Jawaban:

Integral

Penjelasan dengan langkah-langkah:

1. a.

∫ (4x³-9x²+2x-1) dx

4. 1 .x⁽³⁺¹⁾ - 9. 1 x⁽²⁺¹⁾ + 2. 1 x⁽¹⁺¹⁾ - 1x + C

(3+1) (2+1) (1+1)

4. 1 x⁴ - 9. 1 x³ + 2. 1 x² - x + C

(4) 3 2

4 x⁴ - 9x³ + 2x² - x + C

4 3 2

x⁴ - 3x³+x²-x+C

b.

∫ (2x-3)(x+1) dx

(2x²-3x+2x-3) dx

(2x²-x-3) dx

2. 1 x⁽²⁺¹⁾ - 1. x ⁽¹⁺¹⁾ - 3x + C

(2+1) (1+1)

2. 1 x³ - 1. x ² - 3x + C

3 2

2 x³ - 1 x² - 3x + C

3 2

2. a. ∫ ⁶ (4x-1) dx

⁻³

∫⁶ ( 4. 1 x⁽¹⁺¹⁾ - x )

⁻³ (1+1)

₋₃ ∫⁶ ( 4 x² - x )

2

₋₃∫⁶ (2x²-x)

[2x²-x]⁶₋₃

[2(6)²-(6)] - [2(-3)²-(-3)]

[2(36)-6)] - [2(9)+3)]

[72-6] - [ 18+3 ]

[ 66 ] - [ 21]

[ 45 ] Satuan Luas

b. ₋₁∫⁴ (3x²-8x+2) dx

₋₁∫⁴(3. 1 x⁽²⁺¹⁾ - 8. 1 x⁽¹⁺¹⁾ + 2x)

(2+1) (1+1)

(3. 1 x³ - 8. 1 x² + 2x)

3 2

(3 x³ - 8 x² + 2x)

3 2

[x³ - 4x² + 2x]⁴₋₁

[(4)³-4(4)²+2(4)] - [(-1)³-4(-1)²+2(-1)]

[(64)-4(16)+8)] - [(-1)-4(1)-2)]

[(64)-64)+8)] - [(-1)-4-2)]

[(8)] - [(-5-2)]

[(8)] - [(-7)]

[(8)] + [(7)]

[ 15 ] Satuan Luas

c. ₁∫³(2x-1)(3x+4) dx

(6x²-3x+8x-4) dx

(6x²+5x-4) dx

(6. 1 x⁽²⁺¹⁾ + 5. 1 x⁽¹⁺¹⁾ - 4x)

(2+1) (1+1)

(6. 1 x³ + 5. 1 x² - 4x)

3 2

(6 x³ + 5 x² - 4x)

3 2

(2x³ + 5x² - 4x)

2

[(2x³ + 5 x²- 4x)] ³

2 ⁻¹

[(2(3)³ + 5(3)² - 4(3)] - [(2(-1)³ + 5 (-1)²-4(-1)]

2 2

[2(27) + 5 (9) - 12)] - [(2(-1) + 5 (1) + 4)]

2. 2

[54 + 45 - 12] - [ -2 + 5 + 4]

2. 2

[42 + 45 ] - [ 2 + 5 ]

2. 2

[84 + 45 ] - [ 4 + 5 ]

2. 2. 2. 2

[ 129 ] - [ 9 ]

2. 2

[ 120 ]

2

[60] Satuan Luas

Detail Jawaban:

Mapel : Matematika

Kelas : 12 / XII SMA

Materi : Integral

Kode Kategorisasi : -

Kata Kunci : Integral

Demikian

Semoga membantu dan bermanfaat!