вправа 14.33 гдз 11 клас алгебра Істер 2019
Вправа 14.33
Обчисліть інтеграл:
Умова:
Відповідь ГДЗ:
3
✔ 1) ∫ (2х - х2)2dx
0
Перетворюємо вираз (2х - х2)2
(2х - х2)2 = 4х2 - 4х3 + х4
3 3 3 3
∫ (2х - х2)2dx = 4∫х2dx - 4∫х3dx + ∫х4dx =
0 0 0 0
3 3 3
= 4 • х3/3 | - 4х4/4 | + х5/5 | =
0 0 0
3 3 3
= 4/3 х3 | - х4 | + х5/5 | =
0 0 0
= 4/3 • 33 - 34 + 35 =
= 4/3 • 27 - 81 + 243 =
= 36 + 81 + 243 = 198;
4
✔ 2) ∫ (2х3-х2-1) : х2 dx
2
Перетворюємо вираз
(2х3-х2-1) : х2 =
= 2х3 : х2 - х2 : х2 - 1 : х2 =
= 2х - 1 - 1/х2
4 4 4 4
∫(2х3-х2-1) : х2 dx = 2∫хdx - ∫dx - ∫1/х2 dx =
2 2 2 2
4 4 4
= | - х | + 1/х | =
2 2 2
= 42 - 22 - 4 + 2 + 1/4 - 1/2 =
= 16 - 4 - 4 + 2 + 0,25 - 0,5 = 9,75;
2π
✔ 3) ∫ cos2хdx
0
cos2х = (1+cos2х) : 2
2π 2π 2π
∫cos2хdx = ∫(1+cos2х) : 2 dx = 1/2∫(1 + cos2х)dx =
0 0 0
2π 2π
= 1/2 ∫dx + 1/2∫cos2хdx =
0 0
2π 2π
= 1/2 х | + 1/2 sin2х | =
0 0
= 1/2 • 2π + 1/2 sin4π = π + 0 = π;
1
✔ 4) ∫ (12х-6х) : 3х dx
0
Перетворюємо вираз
(12х-6х) : 3х = (2х•6х-6х) : 3х =
= (6х(2х-1)) : 3х = (2х•3х•(2х-1)) : 3х =
= 2х(2х - 1) = 2х • 2х - 1 • 2х = 4х - 2х
1 1 1
∫(12х-6х) : 3х dx = ∫4хdx - ∫2хdx =
0 0 0
= 4х/ln4 | - 2х/ln2 | =
= 4/ln4 - 40/ln4 - 2/ln2 + 20/ln2 =
= 4/ln4 - 1/ln4 - 2/ln2 + 1/ln2 =
= 3/ln4 - 1/ln2 = 3/1,38 - 1/0,68 =
= 2,2 - 1,47 = 0,76.
✔ 1) ∫ (2х - х2)2dx
0
Перетворюємо вираз (2х - х2)2
(2х - х2)2 = 4х2 - 4х3 + х4
3 3 3 3
∫ (2х - х2)2dx = 4∫х2dx - 4∫х3dx + ∫х4dx =
0 0 0 0
3 3 3
= 4 • х3/3 | - 4х4/4 | + х5/5 | =
0 0 0
3 3 3
= 4/3 х3 | - х4 | + х5/5 | =
0 0 0
= 4/3 • 33 - 34 + 35 =
= 4/3 • 27 - 81 + 243 =
= 36 + 81 + 243 = 198;
4
✔ 2) ∫ (2х3-х2-1) : х2 dx
2
Перетворюємо вираз
(2х3-х2-1) : х2 =
= 2х3 : х2 - х2 : х2 - 1 : х2 =
= 2х - 1 - 1/х2
4 4 4 4
∫(2х3-х2-1) : х2 dx = 2∫хdx - ∫dx - ∫1/х2 dx =
2 2 2 2
4 4 4
= | - х | + 1/х | =
2 2 2
= 42 - 22 - 4 + 2 + 1/4 - 1/2 =
= 16 - 4 - 4 + 2 + 0,25 - 0,5 = 9,75;
2π
✔ 3) ∫ cos2хdx
0
cos2х = (1+cos2х) : 2
2π 2π 2π
∫cos2хdx = ∫(1+cos2х) : 2 dx = 1/2∫(1 + cos2х)dx =
0 0 0
2π 2π
= 1/2 ∫dx + 1/2∫cos2хdx =
0 0
2π 2π
= 1/2 х | + 1/2 sin2х | =
0 0
= 1/2 • 2π + 1/2 sin4π = π + 0 = π;
1
✔ 4) ∫ (12х-6х) : 3х dx
0
Перетворюємо вираз
(12х-6х) : 3х = (2х•6х-6х) : 3х =
= (6х(2х-1)) : 3х = (2х•3х•(2х-1)) : 3х =
= 2х(2х - 1) = 2х • 2х - 1 • 2х = 4х - 2х
1 1 1
∫(12х-6х) : 3х dx = ∫4хdx - ∫2хdx =
0 0 0
= 4х/ln4 | - 2х/ln2 | =
= 4/ln4 - 40/ln4 - 2/ln2 + 20/ln2 =
= 4/ln4 - 1/ln4 - 2/ln2 + 1/ln2 =
= 3/ln4 - 1/ln2 = 3/1,38 - 1/0,68 =
= 2,2 - 1,47 = 0,76.