1. Evaluate
sin(4π₯) β 2 sin(2π₯)
π₯β0
π₯3
lim
(1)
2. Find the derivative of
π¦ = β4π₯2 β 1 + cscβ1 2π₯
3π₯ β 7
3. Find
β« 2
ππ₯
π₯ β 2π₯ β 15
2
4. Estimate β« βπ₯ 3 β 1ππ₯
by using trapezoidal rule for π = 6.
1
2 sin(3π₯) β 3 sin(2π₯)
π₯β0
π₯3
1. Evaluate lim
[2]
1
2. Estimate β« β1 β π₯3 ππ₯
by using trapezoidal rule for π = 6.
0
3π₯ + 7
ππ₯
π₯ 2 + 2π₯ β 15
4. Find the derivative of
π¦ = β4π₯ 2 β 1 β sec β1 2π₯
3. Find
β«
2 sin(2π₯) β sin(4π₯)
π₯β0
π₯3
1. Evaluate lim
2. Find the derivative of
3. Find
β«
{3}
π¦ = β9π₯2 β 1 β secβ1 3π₯
3π₯ β 5
ππ₯
π₯ 2 β π₯ β 12
1
4. Estimate β« β1 β π₯ 3 ππ₯
by using Simpsonβ²s rule for π = 6.
0
3 sin(2π₯) β 2 sin(3π₯)
π₯β0
π₯3
1. Evaluate lim
2
2. Estimate β« βπ₯3 β 1ππ₯
by using Simpsonβ²s rule for π = 6.
1
3. Find
β«
3π₯ + 5
ππ₯
π₯ 2 + π₯ β 12
4. Find the derivative of
π¦ = β9π₯ 2 β 1 + csc β1 3π₯
β©4βͺ
sin(4π₯) β 2 sin(2π₯)
4 cos 4π₯ β 4 cos 2π₯
(1)
=
lim
π₯β0
π₯β0
π₯3
3π₯ 2
β16sin(4π₯) + 8 sin(2π₯)
β64 cos 4π₯ + 16 cos 2π₯ β64 + 16
= lim
= lim
=
= β8
π₯β0
π₯β0
6π₯
6
6
1. lim
2. Find the derivative of
π¦ = β4π₯2 β 1 + cscβ1 2π₯
ππ¦
8π₯
2
4π₯
1
4π₯ 2 β 1
β4π₯ 2 β 1
=
β
=
β
=
=
ππ₯ 2β4π₯ 2 β 1 2π₯β4π₯ 2 β 1 β4π₯ 2 β 1 π₯β4π₯ 2 β 1 π₯β4π₯ 2 β 1
π₯
3π₯ β 7
3. Find
β« 2
ππ₯
π₯ β 2π₯ β 15
3π₯ β 7
3π₯ β 7
π΄
π΅
=
=
+
π₯ 2 β 2π₯ β 15 (π₯ β 5)(π₯ + 3) (π₯ β 5) (π₯ + 3)
π₯=5
β«
β¨ π΄=
3π₯ β 7
=1
(π₯ + 3)
πππ
π₯ = β3
β¨ π΅=
3π₯ β 7
=2
(π₯ β 5)
3π₯ β 7
1
2
ππ₯
=
β«
+
(
) ππ₯ = ln(π₯ β 5) + 2 ln(π₯ + 3) + π
(π₯ β 5) (π₯ + 3)
π₯ 2 β 2π₯ β 15
2
4. Estimate β« βπ₯ 3 β 1ππ₯
by using trapezoidal rule for π = 6.
1
π
β« π(π₯)ππ₯ β
π
β=
β
[ π¦0 + 2π¦1 + 2π¦2 + 2π¦3 + 2π¦4 + 2π¦5 + π¦6 ]
2
πβπ 2β1
1
=
=
π
6
6
π₯π
1
π¦π
π¦0 = 0
Factors
1
Product
0
1 + 1 β6
1 + 2 β6
π¦1 = 0.7668
2
1.5336
π¦2 = 1.1706
2
2.3412
1 + 3 β6
1 + 4 β6
π¦3 = 1.5411
2
3.0822
π¦4 = 1.9052
2
3.8104
1 + 5 β6
π¦5 = 2.272
2
4.544
2
π¦6 = 2.6458
Sum
1
2.6458
17.9572
2
β« βπ₯ 3 β 1ππ₯ β
1
1β6
17.9572
× 17.9572 =
= 1.4964
2
12
2 sin(3π₯) β 3 sin(2π₯)
6 cos 3π₯ β 6 cos 2π₯
[2]
lim
π₯β0
π₯β0
π₯3
3π₯ 2
β18sin(3π₯) + 12 sin(2π₯)
β54 cos 3π₯ + 24 cos 2π₯ β54 + 24
= lim
= lim
=
= β5
π₯β0
π₯β0
6π₯
6
6
1. Evaluate lim
1
2. Estimate β« β1 β π₯3 ππ₯
by using trapezoidal rule for π = 6.
0
π
β« π(π₯)ππ₯ β
π
β=
β
[ π¦0 + 2π¦1 + 2π¦2 + 2π¦3 + 2π¦4 + 2π¦5 + π¦6 ]
2
πβπ 1β0
1
=
=
π
6
6
π₯π
0
π¦π
π¦0 = 1
Factors
1
Product
1
0 + 1 β6
0 + 2 β6
π¦1 = 0.9977
2
1.9954
π¦2 = 0.9813
2
1.9626
0 + 3 β6
0 + 4 β6
π¦3 = 0.9354
2
1.8708
π¦4 = 0.8389
2
1.6778
0 + 5 β6
π¦5 = 0.6491
2
1.2982
1
π¦6 = 0
Sum
1
0
9.8048
1
β« β1 β π₯ 3 ππ₯ β
0
1β6
9.8048
× 9.8048 =
= 0.8171
2
12
3π₯ + 7
ππ₯
π₯ 2 + 2π₯ β 15
3π₯ β 7
3π₯ β 7
π΄
π΅
=
=
+
π₯ 2 + 2π₯ β 15 (π₯ + 5)(π₯ β 3) (π₯ + 5) (π₯ β 3)
3. Find
β«
π₯ = β5
β¨ π΄=
β«
3π₯ + 7
=1
(π₯ β 3)
πππ
π₯=3
β¨ π΅=
3π₯ + 7
=2
(π₯ + 5)
3π₯ + 7
1
2
ππ₯
=
β«
+
(
) ππ₯ = ln(π₯ + 5) + 2 ln(π₯ β 3) + π
(π₯ + 5) (π₯ β 3)
π₯ 2 + 2π₯ β 15
4. Find the derivative of
π¦ = β4π₯ 2 β 1 β sec β1 2π₯
ππ¦
8π₯
2
4π₯
1
4π₯ 2 β 1
β4π₯ 2 β 1
=
β
=
β
=
=
ππ₯ 2β4π₯ 2 β 1 2π₯β4π₯ 2 β 1 β4π₯ 2 β 1 π₯β4π₯ 2 β 1 π₯β4π₯ 2 β 1
π₯
2 sin(2π₯) β sin(4π₯)
4 cos 2π₯ β 4 cos 4π₯
{3}
=
lim
π₯β0
π₯β0
π₯3
3π₯ 2
β8 sin(2π₯) + 16sin(4π₯)
β16 cos 2π₯ + 64 cos 4π₯ β16 + 64
= lim
= lim
=
=8
π₯β0
π₯β0
6π₯
6
6
1. lim
2. Find the derivative of
π¦ = β9π₯2 β 1 β secβ1 3π₯
ππ¦
18π₯
3
9π₯
1
9π₯ 2 β 1
β9π₯ 2 β 1
=
β
=
β
=
=
ππ₯ 2β9π₯ 2 β 1 3π₯β9π₯ 2 β 1 β9π₯ 2 β 1 π₯β9π₯ 2 β 1 π₯β9π₯ 2 β 1
π₯
3. Find
π₯2
3π₯ β 5
ππ₯
π₯ 2 β π₯ β 12
3π₯ β 5
3π₯ β 5
π΄
π΅
=
=
+
β π₯ β 12 (π₯ β 4)(π₯ + 3) (π₯ β 4) (π₯ + 3)
π₯=4
β«
β«
β¨ π΄=
3π₯ β 5
=1
(π₯ + 3)
πππ
π₯ = β3
β¨ π΅=
3π₯ β 5
=2
(π₯ β 4)
3π₯ β 5
1
2
ππ₯
=
β«
+
(
) ππ₯ = ln(π₯ β 4) + 2 ln(π₯ + 3) + π
(π₯ β 4) (π₯ + 3)
π₯ 2 β π₯ β 12
1
4. Estimate β« β1 β π₯ 3 ππ₯
by using Simpsonβ²s rule for π = 6.
0
π
β« π(π₯)ππ₯ β
π
β=
β
[ π¦0 + 4π¦1 + 2π¦2 + 4π¦3 + 2π¦4 + 4π¦5 + π¦6 ]
3
πβπ 1β0
1
=
=
π
6
6
π₯π
0
π¦π
π¦0 = 1
Factors
1
Product
1
0 + 1 β6
0 + 2 β6
π¦1 = 0.9977
4
3.9908
π¦2 = 0.9813
2
1.9626
0 + 3 β6
0 + 4 β6
π¦3 = 0.9354
4
3.7416
π¦4 = 0.8389
2
1.6778
0 + 5 β6
π¦5 = 0.6491
4
2.5964
1
π¦6 = 0
Sum
1
0
14.9692
1
β« β1 β π₯ 3 ππ₯ β
0
1β6
14.9692
× 14.9692 =
= 0.8316
3
18
3 sin(2π₯) β 2 sin(3π₯)
6 cos 2π₯ β 6 cos 3π₯
β©4βͺ
=
lim
π₯β0
π₯β0
π₯3
3π₯ 2
β12sin(2π₯) + 18 sin(3π₯)
β24 cos 2π₯ + 54 cos 3π₯ β24 + 54
= lim
= lim
=
=5
π₯β0
π₯β0
6π₯
6
6
1. lim
2
by using Simpsonβ² s rule for π = 6.
2. Estimate β« βπ₯3 β 1ππ₯
1
π
β« π(π₯)ππ₯ β
π
β=
β
[ π¦0 + 4π¦1 + 2π¦2 + 4π¦3 + 2π¦4 + 4π¦5 + π¦6 ]
2
πβπ 2β1
1
=
=
π
6
6
π₯π
1
π¦π
π¦0 = 0
Factors
1
Product
0
1 + 1 β6
1 + 2 β6
π¦1 = 0.7668
4
3.0672
π¦2 = 1.1706
2
2.3412
1 + 3 β6
1 + 4 β6
π¦3 = 1.5411
4
6.1644
π¦4 = 1.9052
2
3.8104
1 + 5 β6
π¦5 = 2.272
4
9.088
2
π¦6 = 2.6458
Sum
1
2.6458
27.117
2
β« βπ₯ 3 β 1ππ₯ β
1
3. Find
β«
1β6
27.117
× 27.117 =
= 1.5065
3
18
3π₯ + 5
ππ₯
π₯ 2 + π₯ β 12
3π₯ + 5
3π₯ + 5
π΄
π΅
=
=
+
π₯ 2 + π₯ β 12 (π₯ + 4)(π₯ β 3) (π₯ + 4) (π₯ β 3)
π₯ = β4
β«
β¨ π΄=
3π₯ + 5
=1
(π₯ β 3)
πππ
π₯=3
β¨ π΅=
3π₯ + 5
=2
(π₯ + 4)
3π₯ + 5
1
2
ππ₯
=
β«
+
(
) ππ₯ = ln(π₯ + 4) + 2 ln(π₯ β 3) + π
(π₯ + 4) (π₯ β 3)
π₯ 2 + π₯ β 12
4. Find the derivative of
π¦ = β9π₯ 2 β 1 + csc β1 3π₯
ππ¦
18π₯
3
9π₯
1
9π₯ 2 β 1
β9π₯ 2 β 1
=
β
=
β
=
=
ππ₯ 2β9π₯ 2 β 1 3π₯β9π₯ 2 β 1 β9π₯ 2 β 1 π₯β9π₯ 2 β 1 π₯β9π₯ 2 β 1
π₯
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