вправа 14.15 гдз 11 клас алгебра Істер 2019
Вправа 14.15
Обчисліть інтеграл:
Умова:
Відповідь ГДЗ:
9
✔ 1) ∫ (х - 4√х)dx
0
9 9 9 9
∫ хdx - ∫ 4√хdx = х2/2 ∫ - 4 ∫ х1/2dx =
0 0 0 0
9 9 9 9
= х2/2 | - (4•х1/2+1):(1/2+1) | = х2/2 | - (4•х3/2):3/2 | =
0 0 0 0
9 9
= х2/2 | - (8•х3/2):3 | = 92/2 - 0 - (8•32•3/2):3 + 0 =
0 0
= 81/2 - (8•279):3 = 40,5 - 72 = -31,5;
4
✔ 2) ∫ (4х + 2/√х)dx
1
4 4 4 4
∫ 4хdx + ∫ 2/√хdx = ∫ 4хdx + ∫ 2х-1/2dx =
1 1 1 1
4 4 4 4
= 4х2/2 | + (2•х-1/2+1):(-1/2+1) | = 2х2 | + (2х1/2):1/2 | =
1 1 1 1
4 4
= 2х2 | + 4х1/2 | = 2 • 42 - 2 • 1 + 4 • 41/2 - 4 • 1 =
1 1
= 32 - 2 + 8 - 4 = 34;
1
✔ 3) ∫ (8х - 3√х)dx
0
1 1 1 1
∫ 8хdx - ∫ х1/3dx = 8х2/2 | - (х1/3+1):(1/3+1) | =
0 0 0 0
1 1 1 1
= 4х2 | - х4/3 : 4/3 | = 4х2 | - (3•х4/3) : 4 | =
0 0 0 0
= 4 • 1 - 0 - (3•1):4 + 0 = 4 - 0,75 = 3,25;
8
✔ 4) ∫ (х - 6/3√х)dx
1
8 8 8 8
∫ хdx - ∫ 6 • х-1/3dx = х2/2 | - (6•х-1/3+1):(-1/3+1) | =
1 1 1 1
8 8 8 8
= х2/2 | - (6•х2/3):2/3 | = х2/2 | - (18•х2/3):2 | =
1 1 1 1
= 82/2 - 12/2 - 9 • 23•2/3 + 9 • 1 =
= 32 - 0,5 - 36 + 9 = 4,5.
✔ 1) ∫ (х - 4√х)dx
0
9 9 9 9
∫ хdx - ∫ 4√хdx = х2/2 ∫ - 4 ∫ х1/2dx =
0 0 0 0
9 9 9 9
= х2/2 | - (4•х1/2+1):(1/2+1) | = х2/2 | - (4•х3/2):3/2 | =
0 0 0 0
9 9
= х2/2 | - (8•х3/2):3 | = 92/2 - 0 - (8•32•3/2):3 + 0 =
0 0
= 81/2 - (8•279):3 = 40,5 - 72 = -31,5;
4
✔ 2) ∫ (4х + 2/√х)dx
1
4 4 4 4
∫ 4хdx + ∫ 2/√хdx = ∫ 4хdx + ∫ 2х-1/2dx =
1 1 1 1
4 4 4 4
= 4х2/2 | + (2•х-1/2+1):(-1/2+1) | = 2х2 | + (2х1/2):1/2 | =
1 1 1 1
4 4
= 2х2 | + 4х1/2 | = 2 • 42 - 2 • 1 + 4 • 41/2 - 4 • 1 =
1 1
= 32 - 2 + 8 - 4 = 34;
1
✔ 3) ∫ (8х - 3√х)dx
0
1 1 1 1
∫ 8хdx - ∫ х1/3dx = 8х2/2 | - (х1/3+1):(1/3+1) | =
0 0 0 0
1 1 1 1
= 4х2 | - х4/3 : 4/3 | = 4х2 | - (3•х4/3) : 4 | =
0 0 0 0
= 4 • 1 - 0 - (3•1):4 + 0 = 4 - 0,75 = 3,25;
8
✔ 4) ∫ (х - 6/3√х)dx
1
8 8 8 8
∫ хdx - ∫ 6 • х-1/3dx = х2/2 | - (6•х-1/3+1):(-1/3+1) | =
1 1 1 1
8 8 8 8
= х2/2 | - (6•х2/3):2/3 | = х2/2 | - (18•х2/3):2 | =
1 1 1 1
= 82/2 - 12/2 - 9 • 23•2/3 + 9 • 1 =
= 32 - 0,5 - 36 + 9 = 4,5.