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