

𝑛→∞

23

𝑛4+ ⋯ + lim

𝑛→∞

1

𝑛) ( lim

𝑛→∞

12

𝑛3+ lim

𝑛→∞

22

𝑛3+ ⋯ + lim

𝑛→∞

𝑛2

𝑛3− lim

𝑛→∞

𝑛2

𝑛3− lim

𝑛→∞

2

𝑛2

+ lim𝑛→∞

1

𝑛3)

(0 + 0 + ⋯ + 0)(0 + 0 + ⋯ + 0 + 0 + 0) = 0

6. 𝐥𝐢𝐦𝒏→∞

𝟓+𝟔+𝟕+⋯+𝒏

𝒏𝟐

lim𝑛→∞

(5

𝑛2+

6

𝑛2+

7

𝑛2+ ⋯ +

𝑛

𝑛2)

lim𝑛→∞

5

𝑛2+ lim

𝑛→∞

6

𝑛2+ lim

𝑛→∞

7

𝑛2+ ⋯ + lim

𝑛→∞(

1

𝑛)

0 + 0 + 0 + ⋯ + 0

7. Simplificar ∑ ∑ 𝟐𝒊𝒌=𝟎

𝒏𝒊=𝟏

= ∑ 2𝑖 =2(𝑛+1)𝑛

2= 𝑛(𝑛 + 1)𝑛

1

GUIA 2

1. ∫ 𝒙𝟐𝒅𝒙𝟏

𝟎 Con 4, 8, 16 y 32 rectángulos.

𝑏 − 𝑎

𝑛=

1 − 0

𝑛=

1

𝑛

= 𝑋𝑖−1 = 0 + (𝑖 − 1)1

𝑛 =

(𝑖−1)

𝑛

= 𝑓(𝑋𝑖−1) = (𝑖−1)2

𝑛2

= ∑ 𝑓(𝑋𝑖−1)

𝑏−𝑎

𝑛

𝑛𝑖=1 = ∑

(𝑖−1)2

𝑛2 (1

𝑛)𝑛

𝑖=1

= ∑(𝑖−1)2

𝑛3=

1

𝑛3∑ (𝑖2 − 2𝑖 + 1)𝑛

𝑖=1𝑛𝑖=1

= 1

𝑛3 (𝑛(𝑛+1)(2𝑛+1)

6−

1

𝑛3(2)

𝑛(𝑛+1)

2+

1

𝑛2)

= 1

𝑛2

(𝑛+1)(2𝑛+1)

6−

1

𝑛2(𝑛 + 1) +

1

𝑛2

= 1

𝑛2 ((𝑛+1)(2𝑛+1)

6− (𝑛 + 1) + 1)

= 1

𝑛2 ((𝑛+1)(2𝑛+1)−6𝑛−6+6

6)

= 1

𝑛2 (2𝑛2+𝑛+2𝑛+1−6𝑛−6+6

6)

= 1

𝑛2 (2𝑛2−3𝑛6+1

6)

𝑃𝑎𝑟𝑎: 𝑛 = 4

∑ 𝑓𝑋𝑖−1

𝑏 − 𝑎

𝑛

4

𝑖=1

= 1

42(

2(4)2 − 3(4) + 1

6)

= 1

16(

32−12+1

6)

= 1

16(

21

6) = 0.21

𝑃𝑎𝑟𝑎: 𝑛 = 8

∑ 𝑓𝑋𝑖−1

𝑏 − 𝑎

8

8

𝑖=1

= 1

82(

2(8)2 − 3(8) + 1

6)

= 1

64(

128−24+1

6)

= 1

64(17.5) = 0.2734

𝑃𝑎𝑟𝑎: 𝑛 = 16

∑ 𝑓𝑋𝑖−1

𝑏 − 𝑎

16

16

𝑖=1

= 1

162(

2(16)2 − 3(16) + 1

6)

= 0.3027

𝑃𝑎𝑟𝑎: 𝑛 = 32

∑ 𝑓𝑋𝑖−1

𝑏 − 𝑎

32

32

𝑖=1

= 1

322(

2(32)2 − 3(32) + 1

6)

= 0.32

2. Calcular ∫ 𝒙𝟑𝒅𝒙𝟏

𝟎 Con 4, 8, 16 y 32 rectángulos.

𝑏 − 𝑎

𝑛=

1 − 0

𝑛=

1

𝑛

𝑋𝑖−1 = 0 + (𝑖 − 1)1

𝑛 =

(𝑖 − 1)

𝑛

𝑓(𝑋𝑖−1) = ( 𝑖 − 1

𝑛)

3

= (𝑖 − 1)3

𝑛3

= ∑ 𝑓(𝑋𝑖−1)

𝑏−𝑎

𝑛

𝑛𝑖=1 = ∑

(𝑖−1)3

𝑛3 (1

𝑛)𝑛

𝑖=1

= ∑(𝑖−1)3

𝑛4𝑛𝑖=1

= 1

𝑛4∑ (𝑖3 − 3𝑖2 + 3𝑖 − 1)𝑛

𝑖=1

= 1

𝑛4 (𝑛2(𝑛+1)2

4) −

3𝑛(𝑛+1)(2𝑛+1)

6+

3𝑛(𝑛+1)

2− 𝑛

= 1

𝑛4 (𝑛(𝑛 + 1)𝑛(𝑛+1)

4−

2𝑛+1

2+

3

2)

= 1

𝑛4 (𝑛(𝑛 + 1) (𝑛(𝑛+1)−4𝑛−2+6

4))

= 1

𝑛4 (𝑛(𝑛 + 1) (𝑛2+𝑛−4𝑛−2+6

4))

= 1

𝑛4(𝑛 + 1) (

𝑛2−3𝑛+4

4)

= (𝑛+1)(𝑛2−3𝑛+4)

4𝑛3

𝑃𝑎𝑟𝑎: 𝑛 = 4

∑ 𝑓𝑋𝑖−1

𝑏 − 𝑎

𝑛

4

𝑖=1

= (4 + 1)(42 − 3(4) + 4

4(4)3

= 0.156

𝑃𝑎𝑟𝑎: 𝑛 = 8

∑ 𝑓𝑋𝑖−1

𝑏 − 𝑎

𝑛

8

𝑖=1

= (9)(82 − 3(8) + 4

4(8)3

= 0.1933

𝑃𝑎𝑟𝑎: 𝑛 = 16

∑ 𝑓𝑋𝑖−1

𝑏 − 𝑎

𝑛

16

𝑖=1

= (17)(162 − 3(16) + 4)

4(16)3

= 0.2199

𝑃𝑎𝑟𝑎: 𝑛 = 32

∑ 𝑓𝑋𝑖−1

𝑏 − 𝑎

𝑛

32

𝑖=1

= (33)(322 − 3(32) + 4)

4(32)3

= 0.2346

3. Demostrar que: ∫ 𝒙𝟑𝒅𝒙 =𝟏

𝟒

𝟏

𝟎 (Use sumas de Riemann)

4. ∫ (𝟐𝒙 + 𝒙𝟐)𝒅𝒙𝟏

𝟎

𝑏 − 𝑎

𝑛=

1 − 0

𝑛=

1

𝑛

𝑋𝑖−1 = 0 + (𝑖 − 1)1

𝑛

𝑓(𝑋𝑖−1) = 2 ((𝑖 − 1)1

𝑛) + ((𝑖 − 1)

1

𝑛)

2

𝑓(𝑋𝑖−1) = 2(𝑖 − 1)

𝑛+

(𝑖 − 1)2

𝑛2

= ∑ 𝑓(𝑋𝑖−1)

𝑏−𝑎

𝑛

𝑛𝑖=1 = ∑ (

2(𝑖−1)

𝑛+

(𝑖−1)2

𝑛2 )1

𝑛

𝑛𝑖=1

= ∑ (2(𝑖−1)

𝑛2+

(𝑖−1)2

𝑛3 )𝑛𝑖=1

= 2

𝑛2∑ 𝑖 − 1𝑛

𝑖=1 +1

𝑛3∑ (𝑖 − 1)2𝑛

= 2

𝑛2

(𝑛2+𝑛)

2−

2

𝑛+

1

𝑛3

(2𝑛3+3𝑛2+𝑛)

6−

2

𝑛3

(𝑛2+𝑛)

2+

1

𝑛2

= 1 +1

𝑛−

2

𝑛+

1

3+

1

2𝑛+

1

6𝑛2−

1

𝑛−

1

𝑛2+

1

𝑛2

= 1 −3

2𝑛+

1

3+

1

6𝑛

𝐴 = ∫ (2𝑥 + 𝑥2)𝑑𝑥 = lim𝑛→∞

∑ 𝑓(𝑥𝑖−1)

𝑏 − 𝑎

𝑛= lim

𝑛→∞(1 −

3

2𝑛+

1

3+

1

6𝑛)

𝑛

𝑖=1

1

0

= (1 +1

3) =

4

3 𝑢2

5. ∫ 𝒆𝒙𝒅𝒙𝟏

𝟎 Con 4, 8 y 16 rectángulos.

𝑏 − 𝑎

𝑛 𝑓(𝑋𝑖−1) = 𝑒

(𝑖−1)𝑛 𝑋𝑖−1 = 0 + (𝑖 − 1)

1

𝑛=

𝑖 − 1

𝑛

∑ 𝑓(𝑋𝑖−1)

𝑏 − 𝑎

𝑛

𝑛

𝑖=1

= ∑ 𝑒(𝑖−1)

𝑛

𝑛

𝑖=1

1

𝑛

= 1

𝑛∑ 𝑒

(𝑖−1)

𝑛𝑛𝑖=1

= 1

𝑛∑ 𝑒

𝑖

𝑛−

1

𝑛 =1

𝑛

𝑛𝑖=1 ∑ 𝑒

𝑖

𝑛𝑛𝑖=1 . 𝑒

1

𝑛

= 1

𝑛𝑒

−1

𝑛 ∑ (𝑒1

𝑛)𝑖

𝑛𝑖=1

Se utiliza la expresión 𝑎 + 𝑎2 + 𝑎3 + ⋯ + 𝑎𝑛 = [(𝑎𝑛−1−1)

(𝑎−1)] − 1

∑ 𝑓(𝑋𝑖−1)

𝑏 − 𝑎

𝑛

𝑛

𝑖=1

=1

𝑛𝑒

−1𝑛 [

[(𝑒1/𝑛)(𝑛+1)

− 1]

𝑒1𝑛 − 1

− 1]

= 1

𝑛𝑒

−1

𝑛 [(𝑒

1 + 1𝑛

𝑒1𝑛

− 1) − 1]

𝑃𝑎𝑟𝑎: 𝑛 = 4

∑ 𝑓(𝑋𝑖−1)

4

𝑖=1

𝑏 − 𝑎

𝑛=

1

𝑛 𝑒−

14 [

𝑒1+14 − 1

𝑒14 − 1

− 1] = 1.51

𝑃𝑎𝑟𝑎: 𝑛 = 8

∑ 𝑓(𝑋𝑖−1) 𝑏 − 𝑎

𝑛=

1

8 𝑒−

18 [

𝑒1+18 − 1

𝑒18 − 1

− 1] = 1.6131

𝑛=8

𝑖=1

𝑃𝑎𝑟𝑎: 𝑛 = 16

∑ 𝑓(𝑋𝑖−1) 𝑏 − 𝑎

𝑛=

1

16 𝑒−

116 [

𝑒1+1

16 − 1

𝑒1

16 − 1− 1] = 1.665

𝑛=16

𝑖=1

GUIA 3

1. ∫ (𝒙𝟑 + 𝟐𝒙 − 𝟏)𝒅𝒙𝟑

−𝟏 = ∫ 𝑥3𝑑𝑥

3

−1+ ∫ 2𝑑𝑥 − ∫ 1𝑑𝑥

3

−1

3

−1

= [𝑥4

4]

−1

3

+ [𝑥2]−1 3 − [𝑥]−1

3

= 34

4−

(−1)4

4+ 32 − (−1)2 − [3 − 1]

= 81

4−

1

4+ 9 − 1 − 3 + 1

=26 𝑢2

2. ∫ 𝑺𝒆𝒏𝑿 𝒅𝒙 = [− cos 𝑥]0

𝜋2⁄

= −𝑐𝑜𝑠𝜋

2⁄ − [− cos 𝜋2⁄ ]

𝝅

𝟐𝟎

= 0 − (−1) = 1 𝑢2

3. ∫ (𝒆𝒙 − 𝟏)𝒅𝒙𝟑

−𝟏

∫ (𝑒𝑥 − 1) = ∫ (𝑒𝑥 − 1) + ∫ (𝑒𝑥 − 1)3

0

0

−1

3

−1

= [𝑒𝑥 − 𝑥]−1 0 + [𝑒𝑥 − 𝑥]

= [𝑒0 − 0] − (𝑒−1 − (−1)) + [𝑒3 − 3] − [𝑒0 − 0]

= −[1 − 0 − (0.37 + 1)] + [20.09 − 3 − 1]

= −[1 − 1.37] + 16.09

= 16.46

4. ∫ 𝒓 √𝒓𝟐 − 𝒙𝟐𝒅𝒙𝟏

𝟎

𝑥 = 𝑟𝑆𝑒𝑛 𝜃 𝑑𝑥 = 𝑟𝐶𝑜𝑠 𝜃 𝑑 𝜃 𝜃 = 𝐴𝑟𝑐𝑆𝑒𝑛𝑥

𝑟

𝑥2𝑟2𝑆𝑒𝑛2𝜃 𝑆𝑒𝑛𝜃 =𝑥

𝑟 𝐶𝑜𝑠𝜃 = √

𝑟2−𝑥2

𝑟

∫ 𝑟√𝑟2 − 𝑥2 = 𝑟 ∫ √𝑟2−𝑟2𝑆𝑒𝑛2𝜃 𝑟 𝐶𝑜𝑠 𝜃 𝑑 𝜃

= 𝑟 ∫ 𝑟√1 − 𝑆𝑒𝑛2 𝜃 𝑟 𝐶𝑜𝑠 𝜃 𝑑 𝜃

= 𝑟 ∫ 𝑟2 √𝐶𝑜𝑠2 𝜃 𝐶𝑜𝑠 𝜃 𝑑 𝜃

= 𝑟3 ∫ 𝐶𝑜𝑠 𝜃 𝐶𝑜𝑠 𝜃 𝑑 𝜃

𝑋 𝑌 = 𝑒𝑥 − 1 -5 -2 -1 0 1 1.5 2 3

-0.99 -0.86 -0.63 0 1.71 3.48 6.38 14.08

= 𝑟3 ∫ 𝐶𝑜𝑠2 𝜃 𝑑 𝜃

= 𝑟3 ∫1

2(1 + 𝐶𝑜𝑠 2 𝜃) 𝑑 𝜃

= 𝑟3

2 [∫ 𝑑 𝜃 + ∫ 𝐶𝑜𝑠 2 𝜃 𝑑 𝜃]

= 𝑟3

2[𝜃 +

1

2 𝑆𝑒𝑛 2 𝜃] + 𝐶

= 𝑟3

2[𝐴𝑟𝑐𝑆𝑒𝑛 (

𝑥

𝑟) +

1

2(2𝑆𝑒𝑛𝜃𝐶𝑜𝑠𝜃)] + 𝐶

= 𝑟3

2[𝐴𝑟𝑐𝑆𝑒𝑛 (

𝑥

𝑟) + (

𝑥

𝑟) (

√𝑟2−𝑥2

𝑟)] + 𝐶

= ∫ 𝑟√𝑟2 − 𝑥21

0 𝑑𝑥 =

𝑟3

2[𝐴𝑟𝑐𝑆𝑒𝑛 (

1

𝑟) +

1

𝑟

√𝑟2−1

𝑟− (𝐴𝑟𝑐𝑆𝑒𝑛(0) +

0

𝑟√

𝑟2

0)]

= 𝑟3

2𝐴𝑟𝑐𝑆𝑒𝑛 (

1

𝑟) +

𝑟√𝑟2−1

2

5. Calcular el Área de la elipse

𝑥2

𝑎2+

𝑦2

𝑏2= 1

𝑦2

𝑏2= 1 −

𝑥2

𝑎2 𝑦2 = 𝑏2 −

𝑏2𝑥2

𝑎2

𝑦2 = 𝑏2 (1 −𝑥2

𝑎2) 𝑦 = 𝑏√1 −

𝑥2

𝑎2 𝑦 = 𝑏√

1

𝑎2(𝑎2 − 𝑥2)

𝑦 =𝑏

𝑎√𝑎2 − 𝑥2

Tomado el intervalo desde [0, a]

𝐴 = 𝐴𝑟𝑒𝑎 𝑑𝑒 𝑙𝑎 𝐸𝑙𝑖𝑝𝑠𝑒

𝐴 = 4 ∫𝑏

𝑎√𝑎2 − 𝑥20

𝑎𝑑𝑥 𝑆𝑒𝑛𝜃

𝑥

𝑎 𝜃 = 𝐴𝑟𝑐𝑆𝑒𝑛

𝑥

𝑎

𝐴 = 4𝑏

𝑎∫ √𝑎2 − 𝑥2𝑎

0𝑑𝑥 𝑥 = 𝑎 𝑆𝑒𝑛 𝜃 𝐶𝑜𝑠 𝜃 = √

𝑥2−𝑎2

𝑎

𝑑𝑥 = 𝑎 𝐶𝑜𝑠𝜃 𝑑 𝜃

Calculando la Integral:

∫ √𝑎2 − 𝑥2 𝑑𝑥 = ∫ √𝑎2 − 𝑎2 𝑆𝑒𝑛2 𝜃 𝑎 𝐶𝑜𝑠 𝜃 𝑑 𝜃

= 𝑎2 ∫ √1 − 𝑆𝑒𝑛2𝜃 𝐶𝑜𝑠 𝜃 𝑑 𝜃

∫ √𝑎2 − 𝑥2 𝑑𝑥 = 𝑎2 ∫ √1 − 𝑆𝑒𝑛2 𝜃 𝐶𝑜𝑠 𝜃 𝑑 𝜃

= 𝑎2 ∫ √𝐶𝑜𝑠2𝜃 𝐶𝑜𝑠 𝜃 𝑑 𝜃

= 𝑎2 ∫ 𝐶𝑜𝑠2𝜃 𝑑 𝜃 = 𝑎2 ∫1

2(1 + 𝐶𝑜𝑠 2 𝜃) 𝑑 𝜃

= 𝑎2

2[𝜃 +

1

2 𝑆𝑒𝑛 2 𝜃] =

𝑎2

2[0 +

1

2(2 𝑆𝑒𝑛𝜃 𝐶𝑜𝑠𝜃)]

= 𝑎2

2[𝐴𝑟𝑐𝑆𝑒𝑛 (

𝑥

𝑎) + (

𝑥

𝑎) (√

𝑥2−𝑎2

𝑎)] + 𝐶

= 𝑎2

2[𝐴𝑟𝑐𝑆𝑒𝑛 (

𝑥

𝑎) +

𝑥√𝑥2−𝑎2

𝑎2 ] + 𝐶

𝐴 =4𝑏

𝑎∫ √𝑎2 − 𝑥2

2

0

𝑑𝑥 = 4𝑏

𝑎(

𝑎2

2) [𝐴𝑟𝑐𝑆𝑒𝑛 (

𝑥

𝑎) +

𝑥√𝑥2 − 𝑎2

𝑎2]

0

𝑎

= 2 𝑏 𝑎 [𝐴𝑟𝑐𝑆𝑒𝑛 (𝑎

𝑎) +

𝑎√𝑎2−𝑎2

𝑎2 ] − (𝐴𝑟𝑐𝑆𝑒𝑛 (0

𝑎) +

0√02−𝑎2

𝑎2 )

= 2 𝑏 𝑎 (𝜋

2)

= 𝑎 𝑏 𝜋

R//A eclipse = 𝑎 𝑏 𝜋