вправа 14.5 гдз 11 клас алгебра Істер 2019
Вправа 14.5
Обчисліть інтеграл:
Умова:
Відповідь ГДЗ:
1 1
✔ 1) ∫ (х - 3)dx = х2/2 - 3х | =
0 0
= 12/2 - 3 • 1 - (02/2 - 3 • 0) =
= 1/2 - 3 = -2,5;
4 4
✔ 2) ∫ (х + 1)dx = х2/2 + х | =
1 1
= 42/2 + 4 - (12/2 + 1) =
= 8 + 4 - 1/2 - 1 = 10,5;
2 2
✔ 3) ∫ (4х + 2)dx = 2х2 + 2х | =
0 0
= 2 • 22 + 2 • 2 - (2 • 02 + 2 • 0) = 12;
3 3
✔ 4) ∫ (2х - 1)dx = х2 - х | =
1 1
= 32 - 3 - (12 - 1) = 9 - 3 - 1 + 1 = 6;
1 1
✔ 5) ∫ (х2 - 1)dx = х3/3 - х | =
0 0
= 13/3 - 1 - (03/3 - 0) = -2/3;
5 5
✔ 6) ∫ (3х2 + х)dx = х3 + х2/2 | =
1 1
= 53 + 52/2 - (13 + 12/2) =
= 125 + 12,5 - 1 - 1/2 = 136;
1 1
✔ 7) ∫ (х2 + 4х - 1)dx = х3/3 + 2х2 - х | =
0 0
= 13/3 + 2 • 12 - 1 - (03/3 + 2 • 02 - 0) = 1 1/3;
2 2
✔ 8) ∫ (9х2 + 2х - 1)dx = 3х2 + х2 - х | =
1 1
= 3 • 23 + 22 - 2 - (3 • 13 + 12 - 1) =
= 24 + 4 - 2 - 3 - 1 + 1 = 23.
✔ 1) ∫ (х - 3)dx = х2/2 - 3х | =
0 0
= 12/2 - 3 • 1 - (02/2 - 3 • 0) =
= 1/2 - 3 = -2,5;
4 4
✔ 2) ∫ (х + 1)dx = х2/2 + х | =
1 1
= 42/2 + 4 - (12/2 + 1) =
= 8 + 4 - 1/2 - 1 = 10,5;
2 2
✔ 3) ∫ (4х + 2)dx = 2х2 + 2х | =
0 0
= 2 • 22 + 2 • 2 - (2 • 02 + 2 • 0) = 12;
3 3
✔ 4) ∫ (2х - 1)dx = х2 - х | =
1 1
= 32 - 3 - (12 - 1) = 9 - 3 - 1 + 1 = 6;
1 1
✔ 5) ∫ (х2 - 1)dx = х3/3 - х | =
0 0
= 13/3 - 1 - (03/3 - 0) = -2/3;
5 5
✔ 6) ∫ (3х2 + х)dx = х3 + х2/2 | =
1 1
= 53 + 52/2 - (13 + 12/2) =
= 125 + 12,5 - 1 - 1/2 = 136;
1 1
✔ 7) ∫ (х2 + 4х - 1)dx = х3/3 + 2х2 - х | =
0 0
= 13/3 + 2 • 12 - 1 - (03/3 + 2 • 02 - 0) = 1 1/3;
2 2
✔ 8) ∫ (9х2 + 2х - 1)dx = 3х2 + х2 - х | =
1 1
= 3 • 23 + 22 - 2 - (3 • 13 + 12 - 1) =
= 24 + 4 - 2 - 3 - 1 + 1 = 23.