1.Evaluate lim sin(4 ) βˆ’ 2sin(2 ) 3 (1) 2.Find the derivative of

1. Evaluate
sin(4π‘₯) βˆ’ 2 sin(2π‘₯)
π‘₯β†’0
π‘₯3
lim
(1)
2. Find the derivative of
𝑦 = √4π‘₯2 βˆ’ 1 + cscβˆ’1 2π‘₯
3π‘₯ βˆ’ 7
3. Find
∫ 2
𝑑π‘₯
π‘₯ βˆ’ 2π‘₯ βˆ’ 15
2
4. Estimate ∫ √π‘₯ 3 βˆ’ 1𝑑π‘₯
by using trapezoidal rule for 𝑛 = 6.
1
2 sin(3π‘₯) βˆ’ 3 sin(2π‘₯)
π‘₯β†’0
π‘₯3
1. Evaluate lim
[2]
1
2. Estimate ∫ √1 βˆ’ π‘₯3 𝑑π‘₯
by using trapezoidal rule for 𝑛 = 6.
0
3π‘₯ + 7
𝑑π‘₯
π‘₯ 2 + 2π‘₯ βˆ’ 15
4. Find the derivative of
𝑦 = √4π‘₯ 2 βˆ’ 1 βˆ’ sec βˆ’1 2π‘₯
3. Find
∫
2 sin(2π‘₯) βˆ’ sin(4π‘₯)
π‘₯β†’0
π‘₯3
1. Evaluate lim
2. Find the derivative of
3. Find
∫
{3}
𝑦 = √9π‘₯2 βˆ’ 1 βˆ’ secβˆ’1 3π‘₯
3π‘₯ βˆ’ 5
𝑑π‘₯
π‘₯ 2 βˆ’ π‘₯ βˆ’ 12
1
4. Estimate ∫ √1 βˆ’ π‘₯ 3 𝑑π‘₯
by using Simpsonβ€²s rule for 𝑛 = 6.
0
3 sin(2π‘₯) βˆ’ 2 sin(3π‘₯)
π‘₯β†’0
π‘₯3
1. Evaluate lim
2
2. Estimate ∫ √π‘₯3 βˆ’ 1𝑑π‘₯
by using Simpsonβ€²s rule for 𝑛 = 6.
1
3. Find
∫
3π‘₯ + 5
𝑑π‘₯
π‘₯ 2 + π‘₯ βˆ’ 12
4. Find the derivative of
𝑦 = √9π‘₯ 2 βˆ’ 1 + csc βˆ’1 3π‘₯
〈4βŒͺ
sin(4π‘₯) βˆ’ 2 sin(2π‘₯)
4 cos 4π‘₯ βˆ’ 4 cos 2π‘₯
(1)
=
lim
π‘₯β†’0
π‘₯β†’0
π‘₯3
3π‘₯ 2
βˆ’16sin(4π‘₯) + 8 sin(2π‘₯)
βˆ’64 cos 4π‘₯ + 16 cos 2π‘₯ βˆ’64 + 16
= lim
= lim
=
= βˆ’8
π‘₯β†’0
π‘₯β†’0
6π‘₯
6
6
1. lim
2. Find the derivative of
𝑦 = √4π‘₯2 βˆ’ 1 + cscβˆ’1 2π‘₯
𝑑𝑦
8π‘₯
2
4π‘₯
1
4π‘₯ 2 βˆ’ 1
√4π‘₯ 2 βˆ’ 1
=
βˆ’
=
βˆ’
=
=
𝑑π‘₯ 2√4π‘₯ 2 βˆ’ 1 2π‘₯√4π‘₯ 2 βˆ’ 1 √4π‘₯ 2 βˆ’ 1 π‘₯√4π‘₯ 2 βˆ’ 1 π‘₯√4π‘₯ 2 βˆ’ 1
π‘₯
3π‘₯ βˆ’ 7
3. Find
∫ 2
𝑑π‘₯
π‘₯ βˆ’ 2π‘₯ βˆ’ 15
3π‘₯ βˆ’ 7
3π‘₯ βˆ’ 7
𝐴
𝐡
=
=
+
π‘₯ 2 βˆ’ 2π‘₯ βˆ’ 15 (π‘₯ βˆ’ 5)(π‘₯ + 3) (π‘₯ βˆ’ 5) (π‘₯ + 3)
π‘₯=5
∫
⇨ 𝐴=
3π‘₯ βˆ’ 7
=1
(π‘₯ + 3)
π‘Žπ‘›π‘‘
π‘₯ = βˆ’3
⇨ 𝐡=
3π‘₯ βˆ’ 7
=2
(π‘₯ βˆ’ 5)
3π‘₯ βˆ’ 7
1
2
𝑑π‘₯
=
∫
+
(
) 𝑑π‘₯ = ln(π‘₯ βˆ’ 5) + 2 ln(π‘₯ + 3) + 𝑐
(π‘₯ βˆ’ 5) (π‘₯ + 3)
π‘₯ 2 βˆ’ 2π‘₯ βˆ’ 15
2
4. Estimate ∫ √π‘₯ 3 βˆ’ 1𝑑π‘₯
by using trapezoidal rule for 𝑛 = 6.
1
𝑏
∫ 𝑓(π‘₯)𝑑π‘₯ β‰…
π‘Ž
β„Ž=
β„Ž
[ 𝑦0 + 2𝑦1 + 2𝑦2 + 2𝑦3 + 2𝑦4 + 2𝑦5 + 𝑦6 ]
2
π‘βˆ’π‘Ž 2βˆ’1
1
=
=
𝑛
6
6
π‘₯𝑛
1
𝑦𝑛
𝑦0 = 0
Factors
1
Product
0
1 + 1 ⁄6
1 + 2 ⁄6
𝑦1 = 0.7668
2
1.5336
𝑦2 = 1.1706
2
2.3412
1 + 3 ⁄6
1 + 4 ⁄6
𝑦3 = 1.5411
2
3.0822
𝑦4 = 1.9052
2
3.8104
1 + 5 ⁄6
𝑦5 = 2.272
2
4.544
2
𝑦6 = 2.6458
Sum
1
2.6458
17.9572
2
∫ √π‘₯ 3 βˆ’ 1𝑑π‘₯ β‰…
1
1⁄6
17.9572
× 17.9572 =
= 1.4964
2
12
2 sin(3π‘₯) βˆ’ 3 sin(2π‘₯)
6 cos 3π‘₯ βˆ’ 6 cos 2π‘₯
[2]
lim
π‘₯β†’0
π‘₯β†’0
π‘₯3
3π‘₯ 2
βˆ’18sin(3π‘₯) + 12 sin(2π‘₯)
βˆ’54 cos 3π‘₯ + 24 cos 2π‘₯ βˆ’54 + 24
= lim
= lim
=
= βˆ’5
π‘₯β†’0
π‘₯β†’0
6π‘₯
6
6
1. Evaluate lim
1
2. Estimate ∫ √1 βˆ’ π‘₯3 𝑑π‘₯
by using trapezoidal rule for 𝑛 = 6.
0
𝑏
∫ 𝑓(π‘₯)𝑑π‘₯ β‰…
π‘Ž
β„Ž=
β„Ž
[ 𝑦0 + 2𝑦1 + 2𝑦2 + 2𝑦3 + 2𝑦4 + 2𝑦5 + 𝑦6 ]
2
π‘βˆ’π‘Ž 1βˆ’0
1
=
=
𝑛
6
6
π‘₯𝑛
0
𝑦𝑛
𝑦0 = 1
Factors
1
Product
1
0 + 1 ⁄6
0 + 2 ⁄6
𝑦1 = 0.9977
2
1.9954
𝑦2 = 0.9813
2
1.9626
0 + 3 ⁄6
0 + 4 ⁄6
𝑦3 = 0.9354
2
1.8708
𝑦4 = 0.8389
2
1.6778
0 + 5 ⁄6
𝑦5 = 0.6491
2
1.2982
1
𝑦6 = 0
Sum
1
0
9.8048
1
∫ √1 βˆ’ π‘₯ 3 𝑑π‘₯ β‰…
0
1⁄6
9.8048
× 9.8048 =
= 0.8171
2
12
3π‘₯ + 7
𝑑π‘₯
π‘₯ 2 + 2π‘₯ βˆ’ 15
3π‘₯ βˆ’ 7
3π‘₯ βˆ’ 7
𝐴
𝐡
=
=
+
π‘₯ 2 + 2π‘₯ βˆ’ 15 (π‘₯ + 5)(π‘₯ βˆ’ 3) (π‘₯ + 5) (π‘₯ βˆ’ 3)
3. Find
∫
π‘₯ = βˆ’5
⇨ 𝐴=
∫
3π‘₯ + 7
=1
(π‘₯ βˆ’ 3)
π‘Žπ‘›π‘‘
π‘₯=3
⇨ 𝐡=
3π‘₯ + 7
=2
(π‘₯ + 5)
3π‘₯ + 7
1
2
𝑑π‘₯
=
∫
+
(
) 𝑑π‘₯ = ln(π‘₯ + 5) + 2 ln(π‘₯ βˆ’ 3) + 𝑐
(π‘₯ + 5) (π‘₯ βˆ’ 3)
π‘₯ 2 + 2π‘₯ βˆ’ 15
4. Find the derivative of
𝑦 = √4π‘₯ 2 βˆ’ 1 βˆ’ sec βˆ’1 2π‘₯
𝑑𝑦
8π‘₯
2
4π‘₯
1
4π‘₯ 2 βˆ’ 1
√4π‘₯ 2 βˆ’ 1
=
βˆ’
=
βˆ’
=
=
𝑑π‘₯ 2√4π‘₯ 2 βˆ’ 1 2π‘₯√4π‘₯ 2 βˆ’ 1 √4π‘₯ 2 βˆ’ 1 π‘₯√4π‘₯ 2 βˆ’ 1 π‘₯√4π‘₯ 2 βˆ’ 1
π‘₯
2 sin(2π‘₯) βˆ’ sin(4π‘₯)
4 cos 2π‘₯ βˆ’ 4 cos 4π‘₯
{3}
=
lim
π‘₯β†’0
π‘₯β†’0
π‘₯3
3π‘₯ 2
βˆ’8 sin(2π‘₯) + 16sin(4π‘₯)
βˆ’16 cos 2π‘₯ + 64 cos 4π‘₯ βˆ’16 + 64
= lim
= lim
=
=8
π‘₯β†’0
π‘₯β†’0
6π‘₯
6
6
1. lim
2. Find the derivative of
𝑦 = √9π‘₯2 βˆ’ 1 βˆ’ secβˆ’1 3π‘₯
𝑑𝑦
18π‘₯
3
9π‘₯
1
9π‘₯ 2 βˆ’ 1
√9π‘₯ 2 βˆ’ 1
=
βˆ’
=
βˆ’
=
=
𝑑π‘₯ 2√9π‘₯ 2 βˆ’ 1 3π‘₯√9π‘₯ 2 βˆ’ 1 √9π‘₯ 2 βˆ’ 1 π‘₯√9π‘₯ 2 βˆ’ 1 π‘₯√9π‘₯ 2 βˆ’ 1
π‘₯
3. Find
π‘₯2
3π‘₯ βˆ’ 5
𝑑π‘₯
π‘₯ 2 βˆ’ π‘₯ βˆ’ 12
3π‘₯ βˆ’ 5
3π‘₯ βˆ’ 5
𝐴
𝐡
=
=
+
βˆ’ π‘₯ βˆ’ 12 (π‘₯ βˆ’ 4)(π‘₯ + 3) (π‘₯ βˆ’ 4) (π‘₯ + 3)
π‘₯=4
∫
∫
⇨ 𝐴=
3π‘₯ βˆ’ 5
=1
(π‘₯ + 3)
π‘Žπ‘›π‘‘
π‘₯ = βˆ’3
⇨ 𝐡=
3π‘₯ βˆ’ 5
=2
(π‘₯ βˆ’ 4)
3π‘₯ βˆ’ 5
1
2
𝑑π‘₯
=
∫
+
(
) 𝑑π‘₯ = ln(π‘₯ βˆ’ 4) + 2 ln(π‘₯ + 3) + 𝑐
(π‘₯ βˆ’ 4) (π‘₯ + 3)
π‘₯ 2 βˆ’ π‘₯ βˆ’ 12
1
4. Estimate ∫ √1 βˆ’ π‘₯ 3 𝑑π‘₯
by using Simpsonβ€²s rule for 𝑛 = 6.
0
𝑏
∫ 𝑓(π‘₯)𝑑π‘₯ β‰…
π‘Ž
β„Ž=
β„Ž
[ 𝑦0 + 4𝑦1 + 2𝑦2 + 4𝑦3 + 2𝑦4 + 4𝑦5 + 𝑦6 ]
3
π‘βˆ’π‘Ž 1βˆ’0
1
=
=
𝑛
6
6
π‘₯𝑛
0
𝑦𝑛
𝑦0 = 1
Factors
1
Product
1
0 + 1 ⁄6
0 + 2 ⁄6
𝑦1 = 0.9977
4
3.9908
𝑦2 = 0.9813
2
1.9626
0 + 3 ⁄6
0 + 4 ⁄6
𝑦3 = 0.9354
4
3.7416
𝑦4 = 0.8389
2
1.6778
0 + 5 ⁄6
𝑦5 = 0.6491
4
2.5964
1
𝑦6 = 0
Sum
1
0
14.9692
1
∫ √1 βˆ’ π‘₯ 3 𝑑π‘₯ β‰…
0
1⁄6
14.9692
× 14.9692 =
= 0.8316
3
18
3 sin(2π‘₯) βˆ’ 2 sin(3π‘₯)
6 cos 2π‘₯ βˆ’ 6 cos 3π‘₯
〈4βŒͺ
=
lim
π‘₯β†’0
π‘₯β†’0
π‘₯3
3π‘₯ 2
βˆ’12sin(2π‘₯) + 18 sin(3π‘₯)
βˆ’24 cos 2π‘₯ + 54 cos 3π‘₯ βˆ’24 + 54
= lim
= lim
=
=5
π‘₯β†’0
π‘₯β†’0
6π‘₯
6
6
1. lim
2
by using Simpsonβ€² s rule for 𝑛 = 6.
2. Estimate ∫ √π‘₯3 βˆ’ 1𝑑π‘₯
1
𝑏
∫ 𝑓(π‘₯)𝑑π‘₯ β‰…
π‘Ž
β„Ž=
β„Ž
[ 𝑦0 + 4𝑦1 + 2𝑦2 + 4𝑦3 + 2𝑦4 + 4𝑦5 + 𝑦6 ]
2
π‘βˆ’π‘Ž 2βˆ’1
1
=
=
𝑛
6
6
π‘₯𝑛
1
𝑦𝑛
𝑦0 = 0
Factors
1
Product
0
1 + 1 ⁄6
1 + 2 ⁄6
𝑦1 = 0.7668
4
3.0672
𝑦2 = 1.1706
2
2.3412
1 + 3 ⁄6
1 + 4 ⁄6
𝑦3 = 1.5411
4
6.1644
𝑦4 = 1.9052
2
3.8104
1 + 5 ⁄6
𝑦5 = 2.272
4
9.088
2
𝑦6 = 2.6458
Sum
1
2.6458
27.117
2
∫ √π‘₯ 3 βˆ’ 1𝑑π‘₯ β‰…
1
3. Find
∫
1⁄6
27.117
× 27.117 =
= 1.5065
3
18
3π‘₯ + 5
𝑑π‘₯
π‘₯ 2 + π‘₯ βˆ’ 12
3π‘₯ + 5
3π‘₯ + 5
𝐴
𝐡
=
=
+
π‘₯ 2 + π‘₯ βˆ’ 12 (π‘₯ + 4)(π‘₯ βˆ’ 3) (π‘₯ + 4) (π‘₯ βˆ’ 3)
π‘₯ = βˆ’4
∫
⇨ 𝐴=
3π‘₯ + 5
=1
(π‘₯ βˆ’ 3)
π‘Žπ‘›π‘‘
π‘₯=3
⇨ 𝐡=
3π‘₯ + 5
=2
(π‘₯ + 4)
3π‘₯ + 5
1
2
𝑑π‘₯
=
∫
+
(
) 𝑑π‘₯ = ln(π‘₯ + 4) + 2 ln(π‘₯ βˆ’ 3) + 𝑐
(π‘₯ + 4) (π‘₯ βˆ’ 3)
π‘₯ 2 + π‘₯ βˆ’ 12
4. Find the derivative of
𝑦 = √9π‘₯ 2 βˆ’ 1 + csc βˆ’1 3π‘₯
𝑑𝑦
18π‘₯
3
9π‘₯
1
9π‘₯ 2 βˆ’ 1
√9π‘₯ 2 βˆ’ 1
=
βˆ’
=
βˆ’
=
=
𝑑π‘₯ 2√9π‘₯ 2 βˆ’ 1 3π‘₯√9π‘₯ 2 βˆ’ 1 √9π‘₯ 2 βˆ’ 1 π‘₯√9π‘₯ 2 βˆ’ 1 π‘₯√9π‘₯ 2 βˆ’ 1
π‘₯