




𝑥
𝑎
𝑓(𝑥)
𝐿
lim 𝑓(𝑥) = 𝐿
𝑥→𝑎
lim 𝑥 + 1 = 2
𝑥→1
𝒇(𝒙)
𝒙
0.5
1.5
0.9
1.9
0.99 1.99
0.999 1.999
1.001 2.001
1.01 2.01
1.1
2.1
1.5
2.5
𝐿"
𝑥
𝑥= 𝑎
𝑎
𝑎




𝑎
𝑏
𝑛
lim 𝑏 = 𝑏
𝑥→𝑎
lim 𝑥 = 𝑎
𝑥→𝑎
lim 𝑥 𝑛 = 𝑎𝑛
𝑥→𝑎
𝑛
𝑛
lim √𝑥 = √𝑎 𝑓𝑜𝑟 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛 𝑎𝑛𝑑 𝑎 > 0
𝑥→𝑎
𝑏
𝑛
𝑓
𝑔
lim {𝑏𝑓(𝑥)} = 𝑏 {lim 𝑓(𝑥)}
𝑥→𝑎
𝑥→𝑎
lim {𝑓(𝑥) ± 𝑔(𝑥)} = lim 𝑓(𝑥) ± lim 𝑔(𝑥)
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
lim {𝑓(𝑥) ∙ 𝑔(𝑥)} = {lim 𝑓(𝑥)} ∙ {lim 𝑔(𝑥)}
𝑥→𝑎
𝑥→𝑎
𝑓(𝑥)
lim {𝑔(𝑥)}
𝑥→𝑎
𝑥→𝑎
lim 𝑓(𝑥)
= 𝑥→𝑎
lim 𝑔(𝑥)
𝑥→𝑎
𝑛
lim {𝑓(𝑥)}𝑛 = {lim 𝑓(𝑥)}
𝑥→𝑎
𝑥→𝑎
4𝑥 2 +2𝑥−1
)
𝑥→∞ 3𝑥 2 +𝑥−2
1−cos 𝑥
lim ( 𝑥 )
𝑥→0
lim (𝑥 2 + 2𝑥 + 5)
𝑥→3
𝑥 2 −4
lim (
𝑥→2
𝑥 2 +𝑥−6
3
lim (
)
3
1
lim ( √3𝑥 2 + 20𝑥 2 )
𝑥→4
𝒍𝒊𝒎(𝒙𝟐 + 𝟐𝒙 + 𝟓)
= 𝑙𝑖𝑚 𝑥 2 + 𝑙𝑖𝑚2𝑥 + 𝑙𝑖𝑚5
𝒙→𝟑
𝑥→3
𝑥→3
= 32 + 2(3) + 5
= 20
𝑥→3
𝒙𝟐 −𝟒
)
𝒙𝟐 +𝒙−𝟔
𝒍𝒊𝒎 (
𝒙→𝟐
0
𝑥=2
0
(𝑥 + 2)(𝑥 − 2)
(𝑥 + 2)
𝑥2 − 4
∴ 𝑙𝑖𝑚 ( 2
) = 𝑙𝑖𝑚
= 𝑙𝑖𝑚
𝑥→2 𝑥 + 𝑥 − 6
𝑥→2 (𝑥 + 3)(𝑥 − 2)
𝑥→2 (𝑥 + 3)
2+2 4
=
=
2+3 5
𝟑
𝟑
𝟏
3
3
1
3
3
𝒍𝒊𝒎 ( √𝟑𝒙𝟐 + 𝟐𝟎𝒙𝟐 ) = √3(4)2 + 20(4)2 = √3(√4) + 20(√4)
𝒙→𝟒
𝑥.
3
3
1
= √3(4)2 + 20(4)2
3
3
= √3(2)3 + 20(2)
1
3
1
= √24 + 40 = √64 = (64)3 = (43 )3 = 4
𝟒𝒙𝟐 +𝟐𝒙−𝟏
𝐥𝐢𝐦 ( 𝟑𝒙𝟐+𝒙−𝟐 )
𝒙→∞
∞
∞
∞
1
𝑥2
4𝑥 2 + 2𝑥 − 1
2 1
2
4+𝑥− 2
4𝑥 + 2𝑥 − 1
2
𝑥
𝑥
lim (
= lim
) = lim
1 2
𝑥→∞ 3𝑥 2 + 𝑥 − 2
𝑥→∞ 3𝑥 2 + 𝑥 − 2
𝑥→∞
3+𝑥− 2
𝑥
𝑥2
2
1
lim 4 + lim − lim 2 4 + 0 − 0 4
𝑥→∞ 𝑥
𝑥→∞ 𝑥
= 𝑥→∞
=
=
1
2
lim 3 + lim 𝑥 − lim 2 3 + 0 − 0 3
𝑥→∞
𝑥→∞
𝑥→∞ 𝑥
1
0

= 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑

1
∞
=0
𝟏−𝐜𝐨𝐬 𝒙
)
𝒙
𝐥𝐢𝐦 (
𝒙→𝟎
(1 − 𝑐𝑜𝑠𝑥)
𝑐𝑜𝑠𝑥)
1−cos 𝑥
lim (
𝑥→0
𝑥
) = lim (
𝑥→0
1−cos 𝑥
𝑥
1−cos2 𝑥
1+cos 𝑥
× 1+𝑐𝑜𝑠 𝑥) = lim 𝑥(1+𝑐𝑜𝑠 𝑥)
𝑥→0
1 − 𝑐𝑜𝑠 2 𝑥 = 𝑠𝑖𝑛2 𝑥
1 − cos2 𝑥
𝑠𝑖𝑛2 𝑥
sin 𝑥
𝑠𝑖𝑛𝑥
∴ lim
= lim
= lim (
×
)
𝑥→0 𝑥(1 + cos 𝑥)
𝑥→0 𝑥(1 + cos 𝑥)
𝑥→0
(1 + cos 𝑥)
𝑥
sin 𝑥
𝑠𝑖𝑛𝑥
𝑠𝑖𝑛0
= lim
× lim
=1×
𝑥→0 𝑥
𝑥→0 (1 + cos 𝑥)
(1 + cos 0)
sin 𝑥
lim
=1
𝑥→0 𝑥
0
=1×
= 1×0= 0
1+1
𝑓(𝑥)
𝑤ℎ𝑒𝑛 𝑓(𝑥) → 0 𝑎𝑛𝑑 𝑔(𝑥) → 0 𝑎𝑠 𝑥 → 𝑎
𝑥→𝑎 𝑔(𝑥)
𝑓(𝑥) 0
lim
=
𝑡ℎ𝑖𝑠 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑎𝑛 𝒊𝒏𝒅𝒆𝒕𝒆𝒓𝒎𝒊𝒏𝒂𝒕𝒆 𝒇𝒐𝒓𝒎
𝑥→𝑎 𝑔(𝑥)
0
𝑓(𝑥)
lim 𝑔(𝑥) 𝑤ℎ𝑒𝑛 𝑓(𝑥) → ∞ 𝑎𝑛𝑑 𝑔(𝑥) → ∞ 𝑎𝑠 𝑥 → 𝑎
lim
𝑥→𝑎
(1 +
𝑓(𝑥) ∞
=
𝑥→𝑎 𝑔(𝑥)
∞
lim
𝑡ℎ𝑖𝑠 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑎𝑛 𝒊𝒏𝒅𝒆𝒕𝒆𝒓𝒎𝒊𝒏𝒂𝒕𝒆 𝒇𝒐𝒓𝒎
𝑓(𝑥)
𝑓 ′ (𝑥)
= lim ′
𝑥→𝑎 𝑔(𝑥)
𝑥→𝑎 𝑔 (𝑥)
lim
ln(1+𝑥 −1 )
𝑥 −1
𝑥→∞
3𝑥 2 −12
)
𝑥−2
𝑥→2
𝑥−1
𝑒
−𝑥
lim
𝑥→1 (𝑥−1)2
𝑒 3𝑥
lim 𝑥 2
𝑥→∞
lim
3𝑠𝑖𝑛5𝑥 (
lim ( 𝑥 )
𝑥→0
lim (
1−𝑐𝑜𝑠4𝑥
)
𝑥2
𝐥𝐧(𝟏+𝒙−𝟏 )
𝒙−𝟏
𝒙→∞
𝐥𝐢𝐦
𝑓(𝑥) = ln(1 + 𝑥 −1 )
lim
𝑥→∞
𝑓(𝑥)
=
𝑔(𝑥)
ln(1+∞−1 )
∞−1
𝑔(𝑥) = 𝑥 −1
0
=0
−𝑥 −2
𝑓 ′ (𝑥) = 1+𝑥 −1
𝑔′ (𝑥) = −𝑥 −2
−𝑥 −2
−1
𝑓 ′ (𝑥)
1
lim ′
= lim 1 + 𝑥−2 = lim (
)
𝑥→∞ 𝑔 (𝑥)
𝑥→∞ −𝑥
𝑥→∞ 1 + 𝑥 −1
1
1
1
=
=
= =1
−1
1+∞
1+0 1
ln(1+𝑥 −1 )
lim 𝑥 −1 = 1
𝑥→∞
𝟏−𝒄𝒐𝒔𝟒𝒙
)
𝒙𝟐
𝟑𝒔𝒊𝒏𝟓𝒙 (
𝐥𝐢𝐦 ( 𝒙 )
𝒙→𝟎
sin 𝑎𝑥
1−cos 𝑎𝑥
𝑎2
l im 𝑥 = 𝑎
lim 𝑥 2 = 2
𝑥→0
𝑥→0
3sin 5𝑥
sin 5𝑥
⟹
lim (
) = 3lim (
) = 3(5) = 15
𝑥→0
𝑥→0
𝑥
𝑥
2
1−𝑐𝑜𝑠4𝑥
4
lim 𝑥 2 = 2 = 8
𝑥→0
3𝑠𝑖𝑛5𝑥 (
∴ lim (
)
𝑥→0
𝑥
1−𝑐𝑜𝑠4𝑥
)
𝑥2
𝟑𝒙𝟐 −𝟏𝟐
)
𝒙−𝟐
𝒙→𝟐
𝐥𝐢𝐦 (
=
lim
3𝑠𝑖𝑛5𝑥 𝑥→0
= (lim
)
𝑥→0
𝑥
3(2)2 −12
2−2
0
=0
3𝑥 2 − 12
6𝑥
∴ lim (
) = lim
= 6(2) = 12
𝑥→2
𝑥→2 1
𝑥−2
1−𝑐𝑜𝑠4𝑥
𝑥2
= 15(8) = 2562890625
3𝑥 2 − 12
3(𝑥 2 − 4)
3(𝑥 − 2)(𝑥 + 2)
∴ lim (
) = lim
= lim
𝑥→2
𝑥→2
𝑥→2
𝑥−2
𝑥−2
𝑥−2
= lim 3(𝑥 + 2) = 3(2 + 2) = 12
𝑥→2
𝒆𝒙−𝟏 −𝒙
𝐥𝐢𝐦 (𝒙−𝟏)𝟐
𝒙→𝟏
𝑓(𝑥) = 𝑒 𝑥−1 − 𝑥
𝑔(𝑥) = (𝑥 − 1)2
𝑒 𝑥−1 −𝑥
𝑥→1 (𝑥−1)2
𝑓(𝑥)
lim 𝑔(𝑥) = lim
𝑥→1
𝑒 0 −1
1−1
= (1−1)2 = 1−1 =
0
0
𝑓 ′ (𝑥) = 𝑒 𝑥−1 − 1
𝑔′ (𝑥) = 2(𝑥 − 1)
𝑓 ′ (𝑥)
𝑒 𝑥−1 − 1
𝑒0 − 1
0
lim ′
= lim
=
=
𝑥→1 𝑔 (𝑥)
𝑥→1 2(𝑥 − 1)
2(1 − 1) 0
𝑓 ′′ (𝑥) = 𝑒 𝑥−1
𝑔′′ (𝑥) = 2
𝑓 ′′ (𝑥)
𝑒 𝑥−1 𝑒 0 1
lim
= lim
=
=
𝑥→1 𝑔′′ (𝑥)
𝑥→1 2
2
2
𝑒 𝑥−1 −𝑥
𝑥→1 (𝑥−1)2
lim
𝒆𝟑𝒙
𝒙→∞ 𝒙𝟐
1
=2
𝐥𝐢𝐦
𝑓(𝑥) = 𝑒 3𝑥
lim
𝑥→∞
𝑓(𝑥)
𝑔(𝑥) = 𝑥 2
= lim
𝑔(𝑥)
𝑒 3𝑥
𝑥→∞ 𝑥 2
𝑒∞
= ∞2 =
∞
∞
𝑓 ′ (𝑥) = 3𝑒 3𝑥
𝑔′ (𝑥) = 2𝑥
𝑓 ′ (𝑥)
𝑒 3𝑥
3𝑒 ∞
∞
lim ′
= lim
=
=
𝑥→∞ 𝑔 (𝑥)
𝑥→∞ 2𝑥
2(∞) ∞
𝑓 ′′ (𝑥) = 9𝑒 3𝑥
𝑔′′ (𝑥) = 2
𝑓 ′′ (𝑥)
9𝑒 3𝑥 9𝑒 ∞ ∞
lim ′′
= lim
=
= =∞
𝑥→∞ 𝑔 (𝑥)
𝑥→∞ 2
2
2
lim
e3x
x→∞ x2
𝑙𝑖𝑚
=∞
𝑙𝑛𝑥
𝑥→∞ √𝑥
𝑡𝑎𝑛3𝑥
𝑙𝑖𝑚 𝑥
𝑥→0
ln(2+𝑒 𝑥 )
𝑙𝑖𝑚 3𝑥
𝑥→∞
(𝑥+1)(2𝑥+1)(𝑥−3)
lim
𝑥→∞
𝑥3





𝑦 = 𝑥2
2𝑥 + 3
𝑦
𝑥
𝑦
(1, 5)
(2, 7)
𝑥
𝑦
𝑦 = 𝑚𝑥 + 𝑏
𝑆𝑙𝑜𝑝𝑒 =
9−5
=2
3−1
𝑦 =
𝑥
𝑦 = 2𝑥 + 3
𝑑𝑦
= 2 (𝑅𝑒𝑎𝑑 𝑑𝑒𝑒 𝑦 𝑑𝑒𝑒 𝑥 𝑒𝑞𝑢𝑎𝑙𝑠 2)
𝑑𝑥
𝑦′ = 2
(𝑅𝑒𝑎𝑑 𝑦 𝑝𝑟𝑖𝑚𝑒 𝑒𝑞𝑢𝑎𝑙𝑠 2)
𝑦 = 2𝑥 + 3
𝒂" 𝑖𝑠 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑑
(𝑎) = 0 𝑤ℎ𝑒𝑟𝑒 𝑎
𝑑𝑥
𝑑
(𝑎𝑥 𝑛 ) = 𝑎 ∙ 𝑛𝑥 𝑛−1
𝑑𝑥
𝑑
(𝑒 𝑥 ) = 𝑒 𝑥
𝑑𝑥
𝑑
(𝑒 𝑎𝑥 ) = 𝑎 ∙ 𝑒 𝑎𝑥
𝑑𝑥
𝑑
(𝑎 𝑥 ) = 𝑎 𝑥 ∙ ln 𝑎
𝑑𝑥
𝑑
1
(ln 𝑥) =
𝑑𝑥
𝑥
𝑑
𝑓′ (𝑥)
𝑖𝑠 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
(ln 𝑓(𝑥)) =
𝑑𝑥
𝑓(𝑥)
𝑑
𝑓′ (𝑥)
(log 𝑎 𝑓(𝑥)) =
𝑑𝑥
𝑓(𝑥)∙ln 𝑎
𝑑
1
(√𝑥) = 2 𝑥
𝑑𝑥
√
𝑑
(𝑠𝑖𝑛𝑎𝑥) = 𝑎𝑐𝑜𝑠𝑎𝑥
𝑑𝑥
𝑑
(𝑐𝑜𝑠𝑎𝑥) = −𝑎𝑠𝑖𝑛𝑎𝑥
𝑑𝑥
sin2 𝑥 + cos 2 𝑥 = 1
1 + tan2 𝑥 = 1
1 + 𝑐𝑜𝑡 2 𝑥 = 𝑐𝑜𝑠𝑒𝑐 2 𝑥
𝑠𝑖𝑛2𝑥 = 2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥
𝑐𝑜𝑠2𝑥 = cos2 𝑥 − sin2 𝑥 = 1 − 2 sin2 𝑥 = 2 cos 2 𝑥 − 1
𝐴+𝐵
𝐴−𝐵
) cos ( 2 )
2
𝐴+𝐵
𝐴−𝐵
2 cos ( 2 ) sin ( 2 )
sin 𝐴 + sin 𝐵 = 2 sin (
sin 𝐴 − sin 𝐵 =
𝑑
(𝑡𝑎𝑛𝑎𝑥) = 𝑎 sec 2 𝑎𝑥
𝑑𝑥
𝑑
(𝑐𝑜𝑠𝑒𝑐𝑎𝑥) = −𝑎𝑐𝑜𝑠𝑒𝑐𝑎𝑥𝑐𝑜𝑡𝑎𝑥
𝑑𝑥
𝑑
(𝑠𝑒𝑐𝑎𝑥) = 𝑠𝑒𝑐𝑎𝑥𝑡𝑎𝑛𝑎𝑥
𝑑𝑥
𝑑
(𝑐𝑜𝑡𝑎𝑥) = −𝑐𝑜𝑠𝑒𝑐 2 𝑎𝑥
𝑑𝑥
𝑑
1
(sin−1 𝑥) =
𝑑𝑥
√1−𝑥 2
𝑑
−1
−1
(𝑐𝑜𝑠 𝑥) =
𝑑𝑥
√1−𝑥 2
𝑑
1
(tan−1 𝑥) =
𝑑𝑥
1+𝑥 2
𝑑
−1
(cot −1 𝑥) =
𝑑𝑥
1+𝑥 2
𝑑
1
(sec −1 𝑥) =
𝑑𝑥
𝑥√𝑥 2 −1
𝑑
−1
(cosec −1 𝑥) =
𝑑𝑥
𝑥√𝑥 2 −1
𝐴+𝐵
𝐴−𝐵
) cos ( 2 )
2
𝐴+𝐵
𝐴−𝐵
−2 sin ( 2 ) sin ( 2 )
cos 𝐴 + cos 𝐵 = 2 cos (
cos 𝐴 − cos 𝐵 =
𝑓(𝑥) = 2𝑥2 − 16𝑥 + 35
𝑔(𝑡) =
2𝑡 5 +𝑡 2 −5
𝑡2
𝜋
𝑦(𝑥) = 𝑥 − 𝑥 √2
𝑦 = 10𝑥
𝒇(𝒙) = 𝟐𝒙𝟐 − 𝟏𝟔𝒙 + 𝟑𝟓
𝑑
𝑓(𝑥) = 𝑓 ′ (𝑥) = 2(2)𝑥 2−1 − 16(1)𝑥 1−1 + 0
𝑑𝑥
= 2(2)𝑥 1 − 16(1)𝑥 0
′
𝑓 (𝑥) = 4𝑥 − 16
𝒙𝟎 = 𝟏
𝒈(𝒕) =
𝟐𝒕𝟓 +𝒕𝟐 −𝟓
𝒕𝟐
2𝑡 5 𝑡 2 5
𝑔(𝑡) = 2 + 2 − 2 = 2𝑡 3 + 1 − 5𝑡 −2
𝑡
𝑡
𝑡
𝑑
𝑔(𝑡) = 𝑔′ (𝑡) = 2(3)𝑡 3−1 + 0 − 5(−2)𝑡 −2−1
𝑑𝑥
(−)
𝑔′ (𝑡) = 6𝑡 2 + 10𝑡 −3
𝒚(𝒙) = 𝒙𝝅 − 𝒙√𝟐
𝑦 ′ (𝑥) = 𝜋𝑥 𝜋−1 − √2𝑥 √2−1
𝒚 = 𝟏𝟎𝒙
𝑥
𝑑𝑦
= 𝑦 ′ = 10𝑥 ∙ ln 10
𝑑𝑥
𝑓(𝑥)
𝑥
𝑓′(𝑥)
𝑑
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
𝑓(𝑥) = 𝑓′(𝑥) = lim
ℎ→0
𝑑𝑥
ℎ
𝑓(𝑥) = 2𝑥 2 − 16𝑥 + 35
𝑑
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
𝑓(𝑥) = 𝑓′(𝑥) = lim
ℎ→0
𝑑𝑥
ℎ
𝑑
2(𝑥 + ℎ)2 − 16(𝑥 + ℎ) + 35 − (2𝑥 2 − 16𝑥 + 35)
𝑓(𝑥) = 𝑓′(𝑥) = lim
ℎ→0
𝑑𝑥
ℎ
ℎ=0
𝑑
2𝑥 2 + 4𝑥ℎ + 2ℎ2 − 16𝑥 − 16ℎ + 35 − 2𝑥 2 + 16𝑥 − 35
𝑓(𝑥) = 𝑓′(𝑥) = lim
ℎ→0
𝑑𝑥
ℎ
4𝑥ℎ + 2ℎ2 − 16ℎ
= lim
ℎ→0
ℎ
ℎ
ℎ
ℎ
ℎ(4𝑥 + 2ℎ − 16)
ℎ→0
ℎ
𝑓′(𝑥) = lim (4𝑥 + 2ℎ − 16)
𝑓′(𝑥) = lim
ℎ→0
= 4𝑥 + 2(0) − 16 = 4𝑥 − 16
𝑑
𝑓(𝑥) = 𝑓 ′ (𝑥) = 4𝑥 − 16
𝑑𝑥
𝑔(𝑡) =
𝑡
𝑡+1
𝑑
𝑔(𝑡 + ℎ) − 𝑔(𝑡)
𝑔(𝑡) = 𝑔′(𝑡) = lim
ℎ→0
𝑑𝑥
ℎ
1
𝑡
𝑡
= lim (
−
)
ℎ→0 ℎ (𝑡 + ℎ) + 1
𝑡+1
ℎ=0
1 (𝑡 + ℎ)(𝑡 + 1) − 𝑡(𝑡 + ℎ + 1)
𝑔′ (𝑡) = lim (
)
ℎ→0 ℎ
((𝑡 + ℎ) + 1)(𝑡 + 1)
1 𝑡 2 + 𝑡 + 𝑡ℎ + ℎ − (𝑡 2 + 𝑡ℎ + 𝑡)
= lim (
)
ℎ→0 ℎ
((𝑡 + ℎ) + 1)(𝑡 + 1)
1
ℎ
= lim (
)
ℎ→0 ℎ ((𝑡 + ℎ) + 1)(𝑡 + 1)
ℎ
ℎ
ℎ
1
𝑔′ (𝑡) = lim (
)
ℎ→0 ((𝑡 + ℎ) + 1)(𝑡 + 1)
1
1
1
=
=
=
((𝑡 + 0) + 1)(𝑡 + 1) (𝑡 + 1)(𝑡 + 1) (𝑡 + 1)2
𝑑
1
𝑔(𝑡) = 𝑔′(𝑡) =
(𝑡 + 1)2
𝑑𝑥
𝑦(𝑥) = 𝑠𝑖𝑛𝑥
𝑑𝑦
𝑦(𝑥 + ℎ) − 𝑦(𝑥)
= 𝑦′(𝑥) = lim
ℎ→0
𝑑𝑥
ℎ
sin(𝑥 + ℎ) − 𝑠𝑖𝑛𝑥
= lim
ℎ→0
ℎ
sin 𝐴 − sin 𝐵 = 2 cos (
𝐴+𝐵
𝐴−𝐵
) sin (
)
2
2
(𝑥 + ℎ) + 𝑥
(𝑥 + ℎ) − 𝑥
1
𝑦′(𝑥) = lim (2 cos (
) sin (
))
ℎ→0 ℎ
2
2
1
2𝑥 + ℎ
ℎ
= lim (2 cos (
) sin ( ))
ℎ→0 ℎ
2
2
2𝑥 + ℎ
2
ℎ
= lim {cos (
)} {lim sin ( )}
ℎ→0
ℎ→0 ℎ
2
2
ℎ
sin (2)
2𝑥 + ℎ
= lim {cos (
)} {lim
}
ℎ
ℎ→0
ℎ→0
2
2
2𝑥 + 0
= cos (
) (1) = 𝑐𝑜𝑠𝑥
2
𝑑𝑦
= 𝑦 ′ (𝑥) = cos 𝑥
𝑑𝑥
𝑦(𝑥) = √5𝑥 − 8
𝑑𝑦
𝑦(𝑥 + ℎ) − 𝑦(𝑥)
= 𝑦′(𝑥) = lim
ℎ→0
𝑑𝑥
ℎ
√5(𝑥 + ℎ) − 8 − √5𝑥 − 8
= lim
ℎ→0
ℎ
√5(𝑥 + ℎ) − 8 − √5𝑥 − 8 √5(𝑥 + ℎ) − 8 + √5𝑥 − 8
∙
ℎ→0
ℎ
√5(𝑥 + ℎ) − 8 + √5𝑥 − 8
𝑦 ′ (𝑥) = lim
5𝑥 + 5ℎ − 8 − (5𝑥 − 8)
= lim
ℎ→0
= lim
ℎ→0
ℎ (√5(𝑥 + ℎ) − 8 + √5𝑥 − 8)
5ℎ
ℎ (√5(𝑥 + ℎ) − 8 + √5𝑥 − 8)
ℎ
ℎ
5
= lim
ℎ→0 √5(𝑥
+ ℎ) − 8 + √5𝑥 − 8
5
=
=
√5𝑥 − 8 + √5𝑥 − 8 2√5𝑥 − 8
5
𝑦 ′ (𝑥) =
5
2√5𝑥 − 8
𝑦 =𝑢∙𝑣
𝑑𝑦
𝑑𝑥
𝑑𝑣
𝑑𝑢
= 𝑢 𝑑𝑥 + 𝑣 𝑑𝑥
𝑦 = (𝑥 2 + 2𝑥 + 1)(𝑥 3 + 3𝑥)
𝑦 = sin2 𝑥
𝑦 = 𝑥 2 𝑐𝑜𝑠𝑥
𝒚 = (𝒙𝟐 + 𝟐𝒙 + 𝟏)(𝒙𝟑 + 𝟑𝒙)
𝑢 = (𝑥 2 + 2𝑥 + 1)
𝑣 = (𝑥 3 + 3𝑥)
𝑡ℎ𝑒𝑛
𝑡ℎ𝑒𝑛
𝑑𝑢
𝑑𝑥
𝑑𝑣
𝑑𝑥
= 2𝑥 + 2
= 3𝑥 2 + 3
𝒚′ = 𝒖 ∙ 𝒗′ + 𝒗 ∙ 𝒖′
=
+ 2𝑥 + 1) ∙ (3𝑥 2 + 3) + (𝑥 3 + 3𝑥) ∙ (2𝑥 + 2)
= 3𝑥 4 + 6𝑥 3 + 3𝑥 2 + 6𝑥 + 3 + 2𝑥 4 + 6𝑥 2 + 2𝑥 3 + 6𝑥
𝑦 ′ = 5𝑥 4 + 8𝑥 3 + 12𝑥 2 + 12𝑥 + 3
(𝑥 2
𝒚 = 𝒔𝒊𝒏𝟐 𝒙
∴ 𝑙𝑒𝑡 𝑢 = 𝑠𝑖𝑛𝑥
𝐴𝑛𝑑 𝑣 = 𝑠𝑖𝑛𝑥
𝑑𝑦
𝒚 = 𝒙𝟐 𝒄𝒐𝒔𝒙
𝑢 = 𝑥2
𝑡ℎ𝑒𝑛 𝑢′ = 2𝑥
𝑑𝑥
𝑦 = (𝑠𝑖𝑛𝑥)(𝑠𝑖𝑛𝑥)
𝑡ℎ𝑒𝑛 𝑢′ = 𝑐𝑜𝑠𝑥
𝑡ℎ𝑒𝑛 𝑣 ′ = 𝑐𝑜𝑠𝑥
𝑦 ′ = 𝑠𝑖𝑛𝑥 ∙ 𝑐𝑜𝑠𝑥 + 𝑠𝑖𝑛𝑥 ∙ 𝑐𝑜𝑠𝑥
= 𝑦 ′ = 2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥
𝑣 = 𝑐𝑜𝑠𝑥
𝑡ℎ𝑒𝑛 𝑣 ′ = −𝑠𝑖𝑛𝑥
𝑦 ′ = 𝑥 2 ∙ (−𝑠𝑖𝑛𝑥) + 𝑐𝑜𝑠𝑥 ∙ (2𝑥)
𝑑𝑦
= 𝑦 ′ = 2𝑥𝑐𝑜𝑠𝑥 − 𝑥 2 𝑠𝑖𝑛𝑥
𝑑𝑥
𝑢
𝑦=𝑣
𝑑𝑦
𝑑𝑥
=
𝑣
𝑑𝑢
𝑑𝑣
−𝑢
𝑑𝑥
𝑑𝑥
𝑣2
𝒙𝟐 +𝟏
𝒚 = 𝟑𝒙𝟐+𝟐
𝒚=
𝒚=
𝒚=
𝒚=
𝟓𝒆𝒙
𝒄𝒐𝒔𝒙
𝟒
𝒙𝟔
𝒙𝟔
𝟓
𝒙𝟐 +𝟏
𝟑𝒙𝟐 +𝟐
𝑢 = 𝑥2 + 1
𝑡ℎ𝑒𝑛
𝑑𝑢
𝑑𝑥
= 2𝑥
𝑑𝑣
= 6𝑥
𝑑𝑥
2
2
(3𝑥 + 2) ∙ (2𝑥) − (𝑥 + 1) ∙ (6𝑥) 6𝑥 3 + 4𝑥 − 6𝑥 3 − 6𝑥
′
∴𝑦 =
=
(3𝑥 2 + 2)2
(3𝑥 2 + 2)2
𝑑𝑦
−2𝑥
= 𝑦′ =
(3𝑥 2 + 2)2
𝑑𝑥
𝑣 = 3𝑥 2 + 2
𝑡ℎ𝑒𝑛
𝟓𝒆𝒙
𝒚 = 𝒄𝒐𝒔𝒙
𝑢 = 5𝑒 𝑥
𝑡ℎ𝑒𝑛
𝑑𝑢
𝑑𝑥
= 5𝑒 𝑥
𝑑𝑣
= −𝑠𝑖𝑛𝑥
𝑑𝑥
𝑥
𝑑𝑦
𝑐𝑜𝑠𝑥 ∙ (5𝑒 ) − 5𝑒 𝑥 ∙ (−𝑠𝑖𝑛𝑥)
′
∴
=𝑦 =
(𝑐𝑜𝑠𝑥)2
𝑑𝑥
𝑥
5𝑒 (𝑐𝑜𝑠𝑥 + 𝑠𝑖𝑛𝑥)
=
cos 2 𝑥
𝑣 = 𝑐𝑜𝑠𝑥
𝟒
𝒚 = 𝒙𝟔
𝑡ℎ𝑒𝑛
𝑢=4
𝑡ℎ𝑒𝑛
𝑑𝑢
𝑑𝑥
=0
𝑑𝑣
= 6𝑥 5
𝑑𝑥
𝑑𝑦
𝑥 6 ∙ (0) − 4 ∙ (6𝑥 5 )
24𝑥 5
24
′
=𝑦 =
=
−
=
−
(𝑥 6 )2
𝑑𝑥
𝑥12
𝑥7
𝑣 = 𝑥6
𝑡ℎ𝑒𝑛
𝑦 = 4𝑥 −6
𝑦 ′ = 4(−6)𝑥 −7 = −24𝑥 −7 = −
𝒚=
24
𝑥 −7
𝒙𝟔
𝟓
1
𝑦 = 𝑥6
5
𝑑𝑦
6
= 𝑦′ = 𝑤 5
𝑑𝑥
5
𝑦(𝑥) = √𝑥
𝑓(𝑡) = 𝑡 50
𝑦(𝑥) = √5𝑥 − 8
𝑓(𝑡) = (2𝑡 3 + 𝑐𝑜𝑠(𝑡))
ℎ(𝑤) = 𝑒 𝑤
ℎ(𝑤) = 𝑒 𝑤
𝑦 = 𝑡𝑎𝑛(𝑥)
4 −3𝑤 2 +9
50
𝑔(𝑥) = 𝑙𝑛𝑥
3
𝑦 = 𝑡𝑎𝑛 ( √3𝑥 2 + 𝑡𝑎𝑛(5𝑥))
𝑔(𝑥) = 𝑙𝑛(𝑥 −4 + 𝑥 4 )
𝑦 ′ (𝑥) =
5
2√5𝑧 − 8
1
1
1
(5𝑧 − 8)−2 =
2
2 √5𝑧 − 8
𝑓(𝑢) = √𝑢
𝑎𝑛𝑑
𝑢 = 𝑔(𝑥) = 5𝑥 − 8
𝑦 = 𝑓(𝑢) 𝑎𝑛𝑑 𝑢 = 𝑔(𝑥)
𝑑𝑦 𝑑𝑦 𝑑𝑢
=
×
𝑑𝑥 𝑑𝑢 𝑑𝑥
𝑦(𝑥) = √5𝑥 − 8
𝑦 = 𝑓(𝑢) = √𝑢
𝑑𝑦
1
= 𝑓 ′ (𝑢) = 2 𝑢
𝑑𝑢
√
𝑢 = 𝑔(𝑥) = 5𝑥 − 8
𝑢′ = 5
𝑑𝑦 𝑑𝑦 𝑑𝑢
=
×
𝑑𝑥 𝑑𝑢 𝑑𝑥
𝑑𝑦
1
1
=
×5=
×5
𝑑𝑥 2√𝑢
2√5𝑥 − 8
𝑑𝑦
5
=
𝑑𝑥 2√5𝑥 − 8
⏟
𝑦(𝑥) =
(5𝑥 − 8)
⏟
𝑖𝑛𝑠𝑖𝑑𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
1
2
1
2
𝑜𝑢𝑡𝑠𝑖𝑑𝑒
𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
1
2
𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓
𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
𝑦(𝑥) =
⏞
1
1
(5𝑥 − 8) −2 ∙
⏟
2 𝑖𝑛𝑠𝑖𝑑𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
𝑙𝑒𝑓𝑡 𝑎𝑙𝑜𝑛𝑒
𝑑𝑦
=
𝑑𝑥
⏟ 𝑓′
(5)
⏟
𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓
𝑖𝑛𝑠𝑖𝑑𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
(𝑔(𝑥))
⏟
𝑔′ (𝑥)
⏟
∙
𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 𝑖𝑛𝑠𝑖𝑑𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
𝑙𝑒𝑓𝑡 𝑎𝑙𝑜𝑛𝑒
𝑡𝑖𝑚𝑒𝑠 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒
𝑜𝑓 𝑖𝑛𝑠𝑖𝑑𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
𝑦(𝑥)
𝑦(2)
𝑓(𝑥) = sin(3𝑥 2 + 𝑥)
50
𝑓(𝑡) = (2𝑡 3 + 𝑐𝑜𝑠(𝑡))
4
2
ℎ(𝑤) = 𝑒 𝑤 −3𝑤 +9
𝑔(𝑥) = ln(𝑥 −4 + 𝑥 4 )
𝑃(𝑡) = cos4 (𝑡) + 𝑐𝑜𝑠(𝑡 4 )
2𝑡+3 3
ℎ(𝑡) = (6−𝑡 2 )
𝒇(𝒙) = 𝐬𝐢𝐧(𝟑𝒙𝟐 + 𝒙)
3𝑥 2 + 𝑥
𝑓 ′ (𝑥) =
𝑐𝑜𝑠
⏟
(3𝑥
⏟ 2 + 𝑥)
(6𝑥
⏟ + 1)
𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓
𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
𝑙𝑒𝑎𝑣𝑒 𝑖𝑛𝑠𝑖𝑑𝑒
𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑎𝑙𝑜𝑛𝑒
𝑡𝑖𝑚𝑒𝑠 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒
𝑜𝑓 𝑖𝑛𝑠𝑖𝑑𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
𝑓′(𝑥) = (6𝑥 + 1)𝑐𝑜𝑠(3𝑥 2 + 𝑥)
𝟓𝟎
𝒇(𝒕) = (𝟐𝒕𝟑 + 𝒄𝒐𝒔(𝒕))
49
𝑓 ′ (𝑡) = 50(2𝑡 3 + 𝑐𝑜𝑠(𝑡)) (6𝑡 2 − sin(𝑡))
= 50(6𝑡 2 − 𝑠𝑖𝑛(𝑡))(2𝑡 3 + 𝑐𝑜𝑠(𝑡))
𝒉(𝒘) = 𝒆𝒘
49
𝟒 −𝟑𝒘𝟐 +𝟗
4
2
ℎ′ (𝑤) = 𝑒 𝑤 −3𝑤 +9 ∙ (4𝑤 3 − 6𝑤)
4
2
= (4𝑤 3 − 6𝑤)𝑒 𝑤 −3𝑤 +9
𝒈(𝒙) = 𝐥𝐧(𝒙−𝟒 + 𝒙𝟒 )
𝑔′ (𝑥) =
1
−4𝑥 −5 + 4𝑥 3
−5
3)
(−4𝑥
∙
+
4𝑥
=
𝑥 −4 + 𝑥 4
𝑥 −4 + 𝑥 4
1
1
𝑥
𝑖𝑛𝑠𝑖𝑑𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
𝑷(𝒕) = 𝐜𝐨𝐬 𝟒 (𝒕) + 𝒄𝒐𝒔(𝒕𝟒 )
cos4 (𝑡) = (cos(𝑡))4
𝑡4
𝑃′ (𝑡) = 4 cos3 (𝑡) (− sin(𝑡)) − sin(𝑡 4 ) (4𝑡 3 )
= −4𝑠𝑖𝑛(𝑡) cos3 (𝑡) − 4𝑡 3 𝑠𝑖𝑛(𝑡 4 )
𝟐𝒕+𝟑 𝟑
𝒉(𝒕) = (𝟔−𝒕𝟐 )
ℎ
2𝑡 + 3 2 𝑑 2𝑡 + 3
= 3(
) ∙
[
]
6 − 𝑡2
𝑑𝑡 6 − 𝑡 2
2𝑡 + 3 2 2(6 − 𝑡 2 ) − (2𝑡 + 3)(−2𝑡)
1𝑄 = 3 (
) [
]
(6 − 𝑡 2 )2
6 − 𝑡2
2𝑡 + 3 2 2𝑡 2 + 6𝑡 + 12
= 3(
) [
]
(6 − 𝑡 2 )2
6 − 𝑡2
′ (𝑡)
𝑦 = 𝑓(𝑥)
𝑦′ 𝑓𝑜𝑟 𝑥𝑦 = 1
𝑦
𝑦=
1
𝑥
⇒
𝑑𝑦
1
= 𝑦′ = − 2
𝑑𝑥
𝑥
𝑦
𝑦
𝑥
𝑦
𝑦 = 𝑦(𝑥)
𝑥𝑦 = 𝑥𝑦(𝑥) = 1
𝑦(𝑥)
𝑦
𝑦
𝑥
𝑥
𝑥
𝑑
𝑑
(1)
(𝑥𝑦(𝑥)) =
𝑑𝑥
𝑑𝑥
𝑥
(1)𝑦(𝑥) + 𝑥
𝑦(𝑥)
𝑑
(𝑦(𝑥)) = 0
𝑑𝑥
𝑑
𝑑𝑦
(𝑦(𝑥)) =
= 𝑦′
𝑑𝑥
𝑑𝑥
𝑦 + 𝑥𝑦 ′ = 0
(𝑥)
𝑦
𝑦
𝑥
𝑦′
𝑦′
𝑦′
𝑦′ = −
𝑦
𝑥
𝑦
1
1
𝑦′ = − 𝑥 = − 2
𝑥
𝑥
𝑦′
3 5
a.
b.
c.
𝑥 𝑦 + 3𝑥 = 8𝑦 3 + 1
𝑥 2 tan(𝑦) + 𝑦10 sec(𝑥) = 2𝑥
𝑒 2𝑥+3𝑦 = 𝑥 2 − 𝑙𝑛(𝑥𝑦 3 )
a.
𝒙𝟑 𝒚𝟓 + 𝟑𝒙 = 𝟖𝒚𝟑 + 𝟏
𝑥
𝑥
𝑦
𝑦(𝑥)
𝑦
𝑦′
3𝑥 2 𝑦 5 + 5𝑥 3 𝑦 4 𝑦 ′ + 3 = 24𝑦 2 𝑦 ′
𝑦′
𝑦′
𝑦′
𝑦′
𝑦′
3𝑥 2 𝑦 5 + 3 = 24𝑦 2 𝑦 ′ − 5𝑥 3 𝑦 4 𝑦 ′
3𝑥 2 𝑦 5 + 3 = (24𝑦 2 − 5𝑥 3 𝑦 4 )𝑦 ′
3𝑥 2 𝑦 5 + 3
𝑦′ =
(24𝑦 2 − 5𝑥 3 𝑦 4 )
𝒙𝟐 𝐭𝐚𝐧(𝒚) + 𝒚𝟏𝟎 𝐬𝐞𝐜(𝒙) = 𝟐𝒙
2𝑥𝑡𝑎𝑛(𝑦) + 𝑥 2 sec 2 (𝑦) 𝑦 ′ + 10𝑦 9 𝑦 ′ 𝑠𝑒𝑐(𝑥) + 𝑦 10 𝑠𝑒𝑐(𝑥) 𝑡𝑎 𝑛(𝑥) = 2
(𝑥 2 sec 2 (𝑦) + 10𝑦 9 𝑠𝑒𝑐(𝑥))𝑦 ′ = 2 − 2𝑥𝑡𝑎𝑛(𝑦) − 𝑦 10 𝑠𝑒𝑐(𝑥) tan(𝑥)
2 − 2𝑥𝑡𝑎𝑛(𝑦) − 𝑦 10 𝑠𝑒𝑐(𝑥) tan(𝑥)
𝑦′ =
𝑥 2 sec 2 (𝑦) + 10𝑦 9 𝑠𝑒𝑐(𝑥)
𝒆𝟐𝒙+𝟑𝒚 = 𝒙𝟐 − 𝒍𝒏(𝒙𝒚𝟑 )
𝑥
𝑦
𝑦
𝑒 2𝑥+3𝑦 ∙ (2 + 3𝑦 ′ ) = 2𝑥 −
𝑦′
𝑦 3 + 3𝑥𝑦 2 𝑦 ′
𝑥𝑦 3
𝑦
𝑦′
𝑦3
3𝑥𝑦 2 𝑦 ′
2𝑒
+ 3𝑦
= 2𝑥 − 3 −
𝑥𝑦
𝑥𝑦 3
1 3𝑦 ′
2𝑥+3𝑦
2𝑒 2𝑥+3𝑦 + 3𝑦 ′𝑒
= 2𝑥 − −
𝑥
𝑦
2𝑥+3𝑦
−1
′
−1
(3𝑒
+ 3𝑦 )𝑦 = 2𝑥 − 𝑥 − 2𝑒 2𝑥+3𝑦
2𝑥 − 𝑥 −1 − 2𝑒 2𝑥+3𝑦
′
𝑦 =
(3𝑒 2𝑥+3𝑦 + 3𝑦 −1 )
2𝑥+3𝑦
′𝑒 2𝑥+3𝑦
𝑓(𝑥) = 5𝑥 3 − 3𝑥 2 + 10𝑥 − 5
𝑓 ′ (𝑥) = 15𝑥 2 − 6𝑥 + 10
′
𝑓 ′′ (𝑥) = (𝑓 ′ (𝑥)) = 30𝑥 − 6
𝑓′(𝑥)
′
𝑓 ′′′ (𝑥) = (𝑓 ′′ (𝑥)) = 30
′
𝑓 (4) (𝑥) = (𝑓 ′′′ (𝑥)) = 0
𝑓 (2) (𝑥) = 𝑓 ′′ (𝑥)
𝑓 2 (𝑥) = [𝑓(𝑥)]2
𝑓 ′ (𝑥) =
𝑓
′′ (𝑥)
𝑑2𝑦
= 2
𝑑𝑥
𝑓
𝑑𝑓
𝑑𝑥
′′′ (𝑥)
𝑑3𝑦
= 3
𝑑𝑥
𝑒𝑡𝑐.
Application of Differentiation
𝑠(𝑡) = 3𝑡 4 − 40𝑡 3 + 126𝑡 2 − 9
3.
If 𝑣 = 4𝑠 2 − 3𝑠 + 1, where v denotes velocity. Determine the acceleration when the distance
s = 2 metres
𝑦 = 2𝑧 4 − 𝑧 3 − 3𝑧 2
4.
Obtain the stationary points of the curve 𝑓(𝑥) = 1+8𝑥 2
6.
𝑓(𝑥) = (1 + 12√𝑥)(4 − 𝑥 2 ) 𝑎𝑡 𝑥 = 9.
The parametric equation of a curve are 𝑥 = 3𝑡 2 , 𝑦 = 3𝑡 − 𝑡 2 . Find the volume generated when
the plane figure bounded by the curve. The x-axis and the ordinates at t = 0 and t = 2 rotate
about the x-axis.
7.
If 𝑥 = 2𝑡 2 𝑎𝑛𝑑 𝑦 = 4𝑡, find 𝑑𝑥 in terms of t.
8.
Given that 𝑦 = 𝑐𝑜𝑠2𝑡 𝑎𝑛𝑑 𝑥 = 𝑠𝑖𝑛𝑡, find 𝑑𝑥 𝑎𝑛𝑑 𝑑𝑥 2
2.
𝑥−𝑥 2
𝑑𝑦
𝑑𝑦
𝑑2 𝑦
Solutions
𝑠 ′ (𝑡) = 12𝑡 3 − 120𝑡 2 + 252𝑡
= 12𝑡(𝑡 − 3)(𝑡 − 7)
𝐼𝑓 𝑠 ′ (𝑡) = 0
𝑡ℎ𝑒𝑛
⇒ 12𝑡(𝑡 − 3)(𝑡 − 7) = 0
𝑡 = 0 𝑜𝑟 𝑡 = 3 𝑜𝑟 𝑡 = 7
If 𝒗 = 𝟒𝒔 − 𝟑𝒔 + 𝟏, then the acceleration is the derivative of velocity function
𝟐
𝑑𝑣
That is, 𝑑𝑠 = 𝑣 ′ = 8𝑠 − 3
When 𝑠 = 2,
𝑑𝑣
𝑑𝑠
= 𝑣 ′ = 8(2) − 3 = 13𝑚/𝑠 2
𝒚 = 𝟐𝒛𝟒 − 𝒛𝟑 − 𝟑𝒛𝟐
3.
𝑑𝑦
𝑑𝑧
𝑑𝑦
= 8𝑧 3 − 3𝑧 2 − 6𝑧
𝑑𝑧
= 𝑧(8𝑧 2 − 3𝑧 − 6)
=0
𝑑𝑦
=0
∴
𝑑𝑧
→
𝑧 = 0,
𝑧=
𝑧(8𝑧 2 − 3𝑧 − 6) = 0
8𝑧 2 − 3𝑧 − 6 = 0
𝑧=0
3 ± √(−3)2 − 4(8)(−6) 3 ± √201
=
2(8)
16
3 + √201
3 − √201
= 1.07359 𝑜𝑟 𝑧 =
= −0.69859
16
16
𝒙−𝒙𝟐
4. Obtain the stationary points of the curve 𝒇(𝒙) = 𝟏+𝟖𝒙𝟐
𝑧 = 0 𝑜𝑟 𝑧 =
𝑑𝑦
𝑑𝑥
=0
𝐿𝑒𝑡 𝑢 = 𝑥 − 𝑥 2
𝑎𝑛𝑑
𝑣 = 1 + 8𝑥 2
𝑑𝑢
𝑑𝑣
= 1 − 2𝑥
𝑎𝑛𝑑
= 16𝑥
𝑑𝑥
𝑑𝑥
(1 − 2𝑥)(1 + 8𝑥 2 ) − (𝑥 − 𝑥 2 )(16𝑥)
∴ 𝑓 ′ (𝑥) =
(1 + 8𝑥 2 )2
1 − 2𝑥 − 8𝑥 2
=
(1 + 8𝑥 2 )2
1 − 2𝑥 − 8𝑥 2 = −(8𝑥 2 + 2𝑥 − 1) = −(4𝑥 − 1)(2𝑥 + 1) = 0
𝑥=−
1
2
1
4
𝒇(𝒙) = (𝟏 + 𝟏𝟐√𝒙)(𝟒 − 𝒙𝟐 ) 𝒂𝒕 𝒙 = 𝟗.
𝑜𝑟
𝑥=
𝐿𝑒𝑡 𝑢 = (1 + 12√𝑥)
𝑎𝑛𝑑
𝑣 = (4 − 𝑥 2 )
1
𝑑𝑢
𝑑𝑣
= 6𝑥 −2
𝑎𝑛𝑑
= −2𝑥
𝑑𝑥
𝑑𝑥
1
𝑓 ′ (𝑥) = (6𝑥 −2 ) (4 − 𝑥 2 ) + (1 + 12√𝑥)(−2𝑥)
6
= ( ) (4 − 𝑥 2 ) − 2𝑥(1 + 12√𝑥)
√𝑥
𝑥=9
𝑓(9) = (37)(−77) = −2849
𝑓′(9) = (2)(−77) − 18(37) = −820
𝑥=9
𝑚 = 𝑓′(9) = −820
(9, 𝑓(9)) = (9, −2849)
6.
7.
𝑦 = 𝑓(9) + 𝑓 ′ (9)(𝑥 − 9)
= −2849 − 820(𝑥 − 9)
𝑦 = −820𝑥 + 4531
The parametric equation of a curve are 𝒙 = 𝟑𝒕𝟐 , 𝒚 = 𝟑𝒕 − 𝒕𝟐 . Find the volume generated
when the plane figure bounded by the curve. The x-axis and the ordinates at t = 0 and t = 2
rotate about the x-axis.
𝑥 = 3𝑡 2 ,
𝑦 = 3𝑡 − 𝑡 2
𝑑𝑥
𝑑𝑦
= 6𝑡
= 3 − 2𝑡
𝑑𝑡
𝑑𝑡
Just like the Chain Rule, we obtain;
𝑑𝑦 𝑑𝑦 𝑑𝑡 𝑑𝑦 𝑑𝑥
=
×
=
÷
𝑑𝑥 𝑑𝑡 𝑑𝑥 𝑑𝑡 𝑑𝑡
𝑑𝑦
1
3 − 2𝑡
= (3 − 2𝑡) × =
𝑑𝑥
6𝑡
6𝑡
𝒅𝒚
𝟐
If 𝒙 = 𝟐𝒕 𝒂𝒏𝒅 𝒚 = 𝟒𝒕, find 𝒅𝒙 in terms of t.
𝑥 = 2𝑡 2 ,
𝑑𝑥
= 4𝑡
𝑑𝑡
Just like the Chain Rule, we obtain;
𝑦 = 4𝑡
𝑑𝑦
=4
𝑑𝑡
8.
𝑑𝑦 𝑑𝑦 𝑑𝑡 𝑑𝑦 𝑑𝑥
=
×
=
÷
𝑑𝑥 𝑑𝑡 𝑑𝑥 𝑑𝑡 𝑑𝑡
𝑑𝑦
1
4
1
=4× =
=
𝑑𝑥
4𝑡 4𝑡 𝑡
𝒅𝒚
𝒅𝟐 𝒚
Given that 𝒚 = 𝒄𝒐𝒔𝟐𝒕 𝒂𝒏𝒅 𝒙 = 𝒔𝒊𝒏𝒕, find 𝒅𝒙 𝒂𝒏𝒅 𝒅𝒙𝟐
𝑦 = cos 2𝑡 ,
𝑥 = sin 𝑡
𝑑𝑦
𝑑𝑥
= −2 sin 2𝑡
= cos 𝑡
𝑑𝑡
𝑑𝑡
Just like the Chain Rule, we obtain;
𝑑𝑦 𝑑𝑦 𝑑𝑡 𝑑𝑦 𝑑𝑥
=
×
=
÷
𝑑𝑥 𝑑𝑡 𝑑𝑥 𝑑𝑡 𝑑𝑡
𝑑𝑦
1
−2 sin 2𝑡
(𝑅𝑒𝑐𝑎𝑙𝑙 𝑡ℎ𝑎𝑡 sin 2𝑡 = 2 sin 𝑡 cos 𝑡)
= −2 sin 2𝑡 ×
=
𝑑𝑥
cos 𝑡
cos 𝑡
𝑑𝑦
2(2 sin 𝑡 cos 𝑡)
=−
= −4 sin 𝑡
𝑑𝑥
cos 𝑡
𝑑2 𝑦
𝑑𝑥 2
𝑑2 𝑦
𝑑 𝑑𝑦
𝑑
𝑑
𝑑𝑡
(−4 sin 𝑡) = (−4 sin 𝑡)
=
( )=
2
𝑑𝑥
𝑑𝑥 𝑑𝑥
𝑑𝑥
𝑑𝑡
𝑑𝑥
𝑑𝑥
𝑑𝑡
1
𝑑
(−4 sin 𝑡) = −4 cos 𝑡
= cos 𝑡
∴
=
𝑎𝑛𝑑
𝑑𝑡
𝑑𝑥 cos 𝑡
𝑑𝑡
𝑑2𝑦
𝑑
𝑑𝑡
1
(−4
(−4
∴
=
sin
𝑡)
=
cos
𝑡)
(
) = −4
𝑑𝑥 2 𝑑𝑡
𝑑𝑥
cos 𝑡








𝑓(𝑥)
𝑓(𝑥)
𝑓(𝑥) = 𝑥 4 + 3𝑥 − 9
𝑓 ′ (𝑥) = 4𝑥 3 + 3
𝑥
𝑥4
𝑥5
𝑥5
5𝑥 4
1
5
𝑥5
𝑥4
3𝑥
3
2
𝑥
2
𝑥
−9𝑥
−9
1
3
𝐹(𝑥) = 𝑥 5 + 𝑥 2 − 9𝑥
5
2
𝐹(𝑥)
𝐹 ′ (𝑥) = 𝑥 4 + 3𝑥 − 9 = 𝑓(𝑥)
𝑓(𝑥)
1
3
𝐹(𝑥) = 𝑥 5 + 𝑥 2 − 9𝑥 + 10
5
2
1 5 3 2
𝐹(𝑥) = 𝑥 + 𝑥 − 9𝑥 − 1954
5
2
1 5 3 2
3469
𝐹(𝑥) = 𝑥 + 𝑥 − 9𝑥 +
5
2
123
𝑒𝑡𝑐.
1
3
𝐹(𝑥) = 𝑥 5 + 𝑥 2 − 9𝑥 + 𝑐, 𝑐 𝑖𝑠 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
5
2
𝑓(𝑥)
𝑓(𝑥)
𝑓(𝑥)
𝐹 (𝑥) = 𝑓(𝑥)
𝐹(𝑥)
′
𝐹(𝑥)
𝑓(𝑥)
∫ 𝑓(𝑥)𝑑𝑥 = 𝐹(𝑥) + 𝑐,
∫
𝑐
𝑓(𝑥)
𝑐 𝑖𝑠 𝑎𝑛𝑦 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑓(𝑥)
𝑥
𝑓(𝑥)
𝒇(𝒙)
𝒙
∫ 𝑥 4 + 3𝑥 − 9𝑑𝑥
1
3
∫ 𝑥 4 + 3𝑥 − 9 𝑑𝑥 = 𝑥 5 + 𝑥 2 − 9𝑥 + 𝑐
5
2
𝑑𝑥
𝑑𝑥
1
3
∫ 𝑥 4 + 3𝑥 − 9𝑑𝑥 = 𝑥 5 + 𝑥 2 − 9𝑥 + 𝑐
5
2
1
3
∫ 𝑥 4 + 3𝑥𝑑𝑥 − 9 = 𝑥 5 + 𝑥 2 + 𝑐 − 9
5
2
1
∫ 𝑥 4 𝑑𝑥 + 3𝑥 − 9 = 𝑥 5 + 𝑐 + 3𝑥 − 9
5
−9
3𝑥 − 9
𝑑𝑥
1
3
∫ 𝑥 4 + 3𝑥 − 9𝑑𝑥 = 𝑥 5 + 𝑥 2 − 9𝑥 + 𝑐
5
2
1
3
∫ 𝑡 4 + 3𝑡 − 9𝑑𝑡 = 𝑡 5 + 𝑡 2 − 9𝑡 + 𝑐
5
2
1
3
∫ 𝑤 4 + 3𝑤 − 9𝑑𝑤 = 𝑤 5 + 𝑤 2 − 9𝑤 + 𝑐
5
2
∫ 2𝑥𝑑𝑥
∫ 2𝑡𝑑𝑥
∫ 2𝑥𝑑𝑥 = 𝑥 2 + 𝑐
𝑥
𝑥
∫ 2𝑡𝑑𝑥 = 2𝑡𝑥 + 𝑐
∫ 𝑘𝑓(𝑥)𝑑𝑥 = 𝑘∫ 𝑓(𝑥)𝑑𝑥 𝑤ℎ𝑒𝑟𝑒 𝑘 𝑖𝑠 𝑎𝑛𝑦 𝑛𝑢𝑚𝑏𝑒𝑟
∫ −𝑓(𝑥)𝑑𝑥 = −∫ 𝑓(𝑥)𝑑𝑥
∫ 𝑓(𝑥) ± 𝑔(𝑥)𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 ± ∫ 𝑔(𝑥)𝑑𝑥.
∫ 𝑓(𝑥)𝑔(𝑥)𝑑𝑥 ≠ ∫ 𝑓(𝑥)𝑑𝑥∫ 𝑔(𝑥)𝑑𝑥
𝑓(𝑥)
∫ 𝑓(𝑥)𝑑𝑥
∫
𝑑𝑥 ≠
𝑔(𝑥)
∫ 𝑔(𝑥)𝑑𝑥
𝑎 𝑎𝑛𝑑 𝑐
∫ 𝑎𝑥 𝑛 𝑑𝑥 =
1
∫ 𝑥 2 𝑑𝑥
1
∫ 𝑥 𝑑𝑥
√
𝑎𝑥 𝑛+1
𝑛+1
−2
= ∫𝑥
+ 𝑐,
𝑑𝑥 =
1
−
2
∫ 𝑒 𝑥 𝑑𝑥 = 𝑒 𝑥 + 𝑐
𝑛 ≠ −1
1
−𝑥
𝑎𝑥
∫ 𝑎 𝑥 𝑑𝑥 = 𝑙𝑛𝑎 + 𝑐
+𝑐
1
∫ 𝑥 𝑑𝑥 = ∫ 𝑥 −1 𝑑𝑥 = ln 𝑥 + 𝑐
= ∫ 𝑥 𝑑𝑥 = 2√𝑥 + 𝑐
𝑓′ (𝑥)
𝑑𝑥
𝑓(𝑥)
1
∫ 𝑥 2 +1 𝑑𝑥
1
∫
∫ 𝑎𝑑𝑥 = 𝑎𝑥 + 𝑐
∫ 𝑠𝑖𝑛𝑥𝑑𝑥 = −𝑐𝑜𝑠𝑥 + 𝑐
∫ 𝑐𝑜𝑠𝑥𝑑𝑥 = 𝑠𝑖𝑛𝑥 + 𝑐
− cos 𝑎𝑥
∫ sin 𝑎𝑥 𝑑𝑥 = 𝑎
+𝑐
sin 𝑎𝑥
∫ cos 𝑎𝑥 𝑑𝑥 = 𝑎 + 𝑐
𝑡𝑎𝑛𝑎𝑥
∫ sec 2 𝑎𝑥 𝑑𝑥 = 𝑎 + 𝑐
𝑠𝑒𝑐𝑎𝑥
∫ sec 𝑎𝑥 tan 𝑎𝑥 𝑑𝑥 = 𝑎 + 𝑐
−𝑐𝑜𝑡𝑎𝑥
∫ 𝑐𝑜𝑠𝑒𝑐 2 𝑎𝑥 𝑑𝑥 = 𝑎 + 𝑐
−𝑐𝑜𝑠𝑒𝑐𝑎𝑥
∫ 𝑐𝑜𝑠𝑒𝑐 𝑎𝑥 cot 𝑎𝑥 𝑑𝑥 =
𝑎
∫ 5𝑡 3 − 10𝑡 −6 + 4𝑑𝑡
∫ 𝑥 8 + 𝑥 −8 𝑑𝑥
7
1
4
∫ 3√𝑥 3 + 𝑥 5 + 6 𝑥 𝑑𝑥
√
= ln 𝑓(𝑥) + 𝑐
= tan−1 𝑥 + 𝑐
∫ √1−𝑥2 𝑑𝑥 = sin−1 𝑥 + 𝑐
1
∫ 𝑥√𝑥 2 −1 𝑑𝑥 = sec −1 𝑥 + 𝑐
+𝑐
∫ 𝑠𝑖𝑛ℎ𝑥𝑑𝑥 = 𝑐𝑜𝑠ℎ𝑥 + 𝑐
∫ 𝑐𝑜𝑠ℎ𝑥𝑑𝑥 = 𝑠𝑖𝑛ℎ𝑥 + 𝑐
∫ sech2 𝑥 𝑑𝑥 = 𝑡𝑎𝑛ℎ𝑥 + 𝑐
∫ 𝑠𝑒𝑐ℎ𝑥𝑡𝑎𝑛ℎ𝑥𝑑𝑥 = −𝑠𝑒𝑐ℎ𝑥 + 𝑐
∫ 𝑐𝑜𝑠𝑒𝑐ℎ2 𝑥𝑑𝑥 = −𝑐𝑜𝑡ℎ𝑥 + 𝑐
∫ 𝑐𝑜𝑠𝑒𝑐ℎ𝑥𝑐𝑜𝑡ℎ𝑥𝑑𝑥 = −𝑐𝑜𝑠𝑒𝑐ℎ𝑥 + 𝑐
∫ 𝑑𝑦
3
∫ (𝑤 + √𝑤 )(4 − 𝑤 2 )𝑑𝑤
∫
4𝑥 10 −2𝑥 4 +15𝑥 2
𝑥3
𝑑𝑥
23
∫ 3𝑒 𝑥 + 5𝑐𝑜𝑠𝑥 − 10𝑠𝑒𝑐 2 𝑥 𝑑𝑥
1
∫ 2𝑠𝑒𝑐𝑤𝑡𝑎𝑛𝑤 + 6𝑤 𝑑𝑤
9
∫ 𝑦2 +1 + 6𝑐𝑜𝑠𝑒𝑐𝑦𝑐𝑜𝑡𝑦 + 𝑦 𝑑𝑦
∫
7−6 sin2 𝜃
𝑑𝜃
sin2 𝜃
∫ 𝟓𝒕𝟑 − 𝟏𝟎𝒕−𝟔 + 𝟒𝒅𝒕
1
1
5
∫ 5𝑡 3 − 10𝑡 −6 + 4𝑑𝑡 = 5 ( ) 𝑡 4 − 10 ( ) 𝑡 −5 + 4𝑡 + 𝑐 = 𝑡 4 + 2𝑡 −5 + 4𝑡 + 𝑐
4
−5
4
−7
𝟖
−𝟖
∫𝒙 +𝒙
𝟒
𝒅𝒙
𝟕
∫ 𝟑√𝒙𝟑 + 𝒙𝟓 + 𝟔
𝟏
√𝒙
1
1 −7
∫ 𝑥 8 + 𝑥 −8 𝑑𝑥 = 𝑥 9 +
𝑥 +𝑐
9
−7
𝒅𝒙
4
3
7
1
1 −1
4 + 7𝑥 −5 + 𝑥 2 𝑑𝑥
+
𝑑𝑥
=
3𝑥
𝑥 5 6√𝑥
6
1
1
1 7 7
1
= 3 𝑥 4 − 𝑥 −4 + ( 1 ) 𝑥 2 + 𝑐
7
4
6 2
4
4 7 7
1 2 1
= 3 𝑥 4 − 𝑥 −4 + ( ) 𝑥 2 + 𝑐
7
4
6 1
12 7 7 −4 1 1
=
𝑥4 − 𝑥 + 𝑥2 + 𝑐
7
4
3
∫ 3 √𝑥 3 +
∫ 𝒅𝒚
∫ 𝑑𝑦 = ∫ 1𝑑𝑦 = 𝑦 + 𝑐
∫ (𝒘 + 𝟑√𝒘)(𝟒 − 𝒘𝟐 )𝒅𝒘
−5
1
3
∫
7
∫ (𝑤 + √𝑤 )(4 − 𝑤 2 )𝑑𝑤 = ∫ 4𝑤 − 𝑤 3 + 4𝑤 3 − 𝑤 3 𝑑𝑤
4
1
3 10
= 2𝑤 2 − 𝑤 4 + 3𝑤 3 −
𝑤3 +𝑐
4
10
𝟐
𝟒𝒙𝟏𝟎 −𝟐𝒙𝟒 +𝟏𝟓𝒙
𝒙𝟑
𝒅𝒙
4𝑥10 − 2𝑥 4 + 15𝑥 2
4𝑥10 2𝑥 4 15𝑥 2
∫
𝑑𝑥 = ∫ 3 − 3 + 3 𝑑𝑥
𝑥3
𝑥
𝑥
𝑥
15
= ∫ 4𝑥 7 − 2𝑥 + 𝑑𝑥
𝑥
1 8
2
= 𝑥 − 𝑥 + 15 ln 𝑥 + 𝑐
2
𝑥
∫
15
𝑑𝑥 = 15 ∫ 1𝑥 𝑑𝑥 = 15 ln 𝑥 + 𝑐
𝑥
∫ 𝟑𝒆𝒙 + 𝟓𝒄𝒐𝒔𝒙 − 𝟏𝟎𝒔𝒆𝒄𝟐 𝒙 𝒅𝒙
∫ 3𝑒 𝑥 + 5𝑐𝑜𝑠𝑥 − 10 sec 2 𝑥 𝑑𝑥 = 3𝑒 𝑥 + 5𝑠𝑖𝑛𝑥 − 10𝑡𝑎𝑛𝑥 + 𝑐
𝟏
∫ 𝟐𝒔𝒆𝒄 𝒘 𝒕𝒂𝒏 𝒘 + 𝟔𝒘 𝒅𝒘
∫ 2𝑠𝑒𝑐𝑤𝑡𝑎𝑛𝑤 +
1
1 1
𝑑𝑤 = 2 ∫ 𝑠𝑒𝑐𝑤𝑡𝑎𝑛𝑤𝑑𝑤 + ∫ 𝑑𝑤
6𝑤
6 𝑤
1
= 2 sec 𝑤 + ln 𝑤 + 𝑐
6
𝟐𝟑
𝟗
∫ 𝒚𝟐+𝟏 + 𝟔𝒄𝒐𝒔𝒆𝒄𝒚𝒄𝒐𝒕𝒚 + 𝒚 𝒅𝒚
∫
∫
23
9
+ 6𝑐𝑜𝑠𝑒𝑐𝑦𝑐𝑜𝑡𝑦 + 𝑑𝑦 = 23 tan−1 𝑦 − 6𝑐𝑜𝑠𝑒𝑐 𝑦 + 9 ln 𝑦 + 𝑐
+1
𝑦
𝑦2
𝟕−𝟔 𝐬𝐢𝐧𝟐 𝜽
𝒅𝜽
𝐬𝐢𝐧𝟐 𝜽
7 − 6 sin2 𝜃
7
∫
𝑑𝜃 = ∫ 2 − 6𝑑𝜃
2
sin 𝜃
sin 𝜃
= ∫ 7𝑐𝑜𝑠𝑒𝑐 2 𝜃 − 6𝑑𝜃
∫
4
∫ √𝑥 𝑑𝑥
7 − 6 sin2 𝜃
𝑑𝜃 = −7𝑐𝑜𝑡𝜃 − 6𝜃 + 𝑐
sin2 𝜃
∫
1
𝑑𝑡
𝑡3
∫ cos 𝑤 𝑑𝑤
∫ 𝑒 𝑦 𝑑𝑦
𝑥
∫ 18𝑥
2 4√
6𝑥 3
+ 5 𝑑𝑥
2𝑡 3 + 1
∫ 4
𝑑𝑡
(𝑡 + 2𝑡)3
∫(8𝑦 − 1) 𝑒 4𝑦
𝑡
1
∫ (1 − ) 𝑐𝑜𝑠(𝑤 − ln 𝑤) 𝑑𝑤
𝑤
2 −𝑦
𝑑𝑦
𝑤
4
∫ 18𝑥 2 √6𝑥 3 + 5 𝑑𝑥
𝑢 = 6𝑥 3 + 5
𝑑𝑢 = 18𝑥 2 𝑑𝑥
𝑥
𝑢
𝑢
4
∫ 18𝑥 2 √6𝑥 3 + 5 𝑑𝑥 = ∫ (6𝑥 3 +
1
5)4
(18𝑥 2 𝑑𝑥)
1
= ∫ 𝑢4 𝑑𝑢
1
5
4 5
4
4
∫ 18𝑥 2 √6𝑥 3 + 5 𝑑𝑥 = ∫ 𝑢4 𝑑𝑢 = 𝑢4 + 𝑐 = (6𝑥 3 + 5)4 + 𝑐
5
5
∫ 𝑓(𝑔(𝑥))𝑔′(𝑥) 𝑑𝑥 = ∫ 𝑓(𝑢) 𝑑𝑢, 𝑤ℎ𝑒𝑟𝑒, 𝑢 = 𝑔(𝑥)
4
∫ √𝑥 𝑑𝑥
𝑥
𝑢
𝑥
𝑥
𝑢
𝑑𝑥
𝑑𝑢
𝑥
𝑥
𝑥
1
∫ (1 − 𝑤) 𝑐𝑜𝑠(𝑤 − ln 𝑤) 𝑑𝑤
2
∫ 3 (8𝑦 − 1) 𝑒 4𝑦 −𝑦 𝑑𝑦
∫ 𝑥 2 (3 − 10𝑥 3 )4 𝑑𝑥
𝑥
𝑑𝑥
∫
2
√1−4𝑥
𝟏
∫ (𝟏 − 𝒘) 𝒄𝒐𝒔(𝒘 − 𝐥𝐧 𝒘) 𝒅𝒘
𝑢 = 𝑤 − ln 𝑤
1
𝑑𝑢 = (1 − ) 𝑑𝑤
𝑤
1
∫ (1 − ) cos(𝑤 − 𝑙𝑛𝑤) 𝑑𝑤 = ∫ cos(𝑢) 𝑑𝑢
𝑤
= sin(𝑢) + 𝑐
= 𝑠𝑖𝑛(𝑤 − 𝑙𝑛𝑤) + 𝑐
𝟐 −𝒚
∫ 𝟑 (𝟖𝒚 − 𝟏) 𝒆𝟒𝒚
𝒅𝒚
𝑢 = 4𝑦 2 − 𝑦
𝑑𝑢 = (8𝑦 − 1)𝑑𝑦
∫ 3(8𝑦 − 1)𝑒 4𝑦
2 −𝑦
𝑑𝑦 = 3∫ 𝑒 𝑢 𝑑𝑢
= 3𝑒 𝑢 + 𝑐 = 3𝑒 4𝑦
2 −𝑦
+𝑐
𝟑 𝟒
𝟐
∫ 𝒙 (𝟑 − 𝟏𝟎𝒙 ) 𝒅𝒙
𝑢 = 3 − 10𝑥 3
𝑑𝑢 = −30𝑥 2 𝑑𝑥
𝑥2
−30
𝑥 2 𝑑𝑥 = −
𝑥 2 𝑑𝑥
1
𝑑𝑢
30
∫ 𝑥 2 (3 − 10𝑥 3 )4 𝑑𝑥 = ∫(3 − 10𝑥 3 )4 𝑥 2 𝑑𝑥
= ∫ 𝑢4 (−
1
) 𝑑𝑢
30
1 1 5
( )𝑢 + 𝑐
30 5
1
(3 − 10𝑥 3 )5 + 𝑐
=−
150
=−
∫
𝒙
√𝟏−𝟒𝒙𝟐
𝒅𝒙
𝑢 = 1 − 4𝑥 2
𝑑𝑢 = −8𝑥𝑑𝑥
∫
𝑥
√1 − 4𝑥 2
1
𝑥𝑑𝑥 = − 𝑑𝑢
8
⇒
1
𝑑𝑥 = ∫ 𝑥(1 − 4𝑥 2 )−2 𝑑𝑥
1
1
= − ∫ 𝑢−2 𝑑𝑢
8
1 1
= − 𝑢2 + 𝑐
4
1
1
= − (1 − 4𝑥 2 )2 + 𝑐
4
4
∫ 𝑠𝑖𝑛(1 − 𝑥)(2 − 𝑐𝑜𝑠(1 − 𝑥)) 𝑑𝑥
∫ 𝑐𝑜𝑠(3𝑧) sin10 (3𝑧) 𝑑𝑧
3
∫ sec 2(4𝑡) (3 − 𝑡𝑎𝑛(4𝑡)) 𝑑𝑡
𝟒
∫ 𝒔𝒊𝒏(𝟏 − 𝒙)(𝟐 − 𝒄𝒐𝒔(𝟏 − 𝒙)) 𝒅𝒙
1−𝑥
1−𝑥
𝑢 = 2 − 𝑐𝑜𝑠(1 − 𝑥)
𝑑𝑢 = −𝑠𝑖𝑛(1 − 𝑥)𝑑𝑥
⇒
𝑠𝑖𝑛(1 − 𝑥)𝑑𝑥 = −𝑑𝑢
−
1−𝑥
1−𝑥
4
∫ 𝑠𝑖𝑛(1 − 𝑥)(2 − 𝑐𝑜𝑠(1 − 𝑥)) 𝑑𝑥 = − ∫ 𝑢4 𝑑𝑢
1
= − 𝑢5 + 𝑐
5
1
5
= − (2 − 𝑐𝑜𝑠(1 − 𝑥)) + 𝑐
5
∫ 𝒄𝒐𝒔(𝟑𝒛) 𝐬𝐢𝐧𝟏𝟎 (𝟑𝒛) 𝒅𝒛
10
sin10(3𝑧) = (𝑠𝑖𝑛(3𝑧))
𝑢 = 𝑠𝑖𝑛(3𝑧)
𝑑𝑢 = 3𝑐𝑜𝑠(3𝑧)𝑑𝑧
⇒
1
𝑐𝑜𝑠(3𝑧)𝑑𝑧 = 𝑑𝑢
3
3𝑧
1
1 1
1
∫ 𝑐𝑜𝑠(3𝑧) sin10 (3𝑧) 𝑑𝑧 = ∫ 𝑢10 𝑑𝑢 = ( ) 𝑢11 + 𝑐 =
sin11 (3𝑧) + 𝑐
3
3 11
33
𝟑
∫ 𝐬𝐞𝐜 𝟐 (𝟒𝒕) (𝟑 − 𝒕𝒂𝒏(𝟒𝒕)) 𝒅𝒕
4𝑡
𝑢 = 3 − 𝑡𝑎𝑛(4𝑡)
𝑑𝑢 = −4 sec 2 (4𝑡) 𝑑𝑡
⇒
1
sec 2 (4𝑡) 𝑑𝑡 = − 𝑑𝑢
4
1
3
∫ sec 2 (4𝑡) (3 − 𝑡𝑎𝑛(4𝑡)) 𝑑𝑡 = − ∫ 𝑢3 𝑑𝑢
4
1
= − 𝑢4 + 𝑐
16
1
4
= − (3 − 𝑡𝑎𝑛(4𝑡)) + 𝑐
16
𝑢
3
∫ 5𝑦+4 𝑑𝑦
3𝑦
∫ 5𝑦2 +4 𝑑𝑦
3𝑦
∫ (5𝑦+4)2 𝑑𝑦
3
∫ 5𝑦2 +4 𝑑𝑦
𝟑
∫ 𝟓𝒚+𝟒 𝒅𝒚
𝑢 = 5𝑦 + 4
𝑑𝑢 = 5𝑑𝑦
∫
⇒
𝑑𝑦 =
1
𝑑𝑢
5
3
3 1
𝑑𝑦 = ∫ 𝑑𝑢
5𝑦 + 4
5 𝑢
3
= ln 𝑢 + 𝑐
5
3
= ln(5𝑦 + 4) + 𝑐
5
𝟑𝒚
∫ 𝟓𝒚𝟐+𝟒 𝒅𝒚
𝑢 = 5𝑦 2 + 4
∫
𝑑𝑢 = 10𝑦𝑑𝑦
⇒
𝑦𝑑𝑦 =
1
𝑑𝑢
10
3𝑦
3 1
3
3
𝑑𝑦
=
∫
𝑑𝑢
=
ln
𝑢
+
𝑐
=
ln(5𝑦 2 + 4) + 𝑐
5𝑦 2 + 4
10 𝑢
10
10
𝟑𝒚
∫ (𝟓𝒚+𝟒)𝟐 𝒅𝒚
𝑢 = 5𝑦 2 + 4
𝑑𝑢 = 10𝑦𝑑𝑦
∫
⇒
𝑦𝑑𝑦 =
1
𝑑𝑢
10
3𝑦
3
𝑑𝑦 =
∫ 𝑢−2 𝑑𝑢
2
(5𝑦 + 4)
10
3
= − 𝑢−1 + 𝑐
10
3
= − (5𝑦 2 + 4)−1 + 𝑐
10
3
= − (5𝑦 2 + 4) + 𝑐
10
𝟑
∫ 𝟓𝒚𝟐+𝟒 𝒅𝒚
𝑦
𝑦2
𝑦
∫
1
𝑑𝑢 = tan−1 𝑢 + 𝑐
1 + 𝑢2
1+
1+
4+
∫
3
1
𝑑𝑦
=
∫
𝑑𝑦
5
5𝑦 2 + 4
4 (4 𝑦 2 + 1)
3
1
= ∫
𝑑𝑦
4 5 𝑦2 + 1
4
3
1
= ∫
𝑑𝑦
2
4
√5𝑦
( 2 ) +1
𝑢2
5𝑦 2
4
𝑢=
√5𝑦
2
∫
𝑑𝑢 =
√5
𝑑𝑦
2
⇒
𝑑𝑦 =
3
3 2
1
𝑑𝑦 = ( ) ∫ 2
𝑑𝑢
+4
4 √5
𝑢 +1
5𝑦 2
2
√5
𝑑𝑢
∫
3
3
1
3
3
√5𝑦
𝑑𝑦 =
∫ 2
𝑑𝑢 =
tan−1(𝑢) + 𝑐 =
tan−1 (
)+𝑐
+4
2
2√5 𝑢 + 1
2√5
2√5
5𝑦 2
2𝑡 3 +1
∫ (𝑡 4 +2𝑡)3 𝑑𝑡
2𝑡 3 +1
∫ 𝑡 4 +2𝑡 𝑑𝑡
𝑥
∫ √1−4𝑥 2 𝑑𝑥
1
∫ √1−4𝑥 2 𝑑𝑥
𝟐𝒕𝟑 +𝟏
∫ (𝒕𝟒+𝟐𝒕)𝟑 𝒅𝒕
𝑢 = 𝑡 4 + 2𝑡
𝟐𝒕𝟑 +𝟏
∫ 𝒕𝟒+𝟐𝒕 𝒅𝒕
𝑑𝑢 = (4𝑡 3 + 2)𝑑𝑡 = 2(2𝑡 3 + 1)𝑑𝑡
⇒
2𝑡 3 + 1
1 1
∫ 4
𝑑𝑡 = ∫ 3 𝑑𝑢
3
(𝑡 + 2𝑡)
2 𝑢
1
1
1
= ∫ 𝑢−3 𝑑𝑢 = (− ) 𝑢−2 + 𝑐
2
2
2
1 4
= − (𝑡 + 2𝑡)−2 + 𝑐
4
1
(2𝑡 3 + 1)𝑑𝑡 = 𝑑𝑢
2
2𝑡 3 + 1
1 1
1
∫ 4
𝑑𝑡 = ∫ 𝑑𝑢 = ln 𝑢 + 𝑐
𝑡 + 2𝑡
2 𝑢
2
1
= ln(𝑡 4 + 2𝑡) + 𝑐
2
∫
𝒙
√𝟏−𝟒𝒙𝟐
𝒅𝒙
𝑢 = 1 − 4𝑥 2
𝑑𝑢 = −8𝑥𝑑𝑥
∫
∫
𝟏
√𝟏−𝟒𝒙𝟐
1
𝑥𝑑𝑥 = − 𝑑𝑢
8
⇒
𝑥
1
1
𝑑𝑥 = − ∫ 𝑢−2 𝑑𝑢
8
√1 − 4𝑥 2
1
1
1
= − (2)𝑢2 + 𝑐 = − √1 − 4𝑥 2 + 𝑐
8
4
𝒅𝒙
𝑥
∫
1
√1 − 𝑢2
𝑑𝑢 = sin−1 𝑢 + 𝑐
𝑥2
4𝑥 2
∫
1
√1 − 4𝑥 2
𝑢 = 2𝑥
∫
𝑑𝑥 = ∫
𝑑𝑢 = 2𝑑𝑥
1
1
√1 − (2𝑥)2
⇒
𝑑𝑥
1
𝑑𝑥 = 𝑑𝑢
2
1
1
𝑑𝑥 = ∫
𝑑𝑢
2 √1 − 𝑢2
√1 − 4𝑥 2
1
= sin−1 𝑢 + 𝑐
2
1 −1
= sin (2𝑥) + 𝑐
2
∫ 𝑒 2𝑡 + sec(2𝑡) tan(2𝑡) 𝑑𝑡
∫ 𝑠𝑖𝑛(𝑡) (4 cos 3(𝑡) + 6 cos2 (𝑡) − 8)𝑑𝑡
𝑥
∫ 𝑥𝑐𝑜𝑠(𝑥 2 + 1) + 𝑥 2 +1 𝑑𝑥
∫ 𝒆𝟐𝒕 + 𝒔𝒆𝒄(𝟐𝒕)𝒕𝒂𝒏(𝟐𝒕)𝒅𝒕
𝑢 = 2𝑡
𝑑𝑢 = 2𝑑𝑡
⇒
𝑑𝑡 =
1
𝑑𝑢
2
1
∫ 𝑒 𝑢 + sec(𝑢) tan(𝑢) 𝑑𝑢
2
1
1
= (𝑒 𝑢 + sec(𝑢)) + 𝑐 = (𝑒 2𝑡 + sec(2𝑡)) + 𝑐
2
2
∫ 𝑒 2𝑡 + sec(2𝑡) tan(2𝑡) 𝑑𝑡 =
1
1
2
2
∫ 𝒔𝒊𝒏(𝒕)(𝟒 𝐜𝐨𝐬 𝟑 (𝒕) + 𝟔 𝐜𝐨𝐬 𝟐 (𝒕) − 𝟖)𝒅𝒕
𝑢 = 𝑐𝑜𝑠(𝑡)
𝑑𝑢 = −𝑠𝑖𝑛(𝑡)𝑑𝑡
⇒
𝑠𝑖𝑛(𝑡)𝑑𝑡 = −𝑑𝑢
sin ∫ (𝑡) (4 cos3 (𝑡) + 6 cos 2 (𝑡) − 8)𝑑𝑡 = −∫ 4𝑢3 + 6𝑢2 − 8𝑑𝑢
= −(𝑢4 + 2𝑢3 − 8𝑢) + 𝑐
= −(cos4 (𝑡) + 2 cos3 (𝑡) − 8𝑐𝑜𝑠(𝑡)) + 𝑐
𝒙
∫ 𝒙𝒄𝒐𝒔(𝒙𝟐 + 𝟏) + 𝒙𝟐+𝟏 𝒅𝒙
∫ 𝑥𝑐𝑜𝑠(𝑥 2 + 1) +
𝑥
1
2
𝑑𝑥
=
∫
𝑥
(cos(𝑥
+
1)
+
) 𝑑𝑥
𝑥2 + 1
𝑥2 + 1
𝑥
𝑢 = 𝑥2 + 1
𝑑𝑢 = 2𝑥𝑑𝑥
∫ 𝑥𝑐𝑜𝑠(𝑥 2 + 1) +
=
=
∫ 𝑥 2 + 𝑒 1−𝑥 𝑑𝑥
∫ 𝑥𝑐𝑜𝑠(𝑥 2 + 1) +
𝑥2
⇒
1
𝑥𝑑𝑥 = 𝑑𝑢
2
𝑥
1
1
𝑑𝑥 = cos ∫ (𝑢) + 𝑑𝑢
+1 2
𝑢
1
(sin(𝑢) + ln 𝑢) + 𝑐
2
1
(𝑠𝑖𝑛(𝑥 2 + 1) + ln(𝑥 2 + 1)) + 𝑐
2
1
𝑑𝑥
𝑥 2 +1
∫ 𝒙𝟐 + 𝒆𝟏−𝒙 𝒅𝒙
∫ 𝑥 2 + 𝑒 1−𝑥 𝑑𝑥 = ∫ 𝑥 2 𝑑𝑥 + ∫ 𝑒 1−𝑥 𝑑𝑥
𝑢 =1−𝑥
𝑑𝑢 = −𝑑𝑥
⇒
𝑑𝑥 = −𝑑𝑢
1
∫ 𝑥 2 + 𝑒 1−𝑥 𝑑𝑥 = ∫ 𝑥 2 𝑑𝑥 − ∫ 𝑒 𝑢 𝑑𝑢 = 𝑥 3 − 𝑒 𝑢 + 𝑐
3
1 3
= 𝑥 − 𝑒 1−𝑥 + 𝑐
3
∫ 𝑥 2 + 𝑒 1−𝑥 𝑑𝑥 = −∫ 𝑥 2 + 𝑒 𝑢 𝑑𝑢
−
𝑥
𝑑𝑢
∫ 𝒙𝒄𝒐𝒔(𝒙𝟐 + 𝟏) +
𝟏
𝒅𝒙
𝒙𝟐 +𝟏
𝑥
𝑢 = 𝑥2 + 1
∫ 𝑥𝑐𝑜𝑠(𝑥 2 + 1) +
𝑑𝑢 = 2𝑥𝑑𝑥
𝑥2
∫ 𝑠𝑖𝑛𝑤√1 − 2𝑐𝑜𝑠𝑤 +
∫
10𝑥+3
𝑑𝑥
𝑥 2 +16
1
𝑥𝑑𝑥 = 𝑑𝑢
2
1
1
𝑑𝑥 = ∫ 𝑥𝑐𝑜𝑠(𝑥 2 + 1) 𝑑𝑥 + ∫ 2
𝑑𝑥
+1
𝑥 +1
1
1
= ∫ cos(𝑢) 𝑑𝑢 + ∫ 2
𝑑𝑥
2
𝑥 +1
1
= sin(𝑢) + tan−1(𝑥) + 𝑐
2
1
= 𝑠𝑖𝑛(𝑥 2 + 1) + tan−1(𝑥) + 𝑐
2
𝑧
∫ 𝑒 −𝑧 + sec 2 (10) 𝑑𝑧
⇒
1
𝑑𝑤
7𝑤+2
𝒛
∫ 𝒆−𝒛 + 𝐬𝐞𝐜 𝟐 (𝟏𝟎) 𝒅𝒛
𝑧
𝑧
∫ 𝑒 −𝑧 + sec 2 ( ) 𝑑𝑧 = ∫ 𝑒 −𝑧 𝑑𝑧 + ∫ sec 2 ( ) 𝑑𝑧
10
10
𝑢 = −𝑧
𝑧
𝑣=
10
𝑑𝑢 = −𝑑𝑧
1
𝑑𝑣 =
𝑑𝑧
10
∫ 𝑒 −𝑧 + sec 2 (
𝟏
⇒
𝑑𝑧 = −𝑑𝑢
⇒
𝑑𝑧 = 10𝑑𝑣
𝑧
) 𝑑𝑧 = −∫ 𝑒 𝑢 𝑑𝑢 + 10∫ sec 2(𝑣) 𝑑𝑣
10
= −𝑒 𝑢 + 10 tan(𝑣) + 𝑐
𝑧
= −𝑒 −𝑧 + 10𝑡𝑎𝑛 ( ) + 𝑐
10
∫ 𝒔𝒊𝒏𝒘√𝟏 − 𝟐𝒄𝒐𝒔𝒘 + 𝟕𝒘+𝟐 𝒅𝒘
∫ 𝑠𝑖𝑛𝑤√1 − 2𝑐𝑜𝑠𝑤 +
1
1
1
𝑑𝑤 = ∫ 𝑠𝑖𝑛𝑤(1 − 2𝑐𝑜𝑠𝑤)2 𝑑𝑤 + ∫
𝑑𝑤
7𝑤 + 2
7𝑤 + 2
𝑢 = 1 − 2𝑐𝑜𝑠(𝑤)
𝑣 = 7𝑤 + 2
𝑑𝑢 = 2𝑠𝑖𝑛(𝑤)𝑑𝑤
𝑑𝑣 = 7𝑑𝑤
∫ 𝑠𝑖𝑛𝑤√1 − 2𝑐𝑜𝑠𝑤 +
∫
𝟏𝟎𝒙+𝟑
𝒅𝒙
𝒙𝟐 +𝟏𝟔
⇒
1
𝑠𝑖𝑛(𝑤)𝑑𝑤 = 𝑑𝑢
2
1
𝑑𝑤 = 𝑑𝑣
7
⇒
1
1
1
1 1
𝑑𝑤 = ∫ 𝑢2 𝑑𝑢 + ∫ 𝑑𝑣
7𝑤 + 2
2
7 𝑣
1 2 3 1
= ( ) 𝑢2 + ln 𝑣 + 𝑐
2 3
7
3
1
1
= (1 − 2𝑐𝑜𝑠𝑤)2 + ln(7𝑤 + 2) + 𝑐
3
7
∫
10𝑥 + 3
10𝑥
3
10𝑥
1
3
𝑑𝑥 = ∫ 2
𝑑𝑥 + ∫ 2
𝑑𝑥 = ∫ 2
𝑑𝑥 + ∫ 2
𝑑𝑥
2
𝑥 + 16
𝑥 + 16
𝑥 + 16
𝑥 + 16
16 𝑥
+1
16
𝑢 = 𝑥 2 + 16
𝑑𝑢 = 2𝑥𝑑𝑥
𝑥
1
𝑣=
𝑑𝑣 = 𝑑𝑥
⇒
4
4
∫
⇒
𝑥𝑑𝑥 =
𝑑𝑥 = 4𝑑𝑣
10𝑥 + 3
1
3
1
𝑑𝑥 = 5 ∫ 𝑑𝑢 + ∫ 2
𝑑𝑣
2
𝑥 + 16
𝑢
4 𝑣 +1
3
= 5 ln 𝑢 + tan−1(𝑣) + 𝑐
4
3
𝑥
= 5 ln(𝑥 2 + 16) + tan−1 ( ) + 𝑐
4
4

𝑥
𝑥


1
𝑑𝑢
2
𝑑𝑥
𝑢
𝑓(𝑥)
[𝑎, 𝑏]
𝑛
𝑥𝑖∗
𝛥𝑥
𝑓(𝑥)
𝒏
𝒃
∫ 𝒇(𝒙) 𝒅𝒙 = 𝐥𝐢𝐦 ∑ 𝒇(𝒙∗𝒊 )𝜟𝒙
𝒏→∞
𝒂
𝒊=𝟏
𝑎
𝑏
𝑎
𝑏
𝑎 𝑎𝑛𝑑 𝑏
𝒃
𝒂
∫𝒂 𝒇(𝒙) 𝒅𝒙 = − ∫𝒃 𝒇(𝒙) 𝒅𝒙
𝒂
∫𝒂 𝒇(𝒙) 𝒅𝒙 = 𝟎
𝒃
𝒃
∫𝒂 𝒄𝒇(𝒙) 𝒅𝒙 = 𝒄 ∫𝒂 𝒇(𝒙) 𝒅𝒙
𝒃
𝒘𝒉𝒆𝒓𝒆 𝒄 𝒊𝒔 𝒂𝒏𝒚 𝒏𝒖𝒎𝒃𝒆𝒓
𝒃
𝒃
∫𝒂 𝒇(𝒙) ± 𝒈(𝒙) 𝒅𝒙 = ∫𝒂 𝒇(𝒙) 𝒅𝒙 ± ∫𝒂 𝒈(𝒙) 𝒅𝒙
𝒃
𝒄
𝒃
∫𝒂 𝒇(𝒙) 𝒅𝒙 = ∫𝒂 𝒇(𝒙) 𝒅𝒙 + ∫𝒄 𝒇(𝒙) 𝒅𝒙
𝒘𝒉𝒆𝒓𝒆 𝒄 𝒊𝒔 𝒂𝒏𝒚 𝒏𝒖𝒎𝒃𝒆𝒓
[𝑎, 𝑐]
𝑐
𝒃
∫𝒂 𝒇(𝒙) 𝒅𝒙
𝒃
=
𝑎
𝒃
∫𝒂 𝒇(𝒕) 𝒅𝒕
7. ∫𝒂 𝒄 𝒅𝒙 = 𝒄(𝒃 − 𝒂),
𝑐 𝑖𝑠 𝑎𝑛𝑦 𝑛𝑢𝑚𝑏𝑒𝑟
𝒃
8. 𝐼𝑓 𝑓(𝑥) ≥ 0 𝑓𝑜𝑟 𝑎 ≤ 𝑥 ≤ 𝑏 𝑡ℎ𝑒𝑛 ∫𝒂 𝒇(𝒙) 𝒅𝒙 ≥ 𝟎
𝑏
[𝑐, 𝑏]
𝒃
𝒃
9. 𝐼𝑓 𝑓(𝑥) ≥ 𝑔(𝑥) 𝑓𝑜𝑟 𝑎 ≤ 𝑥 ≤ 𝑏 𝑡ℎ𝑒𝑛 ∫𝒂 𝒇(𝒙) 𝒅𝒙 ≥ ∫𝒂 𝒈(𝒙) 𝒅𝒙
𝒃
10. 𝐼𝑓 𝑚 ≤ 𝑓(𝑥) ≤ 𝑀 𝑓𝑜𝑟 𝑎 ≤ 𝑥 ≤ 𝑏 𝑡ℎ𝑒𝑛 𝒎(𝒃 − 𝒂) ≤ ∫𝒂 𝒇(𝒙) 𝒅𝒙 ≤ 𝑴(𝒃 − 𝒂)
𝒃
𝒃
11. |∫𝒂 𝒇(𝒙) 𝒅𝒙| ≤ ∫𝒂 |𝒇(𝒙)| 𝒅𝒙

𝑓(𝑥)
𝑥
𝑥
𝑓(𝑥) = 𝑥 2 + 1
[𝑎, 𝑏]
[0,2]
2
∫ 𝑥 2 + 1 𝑑𝑥 =
0

𝑓(𝑥)
14
3
𝑓′(𝑥)
𝑓(𝑥)
𝑏
∫ 𝑓 ′ (𝑥) 𝑑𝑥 = 𝑓(𝑏) − 𝑓(𝑎)
𝑎
𝑓(𝑥)
[𝑎, 𝑏]
𝑓(𝑏) − 𝑓(𝑎)
𝑓(𝑥)
𝑉(𝑡)
𝑡2
∫ 𝑉′(𝑡) 𝑑𝑡 = 𝑉(𝑡2 ) − 𝑉(𝑡1 )
𝑡1
𝑡1
𝑓(𝑥)
[𝑎, 𝑏]
𝑥
𝑔(𝑥) = ∫ 𝑓(𝑡) 𝑑𝑡
𝑎
𝑡2
[𝑎, 𝑏]
𝑔
′ (𝑥)
(𝑎, 𝑏)
= 𝑓(𝑥)
𝑑 𝑥
∫ 𝑓(𝑡) 𝑑𝑡 = 𝑓(𝑥)
𝑑𝑥 𝑎
𝑥
𝑔(𝑥) = ∫−4 𝑒 2𝑡 𝑐𝑜𝑠 2 (1 − 5𝑡) 𝑑𝑡
1 𝑡 4 +1
∫𝑥 2 𝑡 2 +1 𝑑𝑡
𝒙
𝒈(𝒙) = ∫−𝟒 𝒆𝟐𝒕 𝒄𝒐𝒔𝟐 (𝟏 − 𝟓𝒕) 𝒅𝒕
𝑔′(𝑥) = 𝑒 2𝑥 𝑐𝑜𝑠 2 (1 − 5𝑥)
𝟏
𝒕𝟒 +𝟏
∫𝒙𝟐 𝒕𝟐 +𝟏 𝒅𝒕
𝑑
1 𝑡 4 +1
𝑑
𝑥 2 𝑡 4 +1
𝑑𝑡 = 𝑑𝑥 (− ∫1
∫
𝑑𝑥 𝑥 2 𝑡 2 +1
𝑑
𝑥 2 𝑡 4 +1
𝑑𝑡) = − 𝑑𝑥 ∫1
𝑡 2 +1
𝑡 2 +1
𝑑𝑡
𝑥
𝑥
𝑢 = 𝑥2
2
𝑑
𝑑
𝑑𝑢
(𝑔(𝑢)) =
(𝑔(𝑢)) ×
𝑑𝑥
𝑑𝑢
𝑑𝑥
𝑤ℎ𝑒𝑟𝑒 𝑢 = 𝑓(𝑥)
2
𝑑 1 𝑡4 + 1
𝑑 𝑥 𝑡4 + 1
∫
𝑑𝑡 = − ∫ 2
𝑑𝑡
𝑑𝑥 𝑥 2 𝑡 2 + 1
𝑑𝑥 1 𝑡 + 1
𝑑 𝑢 𝑡4 + 1
𝑑𝑢
=−
∫ 2
𝑑𝑢
𝑑𝑢 1 𝑡 + 1
𝑑𝑥
𝑢4 + 1
(2𝑥)
=− 2
𝑢 +1
𝑢4 + 1
= −2𝑥 2
𝑢 +1
𝑥
𝑤ℎ𝑒𝑟𝑒 𝑢 = 𝑥 2
(𝑥 2 )4 + 1
𝑑 1 𝑡4 + 1
∫
𝑑𝑡 = −2𝑥 2 2
(𝑥 ) + 1
𝑑𝑥 𝑥 2 𝑡 2 + 1
𝑥8 + 1
= −2𝑥 4
𝑥 +1

𝑑 𝑢(𝑥)
∫ 𝑓(𝑡) 𝑑𝑡 = 𝑢′ (𝑥)𝑓(𝑢(𝑥))
𝑑𝑥 𝑎

𝑥
𝑑 𝑏
∫ 𝑓(𝑡) 𝑑𝑡 = −𝑣′(𝑥)𝑓(𝑣(𝑥))
𝑑𝑥 𝑣(𝑥)

𝑥
𝑢(𝑥)
∫
𝑎
𝑓(𝑡) 𝑑𝑡 = ∫
𝑣(𝑥)
∴
𝑢(𝑥)
𝑓(𝑡) 𝑑𝑡 + ∫
𝑣(𝑥)
𝑓(𝑡) 𝑑𝑡
𝑎
𝑎
𝑢(𝑥)
𝑑 𝑢(𝑥)
𝑑
∫
𝑓(𝑡) 𝑑𝑡 =
(∫ 𝑓(𝑡) 𝑑𝑡 + ∫ 𝑓(𝑡) 𝑑𝑡)
𝑑𝑥 𝑣(𝑥)
𝑑𝑥 𝑣(𝑥)
𝑎
= −𝑣 ′ (𝑥)𝑓(𝑣(𝑥)) + 𝑢′ (𝑥)𝑓(𝑢(𝑥))
3𝑥
∫ 𝑡 2 𝑠𝑖𝑛(1 + 𝑡 2 ) 𝑑𝑡
√𝑥
𝑑 3𝑥 2
1 1
2
2
∫ 𝑡 𝑠𝑖𝑛(1 + 𝑡 2 ) 𝑑𝑡 = − 𝑥 −2 (√𝑥) 𝑠𝑖𝑛 (1 + (√𝑥) ) + (3)(3𝑥)2 𝑠𝑖𝑛(1 + (3𝑥)2 )
𝑑𝑥 √𝑥
2
1
= − √𝑥𝑠𝑖𝑛(1 + 𝑥) + 27𝑥 2 𝑠𝑖𝑛(1 + 9𝑥 2 )
2
[𝑎, 𝑏]
𝐹(𝑥)
𝑓(𝑥)
𝑏
∫ 𝑓(𝑥) 𝑑𝑥 = 𝐹(𝑥)𝑏𝑎 = 𝐹(𝑏) − 𝐹(𝑎)
𝑎
2
∫ 𝑥 2 + 1 𝑑𝑥
0
𝐹(𝑥) =
1 3
𝑥 +𝑥
3
𝑎𝑛𝑑
𝐹(𝑥) =
1 3
18
𝑥 +𝑥−
3
31
2
2
1
∫ 𝑥 2 + 1 𝑑𝑥 = ( 𝑥 3 + 𝑥)
3
0
0
1
1
= (2)3 + 2 − ( (0)3 + 0)
3
3
14
=
3
2
1
18 2
∫ 𝑥 2 + 1 𝑑𝑥 = ( 𝑥 3 + 𝑥 − )
3
31 0
0
1
18
1
18
= ( (2)3 + 2 − ) − ( (0)3 + 0 − )
3
31
3
31
14 18 18
=
−
+
3 31 31
14
=
3
∫ 𝑦 2 + 𝑦 −2 𝑑𝑦
2
∫1 𝑦 2 + 𝑦 −2 𝑑𝑦
2
∫−1 𝑦 2 + 𝑦 −2 𝑑𝑦
a. ∫ 𝒚𝟐 + 𝒚−𝟐 𝒅𝒚
1
∫ 𝑦 2 + 𝑦 −2 𝑑𝑦 = 𝑦 3 − 𝑦 −1 + 𝑐
3
𝟐
b. ∫𝟏 𝒚𝟐 + 𝒚−𝟐 𝒅𝒚
2
1
1 2
∫ 𝑦 2 + 𝑦 −2 𝑑𝑦 = ( 𝑦 3 − )
3
𝑦 1
1
1
1
1
1
= (2)3 − − ( (1)3 − )
3
2
3
1
8 1 1
= − − +1
3 2 3
17
=
6
𝟐
c. ∫−𝟏 𝒚𝟐 + 𝒚−𝟐 𝒅𝒚
𝑦=0
𝑦=0
𝑦=0




1
∫−3 6𝑥 2 − 5𝑥 + 2 𝑑𝑥
4
∫0 √𝑡(𝑡 − 2) 𝑑𝑡
2 2w5 −w+3
∫1
𝑤2
−10
∫25 𝑑𝑅
𝑑𝑤
𝟏
∫−𝟑 𝟔𝒙𝟐 − 𝟓𝒙 + 𝟐 𝒅𝒙
1
1
5
∫ 6𝑥 2 − 5𝑥 + 2 𝑑𝑥 = (2𝑥 3 − 𝑥 2 + 2𝑥)
2
−3
−3
5
45
= (2 − + 2) − (−54 −
− 6)
2
2
= 84
𝟒
∫𝟎 √𝒕(𝒕 − 𝟐) 𝒅𝒕
4
4 3
1
∫ √𝑡(𝑡 − 2) 𝑑𝑡 = ∫ 𝑡 2 − 2𝑡 2 𝑑𝑡
0
0
2 5 4 3 4
= ( 𝑡 2 − 𝑡 2)
5
3
0
5
3
2
4
= 0 − ( (4)2 − (4)2 )
5
3
=−
32
15
1 5
5
(4)2 = ((4)2 ) = (2)5 = 32
1 3
3
(4)2 = ((4)2 ) = (2)3 = 8
𝟐 𝟐𝐰 𝟓 −𝐰+𝟑
∫𝟏
𝒘𝟐
𝒅𝒘
2
2
2w 5 − w + 3
1
3
∫
𝑑𝑤
=
∫
2𝑤
−
+ 3𝑤 −2 𝑑𝑤
2
𝑤
𝑤
1
1
2
1
= ( 𝑤 4 − ln 𝑤 − 3𝑤)
2
1
3
1
= (8 − 𝑙𝑛2 − ) − ( − 𝑙𝑛1 − 3)
2
2
= 9 − 𝑙𝑛2
𝑙𝑛(1) = 0
−𝟏𝟎
∫𝟐𝟓 𝒅𝑹
−10
∫
−10
𝑑𝑅 = 𝑅25
25
= −10 − 25
= −35
1
3
∫0 4𝑥 − 6√𝑥 2 𝑑𝑥
𝜋
b. ∫03 2𝑠𝑖𝑛𝜃 − 5𝑐𝑜𝑠𝜃𝑑𝜃
𝜋
c. ∫𝜋4 5 − 2𝑠𝑒𝑐𝑧𝑡𝑎𝑛𝑧𝑑𝑧
6
−1
3
1
∫−20 𝑒 −𝑧 − 3𝑧 𝑑𝑧
3
1
∫−2 5𝑡 6 − 10𝑡 + 𝑡 𝑑𝑡
𝟏
𝟑
∫𝟎 𝟒𝒙 − 𝟔√𝒙𝟐 𝒅𝒙
1
1
3
3
∫ 4𝑥 − 6 √𝑥 2 𝑑𝑥 = ∫ 4𝑥 − 6𝑥 2 𝑑𝑥
0
0
= (2𝑥 2 −
=2−
=−
𝝅
8
5
18 5 1
𝑥3)
5
0
18
− (0)
5
b. ∫𝟎𝟑 𝟐𝒔𝒊𝒏𝜽 − 𝟓𝒄𝒐𝒔𝜽𝒅𝜽
𝜋
3
𝜋
∫ 2𝑠𝑖𝑛𝜃 − 5𝑐𝑜𝑠𝜃𝑑𝜃 = (−2𝑐𝑜𝑠𝜃 − 5𝑠𝑖𝑛𝜃)03
0
𝜋
3
𝜋
3
= −2𝑐𝑜𝑠 ( ) − 5𝑠𝑖𝑛 ( ) − (−2𝑐𝑜𝑠0 − 5𝑠𝑖𝑛0)
5√3
+2
2
5√3
=1−
2
= −1 −
𝝅
c. ∫𝝅𝟒 𝟓 − 𝟐𝒔𝒆𝒄𝒛𝒕𝒂𝒏𝒛𝒅𝒛
𝟔
𝜋
4
𝜋
𝜋
𝜋
𝜋
𝜋
∫ 5 − 2𝑠𝑒𝑐𝑧𝑡𝑎𝑛𝑧𝑑𝑧 = (5𝑧 − 2𝑠𝑒𝑐𝑧)𝜋4 = 5 ( ) − 2𝑠𝑒𝑐 ( ) − (5 ( ) − 2𝑠𝑒𝑐 ( ))
𝜋
4
4
6
6
6
6
=
5𝜋
4
− 2√2 +
12
√3
sec 𝑧 =
1
cos 𝑧
−𝟏
𝟑
𝟏
∫−𝟐𝟎 𝒆−𝒛 − 𝟑𝒛 𝒅𝒛
−1
−1
3
1
11
−
𝑑𝑧
=
∫
3𝑒 𝑧 −
𝑑𝑧
−𝑧
3𝑧
3𝑧
−20 𝑒
−20
∫
𝑥 −𝑎 =
1
𝑥𝑎
1
= 𝑥𝑎
𝑥 −𝑎
𝑎𝑛𝑑
−1
11
1
𝑧
∫ 3𝑒 −
𝑑𝑧 = (3𝑒 − ln 𝑧)
3𝑧
3
−20
−20
1
1
= 3𝑒 −1 − ln(−1) − (3𝑒 −20 − ln(−20))
3
3
1
= 3𝑒 −1 − 3𝑒 −20 + ln|20|
3
−1
𝑧
𝟑
𝟏
∫−𝟐 𝟓𝒕𝟔 − 𝟏𝟎𝒕 + 𝒕 𝒅𝒕
𝑡=0
𝑡=0
𝑓(−𝑥) = 𝑓(𝑥)
𝑓(𝑥) = 𝑥 2
𝑓(𝑥) = 𝑐𝑜𝑠(𝑥)
𝑓(−𝑥) = −𝑓(𝑥)
𝑓(𝑥) = 𝑥 3
𝑓(𝑥) = 𝑠𝑖𝑛(𝑥)
[−𝑎, 𝑎]
𝑎
𝑎
∫ 𝑓(𝑥) 𝑑𝑥 = 2 ∫ 𝑓(𝑥) 𝑑𝑥
−𝑎
0
𝑎
∫ 𝑓(𝑥) 𝑑𝑥 = 0
−𝑎
2
∫−2 4𝑥 4 − 𝑥 2 + 1 𝑑𝑥
10
∫−10 𝑥 5 + 𝑠𝑖𝑛(𝑥) 𝑑𝑥
𝟐
∫−𝟐 𝟒𝒙𝟒 − 𝒙𝟐 + 𝟏 𝒅𝒙
2
2
∫ 4𝑥 4 − 𝑥 2 + 1 𝑑𝑥 = 2 ∫ 4𝑥 4 − 𝑥 2 + 1 𝑑𝑥
−2
0
2
4
1
= 2 ( 𝑥 5 − 𝑥 3 + 𝑥)
5
3
0
748
=
15
𝟏𝟎
∫−𝟏𝟎 𝒙𝟓 + 𝒔𝒊𝒏(𝒙) 𝒅𝒙
10
∫ 𝑥 5 + 𝑠𝑖𝑛(𝑥) 𝑑𝑥 = 0
−10
0
∫ 2𝑡 2 √1 − 4𝑡 3 𝑑𝑡
−2
𝑢 = 1 − 4𝑡 3
𝑑𝑢 = −12𝑡 2 𝑑𝑡
0
∫ 2𝑡
−2
2√
1−
4𝑡 3 𝑑𝑡
⇒
𝑡 2 𝑑𝑡 = −
1 0 1
1 3 0
2
= − ∫ 𝑢 𝑑𝑢 = (− 𝑢2 )
6 −2
9
−2
0
3 0
1
∫ 2𝑡 2 √1 − 4𝑡 3 𝑑𝑡 = (− (1 − 4𝑡 3 )2 )
9
−2
−2
3
1
1
= − − (− (33)2 )
9
9
1
= (33√33 − 1)
9
1
𝑑𝑢
12
𝑢 = 1 − 4𝑡 3
𝑑𝑢 = −12𝑡 2 𝑑𝑡
𝑡 = −2
𝑡=0
0
∫ 2𝑡
1−
4𝑡 3 𝑑𝑡
−2
1 0 1
= − ∫ 𝑢2 𝑑𝑢
6 −2
1 3 1
= (− 𝑢2 )
9
33
0
∫ 2𝑡
−2
5
∫−1(1 + 𝑤)(2𝑤 + 𝑤 2 )5 𝑑𝑤
−6
∫−2
4
(1+2𝑥)3
5
− 1+2𝑥 𝑑𝑥
1
∫02 𝑒 𝑦 + 2𝑐𝑜𝑠(𝜋𝑦)𝑑𝑦
0
𝑧
∫𝜋/3 3𝑠𝑖𝑛 (2) − 5𝑐𝑜𝑠(𝜋 − 𝑧)𝑑𝑧
𝟓
𝟓
∫−𝟏(𝟏 + 𝒘)(𝟐𝒘 + 𝒘𝟐 ) 𝒅𝒘
2√
𝑡 2 𝑑𝑡 = −
𝑢 = 1 − 4(−2)3 = 33
𝑢 = 1 − 4(0)3 = 1
⇒
⇒
2√
⇒
1−
4𝑡 3 𝑑𝑡
1 3 1
= (− 𝑢2 )
9
33
3
1
1
= − − (− (33)2 )
9
9
1
= (33√33 − 1)
9
1
𝑑𝑢
12
𝑢 = 2𝑤 + 𝑤 2
𝑑𝑢 = (2 + 2𝑤)𝑑𝑤
𝑤 = −1
𝑤=5
(1 + 𝑤)𝑑𝑤 =
⇒
⇒
⇒
𝑢 = −1
𝑢 = 35
1
𝑑𝑢
2
35
5
1 35
1
∫ (1 + 𝑤)(2𝑤 + 𝑤 2 )5 𝑑𝑤 = ∫ 𝑢5 𝑑𝑢 = ( 𝑢6 ) = 153188802
2 −1
12
−1
−1
−𝟔
∫−𝟐
𝟒
(𝟏+𝟐𝒙)𝟑
𝟓
− 𝟏+𝟐𝒙 𝒅𝒙
𝑢 = 1 + 2𝑥
𝑑𝑢 = 2𝑑𝑥
𝑥 = −2 ⇒
𝑥 = −6 ⇒
−6
∫
−2
𝟏
⇒
𝑑𝑥 =
𝑢 = −3
𝑢 = −11
1
𝑑𝑢
2
4
5
1 −11 −3 5
−
𝑑𝑥
=
∫
4𝑢 − 𝑑𝑢
(1 + 2𝑥)3 1 + 2𝑥
2 −3
𝑢
1
= (−2𝑢−2 − 5 ln(𝑢)) −11
−3
2
1
2
1
2
= (−
− 5𝑙𝑛11) − (− − 5𝑙𝑛3)
2
121
2
9
5
5
= 1089 − 𝑙𝑛11 + 𝑙𝑛3
2
2
∫𝟎𝟐 𝒆𝒚 + 𝟐𝒄𝒐𝒔(𝝅𝒚)𝒅𝒚
1
2
1
2
1
2
∫ 𝑒 𝑦 + 2 cos(𝜋𝑦) 𝑑𝑦 = ∫ 𝑒 𝑦 𝑑𝑦 + ∫ 2 cos(𝜋𝑦) 𝑑𝑦
0
𝑢 = 𝜋𝑦
0
𝑑𝑢 = 𝜋𝑑𝑦
𝑦=0
1
𝑦=
2
1
2
⇒
⇒
0
𝑑𝑦 =
𝑢=0
𝜋
𝑢=
2
⇒
1
2
𝜋
1
𝑑𝑢
𝜋
2 2
∫ 𝑒 𝑦 + 2 cos(𝜋𝑦) 𝑑𝑦 = ∫ 𝑒 𝑦 𝑑𝑦 + ∫ cos(𝑢) 𝑑𝑢
𝜋 0
0
0
𝜋
2
2
=
+ ( sin 𝑢)
𝜋
0
1
2
𝜋
2
= (𝑒 2 − 𝑒 0 ) + 𝑠in ( ) − sin(0)
𝜋
2
𝜋
(𝑒 𝑦 )1/2
0
1
= 𝑒2 − 1 +
𝟎
𝒛
2
𝜋
∫𝝅/𝟑 𝟑𝒔𝒊𝒏 (𝟐) − 𝟓𝒄𝒐𝒔(𝝅 − 𝒛)𝒅𝒛
0
∫
𝜋/3
0
0
𝑧
𝑧
3𝑠𝑖𝑛 ( ) − 5𝑐𝑜𝑠(𝜋 − 𝑧)𝑑𝑧 = ∫ 3𝑠𝑖𝑛 ( ) 𝑑𝑧 − ∫ 5𝑐𝑜𝑠(𝜋 − 𝑧)𝑑𝑧
2
2
𝜋/3
𝜋/3
1
𝑑𝑢 = 𝑑𝑧 ⇒
𝑑𝑧 = 2𝑑𝑢
2
𝜋
𝜋
𝑧=
⇒ 𝑢=
3
6
𝑧=0
⇒
𝑢=0
𝑣 = 𝜋 − 𝑧 𝑑𝑣 = −𝑑𝑧
⇒
𝑑𝑧 = −𝑑𝑣
𝜋
2𝜋
𝑧=
⇒
𝑣=
3
3
𝑧=0 ⇒
𝑣=𝜋
𝑢=
𝑧
2
0
0
𝜋
𝑧
∫ 3𝑠𝑖𝑛 ( ) − 5cos(𝜋 − 𝑧)𝑑𝑧 = 6 ∫ 𝑠𝑖𝑛(𝑢)𝑑𝑢 − 5 ∫ cos(𝑣) 𝑑𝑣
𝜋
𝜋
2𝜋
2
3
6
3
= −6 cos(𝑢)0𝜋 + 5 sin(𝑣)𝜋2𝜋
6
3
5√3
= 3√3 − 6 + (−
)
2
=
√3
−6
2
∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢
𝑢
𝑑𝑣
𝑣
𝑑𝑣
𝑣
𝑑𝑣
𝑣 = ∫ 𝑑𝑣
𝑢
𝑑𝑢
𝑢
𝑑𝑣
∫ 𝑥𝑒 6𝑥 𝑑𝑥
𝑥
𝑢=𝑥
𝑥
𝑢
𝑢
𝑑𝑣
𝑑𝑣
𝑑𝑢
𝑣
𝑢=𝑥
𝑑𝑣 = 𝑒 6𝑥 𝑑𝑥
1
𝑣 = ∫ 𝑒 6𝑥 𝑑𝑥 = 𝑒 6𝑥
6
𝑑𝑢 = 𝑑𝑥
𝑥 6𝑥
1
𝑒 − ∫ 𝑒 6𝑥 𝑑𝑥
6
6
𝑥 6𝑥
1 6𝑥
= 𝑒 −
𝑒 +𝑐
6
36
∫ 𝑥𝑒 6𝑥 𝑑𝑥 =
𝑢
𝑏
𝑑𝑣
𝑏
∫ 𝑢𝑑𝑣 = (𝑢𝑣)𝑏𝑎 − ∫ 𝑣𝑑𝑢
(𝑢𝑣)𝑏𝑎
𝑎
𝑎
𝑢𝑣
𝑎
𝑏
2
∫ 𝑥𝑒 6𝑥 𝑑𝑥
−1
𝑢
2
2
𝑥
1 2
∫ 𝑥𝑒 6𝑥 𝑑𝑥 = ( 𝑒 6𝑥 ) − ∫ 𝑒 6𝑥 𝑑𝑥
6
6 −1
−1
−1
𝑥 6𝑥 2
1 6𝑥 2
=( 𝑒 ) −( 𝑒 )
6
36
−1
−1
11 12
7 −6
=
𝑒 + 𝑒
36
36
∫ 𝑥𝑒 6𝑥 𝑑𝑥 =
𝑥 6𝑥
1
𝑒 − 𝑒 6𝑥 + 𝑐
6
36
𝑥 6𝑥
1 6𝑥 2
∫ 𝑥𝑒 𝑑𝑥 = ( 𝑒 − 𝑒 )
6
36
−1
−1
1 12
1 12
1
1
=( 𝑒 −
𝑒 ) − (− 𝑒 −6 − 𝑒 −6 )
3
36
6
36
11 12
7 −6
=
𝑒 + 𝑒
36
36
2
6𝑥
𝑑𝑣



𝑑𝑦
𝑑𝑥
= 5𝑥 + 3
𝐹 = 𝑚𝑎
𝑎=
𝑑𝑣
𝑑𝑡
𝑂𝑅 𝑎 =
𝑑2 𝑢
𝑑𝑡 2
𝑣
𝑢
𝑡
𝐹
𝑑𝑣
𝑚 𝑑𝑡 = 𝐹(𝑡, 𝑣)
𝑑2 𝑢
𝑑𝑢
𝑚 𝑑𝑡 2 = 𝐹(𝑡, 𝑢, 𝑑𝑡 )
𝑎𝑦 ′′ + 𝑏𝑦 ′ + 𝑐𝑦 = 𝑔(𝑡)
sin(𝑦)
𝑑2 𝑦
𝑑𝑥 2
= (1 − 𝑦)
𝑑𝑦
𝑑𝑥
+ 𝑦 2 𝑒 −5𝑦
𝑦 (4) + 10𝑦 ′′′ − 4𝑦 ′ + 2𝑦 = cos(𝑡)
𝜕2 𝑢
𝜕𝑢
𝛼 2 𝜕𝑥 2 = 𝜕𝑡
𝑎2 𝑢𝑥𝑥 = 𝑢𝑡𝑡
𝜕3 𝑢
𝜕𝑥 2 𝜕𝑡
𝜕𝑢
= 1 + 𝜕𝑡
𝑢
𝑥 𝑎𝑛𝑑 𝑡
𝑦
𝑥
1
𝑦(𝑥) = 𝑥 −2
3
1
𝑦(𝑥) = −9𝑥 −2 + 7𝑥 −2
𝑦(𝑥0 ) = 𝑦0
4𝑥 2 𝑦 ′′ + 12𝑥𝑦 ′ + 3𝑦 = 0;
𝑦 (𝑘) (𝑥0 ) = 𝑦𝑘
1
3
𝑦(4) = , 𝑦 ′ (4) = −
8
64
𝑦 = 𝑦(𝑥)
𝑦
𝑦(𝑥)
′
𝑦 = 𝑓( 𝑥, 𝑦)
𝑦
′
𝑦′
𝑓(𝑥, 𝑦)
𝑀(𝑥, 𝑦)
𝑑𝑦
−𝑁(𝑥, 𝑦)
𝑑𝑥
𝑀(𝑥,𝑦)
= −𝑁(𝑥,𝑦),
𝑀(𝑥, 𝑦)𝑑𝑥 + 𝑁(𝑥, 𝑦)𝑑𝑦 = 0
𝑓(𝑥, 𝑦)
−𝑝(𝑥)𝑦 + 𝑞(𝑥)
𝑥
𝑦
𝑥
𝑓(𝑥, 𝑦) + 𝑝(𝑥)𝑦 = 𝑞(𝑥)
𝑦
𝒚
𝑓(𝑡𝑥, 𝑡𝑦) = 𝑓(𝑥, 𝑦)
𝑀(𝑥, 𝑦) = 𝐴(𝑥)
𝑥
𝑁(𝑥, 𝑦) = 𝐵(𝑦)
𝑦
𝜕𝑀(𝑥,𝑦)
𝜕𝑥
=
𝜕𝑁(𝑥,𝑦)
𝜕𝑦
𝑓(𝑥, 𝑦) =






𝑑𝑦
𝑑𝑥
= 𝑓(𝑥, 𝑦)
𝑦 = 𝑦(𝑥)
𝑑𝑦
𝑑𝑥
𝑝(𝑥)
+ 𝑝(𝑥)𝑦 = 𝑞(𝑥)
𝑔(𝑥)
𝐼(𝑥)
𝐼(𝑥) =
𝑒 ∫ 𝑝(𝑥)𝑑𝑥
𝐼(𝑥)
𝐼(𝑥)
(𝐼(𝑥)𝑦(𝑥))
′
𝑦(𝑥)
𝑑𝑦
= 9.8 − 0.196𝑦
𝑑𝑥
𝑑𝑦
+ 0.196𝑦 = 9.8
𝑑𝑥
𝑝(𝑥) = 0.196,
0.196𝑑𝑥
∴ 𝐼(𝑥) = 𝑒 ∫
= 𝑒 0.196𝑥
𝑞(𝑥) = 9.8
𝑐
𝑑𝑦
+ 0.196𝑒 0.196𝑥 𝑦 = 9.8𝑒 0.196𝑥
𝑑𝑥
𝑑(𝑒 0.196𝑥 𝑦)
= (𝑒 0.196𝑥 𝑦)′ = 9.8𝑒 0.196𝑥
𝑑𝑥
𝑥,
0.196𝑥
𝑒
𝑦 = 9.8(0.196𝑒 0.196𝑥 ) + 𝑐
= 50𝑒 0.196𝑥 + 𝑐
𝑒 0.196𝑥
𝑒 0.196𝑥 ,
𝑦(𝑥) = 50 +
𝑐
𝑒 0.196𝑥
= 50 + 𝑐𝑒 −0.196𝑥
𝑑𝑦
= 9.8 − 0.196𝑦
𝑑𝑥
𝑦(0) = 48
𝑦(𝑥) = 50 + 𝑐𝑒 −0.196𝑥
𝑐
48 = 𝑦(0) = 50 + 𝑐𝑒 −0.196(0)
48 = 50 + 𝑐
𝑐 = −2
𝑦(𝑥) = 50 − 2𝑒 −0.196𝑥
cos(𝑥) 𝑦 ′ + sin(𝑥) 𝑦 = 2 cos3 (𝑥) sin(𝑥) − 1
𝜋
𝜋
𝑦 ( ) = 3√2, 0 ≤ 𝑥 <
4
2
sin(𝑥)
1
𝑦 = 2cos 2 (𝑥) sin(𝑥) −
cos(𝑥)
cos(𝑥)
′
2
𝑦 + tan(𝑥)𝑦 = 2cos (𝑥) sin(𝑥) − sec(𝑥)
𝑦′ +
𝐼(𝑥) = 𝑒 ∫ tan(𝑥)𝑑𝑥 = 𝑒 ln sec(𝑥) = sec(𝑥)
𝑒 ln 𝑓(𝑥) = 𝑓(𝑥)
sec(𝑥)𝑦 ′ + sec(𝑥)tan(𝑥)𝑦 = 2sec(𝑥)cos2 (𝑥) sin(𝑥) − sec 2 (𝑥)
(sec(𝑥)𝑦)′ = 2𝑐𝑜𝑠(𝑥) sin(𝑥) − sec 2(𝑥)
sec(𝑥)𝑦 = ∫ sin(2𝑥) − sec 2 (𝑥)
1
sec(𝑥) 𝑦 = − 𝑐𝑜𝑠(2𝑥) − tan(𝑥) + 𝑐
2
sin(2𝑥) = 2𝑐𝑜𝑠(𝑥) sin(𝑥)
1
𝑦(𝑥) = − cos(𝑥)𝑐𝑜𝑠(2𝑥) − cos(𝑥) tan(𝑥) + 𝑐𝑐𝑜𝑠(𝑥)
2
1
= − cos(𝑥) 𝑐𝑜𝑠(2𝑥) − 𝑠𝑖𝑛(𝑥) + 𝑐𝑐𝑜𝑠(𝑥)
2
𝑐
𝜋
1
𝜋
𝜋
𝜋
𝜋
3√2 = 𝑦 ( ) = − cos ( ) 𝑐𝑜𝑠 (2 ∙ ) − 𝑠𝑖𝑛 ( ) + 𝑐𝑐𝑜𝑠 ( )
4
2
4
4
4
4
√2
√2
3√2 = −
+𝑐
2
2
𝑐=7
1
𝑦(𝑥) = − cos(𝑥) 𝑐𝑜𝑠(2𝑥) − 𝑠𝑖𝑛(𝑥) + 7𝑐𝑜𝑠(𝑥)
2
𝑁(𝑦)𝑑𝑦 = 𝑀(𝑥)𝑑𝑥
∫ 𝑁(𝑦)𝑑𝑦 = ∫ 𝑀(𝑥)𝑑𝑥
𝑦(𝑥)
𝑦 = 𝑦(𝑥)
𝑑𝑦
𝑑𝑥
= 6𝑦 2 𝑥,
1
𝑦(1) = 25
𝑑𝑦
= 6𝑥𝑑𝑥
𝑦2
∫ 𝑦 −2 𝑑𝑦 = ∫ 6𝑥𝑑𝑥
−𝑦
−
−1
6𝑥 2
=
+𝑐
2
1
= 3𝑥 2 + 𝑐
𝑦
1
𝑦(1) = 25
−
1
= 3(1)2 + 𝑐
1
25
−25 − 3 = 𝑐
𝑐 = −28
1
= 3𝑥 2 − 28
𝑦
1
𝑦(𝑥) =
28 − 3𝑥 2
−
𝑑𝑦
= 𝑒 𝑦−𝑡 𝑠𝑒𝑐(𝑦)(1 + 𝑡 2 ),
𝑑𝑡
𝑦(0) = 0
𝑑𝑦 𝑒 𝑦 ∙ 𝑒 −𝑡
(1 + 𝑡 2 )
=
𝑑𝑡
cos(𝑦)
𝑒 −𝑦 cos(𝑦) 𝑑𝑦 = 𝑒 −𝑡 (1 + 𝑡 2 )𝑑𝑡
∫ 𝑒 −𝑦 cos(𝑦) 𝑑𝑦 = ∫ 𝑒 −𝑡 (1 + 𝑡 2 )𝑑𝑡
𝑒 −𝑦
(sin(𝑦) − cos(𝑦)) = −𝑒 −𝑡 (𝑡 2 + 2𝑡 + 3) + 𝑐
2
1
(−1) = −(3) + 𝑐
2
∴𝑐=
5
2
∴ The implicit solution will be
𝑒 −𝑦
5
(sin(𝑦) − cos(𝑦)) = −𝑒 −𝑡 (𝑡 2 + 2𝑡 + 3) +
2
2
𝑦′ =
3𝑥 2 + 4𝑥 − 4
2𝑦 − 4
𝑦(1) = 3
(2𝑦 − 4)𝑑𝑦 = (3𝑥 2 + 4𝑥 − 4)𝑑𝑥
∫(2𝑦 − 4)𝑑𝑦 = ∫(3𝑥 2 + 4𝑥 − 4)𝑑𝑥
𝑦 2 − 4𝑦 = 𝑥 3 + 2𝑥 2 − 4𝑥 + 𝑐
(3)2 − 4(3) = (1)3 + 2(1)2 − 4(1) + 𝑐
𝑐 = −2
𝑦 2 − 4𝑦 = 𝑥 3 + 2𝑥 2 − 4𝑥 − 2
𝑦 2 − 4𝑦 − (𝑥 3 + 2𝑥 2 − 4𝑥 − 2) = 0
𝑦
𝑥
𝑦=
−𝑏± √𝑏2 −4𝑎𝑐
𝑦(𝑥) =
2𝑎
, 𝑤ℎ𝑒𝑟𝑒 𝑎 = 1, 𝑏 = −4, 𝑐 = −(𝑥 3 + 2𝑥 2 − 4𝑥 − 2)
−(−4)± √(−4)2 +4(1)(𝑥 3 +2𝑥 2 −4𝑥−2)
2(1)
=
4± √16+4(𝑥 3 +2𝑥 2 −4𝑥−2)
2
4 ± 2√4 + (𝑥 3 + 2𝑥 2 − 4𝑥 − 2)
𝑦(𝑥) =
2
𝑦(𝑥) = 2 ± √4 + (𝑥 3 + 2𝑥 2 − 4𝑥 − 2)
±
3 = 𝑦(1) = 2 ± √4 + ((1)3 + 2(1)2 − 4(1) − 2) = 2 ± 1 = 3 𝑜𝑟 1
𝑦(𝑥) = 2 + √4 + (𝑥 3 + 2𝑥 2 − 4𝑥 − 2)
𝑑𝑦
𝑀(𝑥, 𝑦) + 𝑁(𝑥, 𝑦) 𝑑𝑥 = 0
“ = 0”
“+”
Ψ(𝑥, 𝑦)
Ψ𝑥 = 𝑀(𝑥, 𝑦) 𝑎𝑛𝑑
Ψ𝑦 = 𝑁(𝑥, 𝑦)
𝑑𝑦
Ψ𝑥 + Ψ𝑦 𝑑𝑥 = 0
𝑑
(Ψ(𝑥, 𝑦(𝑥))) = 0
𝑑𝑥
Ψ(𝑥, 𝑦) = 𝑐
Ψ(𝑥, 𝑦)
Ψ(𝑥, 𝑦)
Ψ(𝑥, 𝑦)
Ψ(𝑥, 𝑦)
Ψ(𝑥, 𝑦)
Ψ𝑥 = 𝑀
Ψ𝑦 = 𝑁
Ψ(𝑥, 𝑦)
Ψ𝑥𝑦 = Ψ𝑦𝑥
Ψ𝑥𝑦 = (𝛹𝑥 )𝑦 = (𝑀)𝑦 = 𝑀𝑦
Ψ𝑦𝑥 = (𝛹𝑦 )𝑥 = (𝑁)𝑥 = 𝑁𝑥
Ψ(𝑥, 𝑦)
𝑀𝑦 = 𝑁𝑥
Ψ(𝑥, 𝑦)
2𝑥𝑦 − 9𝑥 2 + (2𝑦 + 𝑥 2 + 1)
𝑀 = 2𝑥𝑦 − 9𝑥 2
𝑁 = 2𝑦 + 𝑥 2 + 1
𝑑𝑦
=0
𝑑𝑥
𝑀𝑦 = 2𝑥
𝑁𝑥 = 2𝑥
Ψ(𝑥, 𝑦)
Ψ𝑥 = 𝑀
Ψ𝑦 = 𝑁
Ψ(𝑥, 𝑦)
Ψ = ∫ 𝑀 𝑑𝑥
Ψ = ∫ 𝑁 𝑑𝑦
𝑦(0) = −3
Ψ(𝑥, 𝑦) = ∫ 2𝑥𝑦 − 9𝑥 2 𝑑𝑥 = 𝑥 2 𝑦 − 3𝑥 3 + ℎ(𝑦)
𝑦
𝑥
𝑦
𝑦
𝑐
Ψ(𝑥, 𝑦)
Ψ𝑥 = 𝑀
Ψ(𝑥, 𝑦)
ℎ(𝑦)
Ψ(𝑥, 𝑦)
𝑦
Ψ𝑦 = 𝑁
𝑁
ℎ(𝑦)
Ψ𝑦 = 𝑥 2 + ℎ′ (𝑦) = 2𝑦 + 𝑥 2 + 1 = 𝑁
ℎ′ (𝑦) = 2𝑦 + 1
𝑦
ℎ(𝑦)
𝑥
ℎ(𝑦)
ℎ(𝑦) = ∫ 2𝑦 + 1 𝑑𝑦 = 𝑦 2 + 𝑦
Ψ(𝑥, 𝑦)
Ψ(𝑥, 𝑦) = 𝑥 2 𝑦 − 3𝑥 3 + 𝑦 2 + 𝑦 = 𝑦 2 + (𝑥 2 + 1)𝑦 − 3𝑥 3
Ψ(𝑥, 𝑦) = 𝑦 2 + (𝑥 2 + 1)𝑦 − 3𝑥 3 = 𝑐
𝑦(0) = −3
2
2
(−3) + ((0) + 1)(−3) − 3(0)3 = 𝑐
⟹
𝑐=6
𝑦 2 + (𝑥 2 + 1)𝑦 − 3𝑥 3 − 6 = 0
𝑦
𝑦(𝑥)
−(𝑥 2 + 1) ± √(𝑥 2 + 1)2 + 4(1)(−3𝑥 3 − 6)
2(1)
2
4
−(𝑥 + 1) ± √𝑥 + 12𝑥 3 + 2𝑥 2 + 25
=
2
𝑦(𝑥) =
±
ℎ(𝑦)
−3 = 𝑦(0) =
“−”
−(1) ± √25 −1 ± 5
=
= 2 𝑜𝑟 − 3
2
2
−(𝑥 2 + 1) − √𝑥 4 + 12𝑥 3 + 2𝑥 2 + 25
𝑦(𝑥) =
2
2𝑥𝑦 2 + 4 = 2(3 − 𝑥 2 𝑦)𝑦 ′
𝑦(−1) = 8
2𝑥𝑦 2 + 4 − 2(3 − 𝑥 2 𝑦)𝑦 ′ = 0
2𝑥𝑦 2 + 4 + 2(𝑥 2 𝑦 − 3)𝑦 ′ = 0
𝑀 = 2𝑥𝑦 2 + 4
𝑁 = 2(𝑥 2 𝑦 − 3)
𝑁
𝑀𝑦 = 4𝑥𝑦
𝑁𝑥 = 4𝑥𝑦
𝑀
𝑥
𝑦
𝑁
Ψ(𝑥, 𝑦) = ∫ 2(𝑥 2 𝑦 − 3) 𝑑𝑦 = 𝑥 2 𝑦 2 − 6𝑦 + ℎ(𝑥)
𝑥
𝑦
Ψ𝑥 = 2𝑥𝑦 2 + ℎ′ (𝑥) = 2𝑥𝑦 2 + 4 = 𝑀
ℎ′ (𝑥) = 4
ℎ(𝑥)
ℎ(𝑥) = ∫ 4 𝑑𝑥 = 4𝑥
Ψ(𝑥, 𝑦)
Ψ(𝑥, 𝑦) = 𝑥 2 𝑦 2 − 6𝑦 + 4𝑥
Ψ(𝑥, 𝑦) = 𝑥 2 𝑦 2 − 6𝑦 + 4𝑥 = 𝑐
𝑦(−1) = 8
64 − 48 − 4 = 𝑐
⟹
𝑐 = 12
𝑥 2 𝑦 2 − 6𝑦 + 4𝑥 − 12 = 0
6 ± √(−6)2 − 4(𝑥 2 )(4𝑥 − 12)
𝑦(𝑥) =
2𝑥 2
6 ± √36 + 48𝑥 2 − 16𝑥 3
=
2𝑥 2
𝑥
=
3 ± √9 + 12𝑥 2 − 4𝑥 3
𝑥2
𝑦(𝑥) =
3 ± √9 + 12𝑥 2 − 4𝑥 3
𝑥2
𝑦
𝑦′ = 𝐹 ( )
𝑥
𝑣(𝑥) =
𝑦
𝑥
𝑦 = 𝑥𝑣
𝑦
𝑣
𝑥
𝑑𝑦
𝑑𝑣
=𝑣+𝑥
𝑑𝑥
𝑑𝑥
𝑣+𝑥
𝑥
𝑑𝑣
= 𝐹(𝑣) − 𝑣
𝑑𝑥
𝑑𝑣
= 𝐹(𝑣)
𝑑𝑥
⇒
𝑥𝑦𝑦 ′ + 4𝑥 2 + 𝑦 2 = 0
𝑑𝑣
𝑑𝑥
=
𝐹(𝑣) − 𝑣
𝑥
𝑦(2) = −7,
𝑥2
𝑦 ′
𝑦 2
𝑦 = −4 − ( )
𝑥
𝑥
𝑦
𝑥
𝑣(𝑣 + 𝑥𝑣 ′ ) = −4 − 𝑣 2
𝑥>0
𝑣𝑥𝑣 ′ = −4 − 2𝑣 2
4 + 2𝑣 2
𝑥𝑣 ′ = −
𝑣
𝑣
1
𝑑𝑣 = − 𝑑𝑥
2
4 + 2𝑣
𝑥
1
𝑙𝑛(4 + 2𝑣 2 ) = − 𝑙𝑛(𝑥) + 𝑐
4
1
𝑙𝑛(4 + 2𝑣 2 )4 = ln(𝑥)−1 + ln(𝐴)
1
𝑙𝑛(4 + 2𝑣 2 )4 = ln 𝐴(𝑥)−1
1
−1
(4 + 2𝑣 2 )4 = 𝑒 ln 𝐴(𝑥)
1
𝐴 𝑐
(4 + 2𝑣 2 )4 = 𝐴(𝑥)−1 = =
𝑥 𝑥
𝑙𝑛 (𝐴)
𝑐
𝑣
𝑐 4 𝑐4
𝑐
4 + 2𝑣 2 = ( ) = 4 = 4
𝑥
𝑥
𝑥
1
𝑐
𝑣 2 = ( 4 − 4)
2 𝑥
𝑦 2 1 𝑐
( ) = ( 4 − 4)
𝑥
2 𝑥
1
𝑐
𝑐 − 4𝑥 4
𝑦 2 = 𝑥 2 ( 4 − 4) =
2
𝑥
2𝑥 2
49 =
𝑐
𝑐 − 4(16)
2(4)
⇒
𝑐 = 456
𝑦
𝑦2 =
456 − 4𝑥 4
2𝑥 2
⇒
228 − 2𝑥 4
𝑦(𝑥) = ±√
𝑥2
𝑐
228 − 2𝑥 4
√
𝑦(𝑥) = −
𝑥2
(𝑦 2 + 2𝑥𝑦)𝑑𝑥 − 𝑥 2 𝑑𝑦 = 0
𝑦(1) = 1,
𝑥 2 𝑑𝑥
𝑑𝑦 𝑦 2 + 2𝑥𝑦
=
𝑑𝑥
𝑥2
𝑓(𝑡𝑥, 𝑡𝑦) =
𝑦 2 +2𝑥𝑦
𝑥2
𝑑𝑦
= 𝑓(𝑡𝑥, 𝑡𝑦)
𝑑𝑥
(𝑡𝑦)2 + 2(𝑡𝑥)(𝑡𝑦)
𝑓(𝑡𝑥, 𝑡𝑦) =
(𝑡𝑥)2
𝑡 2 𝑦 2 + 2𝑡 2 𝑥𝑦
=
𝑡2𝑥2
2
𝑡 (𝑦 2 + 2𝑥𝑦)
=
𝑡2𝑥2
∴ 𝐼𝑡 𝑖𝑠 ℎ𝑜𝑚𝑜𝑔𝑒𝑛𝑜𝑢𝑠𝑜𝑓 𝑑𝑒𝑔𝑟𝑒𝑒 2
𝑑𝑦
= 𝑑𝑥
𝑦
𝑣=𝑥
𝑑𝑦 𝑦 2 + 2𝑥𝑦
=
𝑑𝑥
𝑥2
2
𝑑𝑦 𝑦
2𝑥𝑦
𝑦 2 2𝑦
= 2+ 2 =( ) +
𝑑𝑥 𝑥
𝑥
𝑥
𝑥
𝑑𝑦
′
𝑎𝑛𝑑 𝑑𝑥 = 𝑣 + 𝑥𝑣
(𝑣 + 𝑥𝑣 ′ ) = 𝑣 2 + 2𝑣
𝑥
𝑑𝑣
= 𝑣 2 + 𝑣 = 𝑣(𝑣 + 1)
𝑑𝑥
1
1
𝑑𝑣 = 𝑑𝑥
𝑣(𝑣 + 1)
𝑥
𝑙𝑛(𝑣) −𝑙𝑛(𝑣 + 1) = 𝑙𝑛(𝑥) + 𝑐
𝑣
ln (
) = ln(𝑥) + ln(𝑐)
𝑣+1
𝑣
ln (
) = ln(𝑐𝑥)
𝑣+1
𝑣
= 𝑒 ln(𝑐𝑥)
𝑣+1
𝑣
= 𝑐𝑥
𝑣+1
𝑥>0
𝑦
𝑥 = 𝑐𝑥
𝑦
𝑥+1
𝑦
𝑦
= 𝑐𝑥 ( + 1) = 𝑐(𝑦 + 𝑥)
𝑥
𝑥
1 = 𝑐(2)
𝑐
⇒
𝑐=
𝑦
𝑦 1
= (𝑦 + 𝑥)
𝑥 2
2𝑦 = 𝑥(𝑦 + 𝑥) = 𝑥𝑦 + 𝑥 2
2𝑦 − 𝑥𝑦 = 𝑥 2
𝑥2
𝑦(𝑥) =
2−𝑥
1
2