вправа 14.21 гдз 11 клас алгебра Істер 2019
Вправа 14.21
Обчисліть інтеграл:
Умова:
Відповідь ГДЗ:
1
✔ 1) ∫ (х + 1)3dx
-1
перетворюємо (х + 1)3
(х + 1)3 = х3 + 3х2 + 3х + 1
1
∫ (х3 + 3х2 + 3х + 1)dx =
-1
1 1 1 1
∫ х3dx + ∫ 3х2dx + ∫ 3хdx + ∫ dx =
-1 -1 -1 -1
1 1 1 1
х4/4 | + 3х3/3 | + 3х2/2 | + х | =
-1 -1 -1 -1
= 1/4 - (-1)4/4 + 1 - (-1)3 + 3/2 -
- (3•(-1)2):2 + 1 - (-1) = 1/4 - 1/4 + 1 + 3/2 - 3/2 + 1 + 1 = 3;
2π 2π
✔ 2) ∫ sin х/4 dx = -4cos х/4 | =
0 0
= -4cos 2π/4 + 4cos 0/4 = 0 + 4 • 1 = 4;
1
✔ 3) ∫ (1 - 2х)4dx
0
перетворюємо вираз: (1 - 2х)4
(1 - 2х)4 = 1 - 4 • 2х + 6 • 4х2 - 4,8х3 + 16х4 =
= 1 - 8х + 24х2 + 32х3 + 16х4
1
∫ (1 - 8х + 24х2 - 32х3 + 16х4)dx =
0
1 1 1 1 1
= ∫ dx - 8∫хdx + 24∫х2dx - 32∫х3dx + 16∫х4dx =
0 0 0 0 0
1 1 1 1 1
= х | - 8х2/2 | + 24х3/3 | - 32х4/4 | + 16х5/5 | =
0 0 0 0 0
1 1 1 1 1
= х | - 4х2 | + 8х3 | - 8х4 | + 16х5/5 | =
0 0 0 0 0
1 - 4 + 8 - 8 + 16/5 = -3 + 3 1/5 = 0,2;
π/3 π/3
✔ 4) ∫ dx : (cos2(х/2-π/6)) = 2tg(х/2-π/6) | =
0 0
= 2tg(π/6 - π/6) = 2tg0 = 0;
8 8 8
✔ 5) ∫ √х+1 dx = ∫ (х + 1)1/2dx = ((х+1)1/2+1):(1/2+1) | =
3 3 3
8 8
= (х+1)3/2 : 3/2 | = (2(х+1)3/2):3 | =
3 3
= (2(8+1)3/2):3 - (2(3+1)3/2):3 =
= (2•93/2):3 - (2•43/2):3 - (2(3+1)3/2):3 =
= (2•93/2):3 - (2•43/2):3 =
= (2•32•3/2):3 - (2•22•3/2):3 =
= (2•33):3 - (2•23):3 = (2•27):3 - (2•8):3 =
= 54/3 - 16/3 = 38/3 = 12 2/3;
3 3
✔ 6) ∫ dx:4√1+5х = ∫ (1 + 5х)1/4dx =
0 0
3 3
= 1/5 • (1+5х)1/4+1:(1/4+1) | = 1/5 • (1+5х)5/4:5/4 | =
0 0
3
= (4•(1+5х)5/4):25 | = (4•(1+5•3)5/4):25 =
0
= (4•165/4):25 = (4•24•5/4):25 = (4•25):25 =
= (4•32):25 = 128/25 = 5,12.
✔ 1) ∫ (х + 1)3dx
-1
перетворюємо (х + 1)3
(х + 1)3 = х3 + 3х2 + 3х + 1
1
∫ (х3 + 3х2 + 3х + 1)dx =
-1
1 1 1 1
∫ х3dx + ∫ 3х2dx + ∫ 3хdx + ∫ dx =
-1 -1 -1 -1
1 1 1 1
х4/4 | + 3х3/3 | + 3х2/2 | + х | =
-1 -1 -1 -1
= 1/4 - (-1)4/4 + 1 - (-1)3 + 3/2 -
- (3•(-1)2):2 + 1 - (-1) = 1/4 - 1/4 + 1 + 3/2 - 3/2 + 1 + 1 = 3;
2π 2π
✔ 2) ∫ sin х/4 dx = -4cos х/4 | =
0 0
= -4cos 2π/4 + 4cos 0/4 = 0 + 4 • 1 = 4;
1
✔ 3) ∫ (1 - 2х)4dx
0
перетворюємо вираз: (1 - 2х)4
(1 - 2х)4 = 1 - 4 • 2х + 6 • 4х2 - 4,8х3 + 16х4 =
= 1 - 8х + 24х2 + 32х3 + 16х4
1
∫ (1 - 8х + 24х2 - 32х3 + 16х4)dx =
0
1 1 1 1 1
= ∫ dx - 8∫хdx + 24∫х2dx - 32∫х3dx + 16∫х4dx =
0 0 0 0 0
1 1 1 1 1
= х | - 8х2/2 | + 24х3/3 | - 32х4/4 | + 16х5/5 | =
0 0 0 0 0
1 1 1 1 1
= х | - 4х2 | + 8х3 | - 8х4 | + 16х5/5 | =
0 0 0 0 0
1 - 4 + 8 - 8 + 16/5 = -3 + 3 1/5 = 0,2;
π/3 π/3
✔ 4) ∫ dx : (cos2(х/2-π/6)) = 2tg(х/2-π/6) | =
0 0
= 2tg(π/6 - π/6) = 2tg0 = 0;
8 8 8
✔ 5) ∫ √х+1 dx = ∫ (х + 1)1/2dx = ((х+1)1/2+1):(1/2+1) | =
3 3 3
8 8
= (х+1)3/2 : 3/2 | = (2(х+1)3/2):3 | =
3 3
= (2(8+1)3/2):3 - (2(3+1)3/2):3 =
= (2•93/2):3 - (2•43/2):3 - (2(3+1)3/2):3 =
= (2•93/2):3 - (2•43/2):3 =
= (2•32•3/2):3 - (2•22•3/2):3 =
= (2•33):3 - (2•23):3 = (2•27):3 - (2•8):3 =
= 54/3 - 16/3 = 38/3 = 12 2/3;
3 3
✔ 6) ∫ dx:4√1+5х = ∫ (1 + 5х)1/4dx =
0 0
3 3
= 1/5 • (1+5х)1/4+1:(1/4+1) | = 1/5 • (1+5х)5/4:5/4 | =
0 0
3
= (4•(1+5х)5/4):25 | = (4•(1+5•3)5/4):25 =
0
= (4•165/4):25 = (4•24•5/4):25 = (4•25):25 =
= (4•32):25 = 128/25 = 5,12.