![Determinants of Matrix
To every a square matrix that is assigned a specific number called the determinant of matrix
the determinant of a matrix A is denoted by 𝐴
If A is (1*1) matrix, A=(a) :then 𝐴 = 𝑎 = 𝑎
If A is (2*2) matrix , A=
𝑎 𝑏
𝑐 𝑑
: then 𝐴 =
𝑎 𝑏
𝑐 𝑑
=ad-bc
1f A is (3*3) matrix ,A=
𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
𝑎31 𝑎32 𝑎33
then
𝐴 =
𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
𝑎31 𝑎32 𝑎33
= 𝑎11
𝑎22 𝑎22
𝑎32 𝑎33
- 𝑎12
𝑎21 𝑎23
𝑎31 𝑎33
+ 𝑎13
𝑎21 𝑎22
𝑎31 𝑎32
or
𝐴 =
𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
𝑎31 𝑎32 𝑎33
=[𝑎11𝑎22𝑎33+𝑎12𝑎23𝑎31 + 𝑎13𝑎21𝑎32] − [𝑎12𝑎21𝑎33 + 𝑎11𝑎23𝑎32
+ 𝑎13𝑎22𝑎31]](https://image.slidesharecdn.com/math1-compressed-231111211222-79beb6da/85/math1-compressed-pdf-19-320.jpg)
![The Definite Integral :
Theorem: (Fundamental Theorem I):
𝑎
𝑏
𝑓 𝑥 𝑑𝑥 = 𝐹 𝑥 ]𝑎
𝑏 = 𝐹 𝑏 − 𝐹 𝑎 ,
𝑑
𝑑𝑥
𝐹 𝑥 = 𝑓(𝑥)
EX:
1. Find 2
3
𝑥2
𝑑𝑥 =
𝑥3
3
]2
3
=
33
3
−
23
3
=
27
3
−
8
3
=
19
3
2. 0
𝜋
sin2 𝑥 𝑑𝑥 = 0
𝜋 1
2
(1 − cos 2𝑥)𝑑𝑥 =
1
2
𝑥 −
sin 2𝑥
2
]0
𝜋
=
1
2
𝜋 −
sin 2𝜋
2
−
1
2
0](https://image.slidesharecdn.com/math1-compressed-231111211222-79beb6da/85/math1-compressed-pdf-95-320.jpg)
![2. Area between two curves :
Case 1(No intersecting of two curves in [a,b])
If 𝑦1 = 𝑓 𝑥 , 𝑦2 = 𝑔 𝑥 then the area is defined as
𝐴𝑎
𝑏 = 𝑎
𝑏
𝑦1 − 𝑦2 𝑑𝑥
Case 2(An intersecting point exists in [a,b])
The area between two curves 𝑦1, 𝑦2 from x=a to x=b is defined as
𝐴𝑎
𝑏 =
𝑎
𝑐
𝑦2 − 𝑦1 𝑑𝑥 +
𝑐
𝑏
(𝑦1 − 𝑦2) 𝑑𝑥](https://image.slidesharecdn.com/math1-compressed-231111211222-79beb6da/85/math1-compressed-pdf-100-320.jpg)
math1مرحلة اولى -compressed.pdf
- 1. Math 1 Lecturer Wala’a Abd ul_Mageed
- 2. Matrix
- 3. The location of -5 is 𝑎12 ,,, and location of 3 is 𝑎24
- 5. B= 𝑏11 𝑏12 𝑏13 𝑏21 𝑏22 𝑏23 Equal Matrices :
- 6. By symbols 𝐴𝑚×𝑛 and 𝐵𝑚×𝑛 are two matrices then 𝐴𝑚×𝑛= 𝐵𝑚×𝑛 if and only if 𝑎𝑖𝑗 = 𝑏𝑖𝑗 i=1,2….m , j=1,2…n EX: 1. A= 1 −3 4 0 5 −2 B= 1 −3 4 0 5 −2 A=B 1. If A= 𝑎 𝑏 𝑐 𝑑 𝑒 𝑓 = B= 1 −2 16 4 5 11 Then a= 1 , b=-2 , c= 16 , d=4 , e= 5 , f = 11
- 7. The Addition & Subtraction of Matrices : 𝐴𝑚×𝑛 and 𝐵𝑚×𝑛 are two matrices then 𝐴𝑚×𝑛+ 𝐵𝑚×𝑛 is 𝑎𝑖𝑗 + 𝑏𝑖𝑗 And 𝐴𝑚×𝑛- 𝐵𝑚×𝑛 is 𝑎𝑖𝑗 − 𝑏𝑖𝑗 Ex:
- 8. Theorem: The Multiplication: 1. Scalar Multiplication :
- 9. 𝐴𝑚×𝑛 and 𝐾 be a scalar then 𝐾(𝐴𝑚×𝑛) = (𝐾 𝑎𝑖𝑗) EX: A= 3 6 0 5 k=-2 then kA = -2. 3 6 0 5 = −6 −12 0 −10 1. Two matrices Multiplication:
- 13. Types of Matrices :
- 17. Transpose of Matrix :
- 18. EX: If A= 1 8 1.5 7 −66 3.3 then 𝐴𝑇 = 1 7 8 −66 1.5 3.3
- 19. Determinants of Matrix To every a square matrix that is assigned a specific number called the determinant of matrix the determinant of a matrix A is denoted by 𝐴 If A is (1*1) matrix, A=(a) :then 𝐴 = 𝑎 = 𝑎 If A is (2*2) matrix , A= 𝑎 𝑏 𝑐 𝑑 : then 𝐴 = 𝑎 𝑏 𝑐 𝑑 =ad-bc 1f A is (3*3) matrix ,A= 𝑎11 𝑎12 𝑎13 𝑎21 𝑎22 𝑎23 𝑎31 𝑎32 𝑎33 then 𝐴 = 𝑎11 𝑎12 𝑎13 𝑎21 𝑎22 𝑎23 𝑎31 𝑎32 𝑎33 = 𝑎11 𝑎22 𝑎22 𝑎32 𝑎33 - 𝑎12 𝑎21 𝑎23 𝑎31 𝑎33 + 𝑎13 𝑎21 𝑎22 𝑎31 𝑎32 or 𝐴 = 𝑎11 𝑎12 𝑎13 𝑎21 𝑎22 𝑎23 𝑎31 𝑎32 𝑎33 =[𝑎11𝑎22𝑎33+𝑎12𝑎23𝑎31 + 𝑎13𝑎21𝑎32] − [𝑎12𝑎21𝑎33 + 𝑎11𝑎23𝑎32 + 𝑎13𝑎22𝑎31]
- 20. Note that if 𝐴 = 0 then a matrix A is called a singular The Properties of Determinants:
- 21. The abjoint of Matrix : Minor of matrix : let A= 𝑎11 𝑎12 … . . 𝑎1𝑛 𝑎21 𝑎22 … . . 𝑎2𝑛 … … 𝑎 𝑛1 𝑎𝑛2 … . . 𝑎𝑛𝑛 (*) Cofactor of matrix: let A which defined in (*) with 𝑀𝑖𝑗 (minor of A) ,then the Cofactor of matrix is 𝐶𝑖𝑗= (−1)𝑖+𝑗 𝑀𝑖𝑗
- 22. ad joint of matrix : the ad joint of matrix A in(*) is the transpose of matrix Cofactor of A i.e adj(A) = 𝐶𝑇 EX: let A = 2 3 5 4 1 6 1 4 0 find the ad joint of A ?
- 24. adj(A) = 𝐶𝑇 = −24 20 13 6 −5 8 15 −5 −10 Inverse of a square matrix : denoted by 𝐴−1 is 𝐴−1 = 1 𝐴 adj(A) , 𝐴 ≠ 0 EX: let A= 1 2 3 4 1 5 6 0 2 find 𝐴−1 ?
- 25. C= 2 22 −6 −4 −16 12 7 7 −7 And adj(A) = 𝐶𝑇 = 2 −4 7 22 −16 7 −6 12 −7 ∴ 𝐴−1 = 1 𝐴 adj(A)
- 26. = 1 28 2 −4 7 22 −16 7 −6 12 −7 = 2 28 −4 28 7 28 22 28 −16 28 7 28 −6 28 12 28 −7 28 The properties of multiplication
- 28. Solution the System of Linear Equations;
- 30. EX; solve the following system: A= 1 2 1 3 −4 −2 5 3 5 x= 𝑥1 𝑥2 𝑥3 b= 4 2 −1 x= 𝐴−1.b we find 𝐴−1
- 31. Grammer’s Rule :
- 32. If D = 𝑎11 𝑎12 … … 𝑎1𝑛 𝑎 21 …….. 𝑎22 … . . 𝑎2𝑛 𝑎𝑛1 𝑎𝑛2 … … 𝑎𝑛𝑛 ≠ 0 Then the system has a unique solution by Grammer’s Rule is 𝑥1 = 1 𝐷 𝑏1 𝑎12 … . . 𝑎𝑛1 𝑏2 ⋮ 𝑎22……… ⋮ 𝑎𝑛2 𝑏𝑛 𝑎2𝑛 … … 𝑎𝑛𝑛 = 𝐷𝑥1 𝐷 ,,𝑥2 = 1 𝐷 𝑎11 𝑏1 … . . 𝑎𝑛1 𝑎21 ⋮ 𝑏2……… ⋮ 𝑎𝑛2 𝑎𝑛1 𝑏𝑛 … … 𝑎𝑛𝑛 = 𝐷𝑥2 𝐷 ……… 𝑥𝑛 = 1 𝐷 𝑎11 𝑎12 … … 𝑏1 𝑎 21 …….. 𝑎22 … . . 𝑏2 ⋮ 𝑎𝑛1 𝑎𝑛2 … … 𝑏𝑛 = 𝐷𝑥𝑛 𝐷 EX: solve the system :
- 33. 3𝑥1 − 𝑥2 = 9 𝑥1 + 2𝑥2 = −4 D= 3 −1 1 2 =6+1=7 𝑥1 = 1 7 9 −1 −4 2 = 1 7 18 − 4 = 14 7 = 2 𝑥2 = 1 7 3 9 1 −4 = 1 7 −12 − 9 = −21 7 = −3
- 34. EX: solve the system : 𝑋1 = 11 3 −2 −15 −2 1 3 4 −1 −25 = 50 −25 = −2
- 35. 𝑋2 = 1 11 −2 4 −15 1 3 3 −1 −25 = −25 −25 = 1 𝑋3 = 1 3 11 4 −2 −15 3 4 3 −25 = 125 −25 = −3 Vectors: A vector in the plane is a directed line segment . A directed line segment 𝐴𝐵 and the length 𝑖s denoted by 𝐴𝐵
- 36. Two equal vectors : Vector from two points : Let 𝑝1 = 𝑥1, 𝑦1 𝑎nd 𝑝2 = 𝑥2, 𝑦2 then the vector 𝑝1𝑝2 is 𝑝1𝑝2 = (𝑥2 − 𝑥1,𝑦2 − 𝑦1) And the length of 𝑝1𝑝2 𝑖𝑠 (𝑥2 − 𝑥1)2+(𝑦2−𝑦1)2 ( magnitude of vector)
- 37. EX: let A= (2,3) and B=(1,-2) , find 𝐴𝐵 . 𝐴𝐵 = 1 − 2, −2 − 3 = −1, −5 𝑎𝑛𝑑 𝐴𝐵 = (1 − 2)2+(−2 − 3)2= 1 + 25 = 26 Note: in two dimension then the components of a vector a is ( 𝑎1,𝑎2) and in three dimension is ( 𝑎1,𝑎2, 𝑎3) The Addition of Two Vectors :The Addition of Two Vectors :
- 38. Theorem: Let a=(𝑎1, 𝑎2), 𝑏 = (𝑏1, 𝑏2)𝑖𝑛 𝑅2 , 𝑎𝑛𝑑 𝐾 𝑏𝑒 𝑎 𝑠𝑐𝑎𝑙𝑎𝑟 . 𝑡ℎ𝑒𝑛 a. a+ b = (𝑎1 + 𝑏1, 𝑎2 + 𝑏2) b- ka =(K𝑎1, 𝐾𝑎2) and if a=(𝑎1, 𝑎2, 𝑎3), 𝑏 = (𝑏1, 𝑏2, 𝑏3)𝑖𝑛 𝑅3 , 𝑎𝑛𝑑 𝐾 𝑏𝑒 𝑎 𝑠𝑐𝑎𝑙𝑎𝑟 . 𝑡ℎ𝑒𝑛 a. a+ b = (𝑎1 + 𝑏1, 𝑎2 + 𝑏2, 𝑎3 + 𝑏3) b- ka =(K𝑎1, 𝐾𝑎2, 𝐾𝑎3 EX: 𝑣 = 4, −1 , 𝑤 = 3,2 be are vectors find 𝑣 + 𝑤 𝑣 + 𝑤 = 4 + 3, −1 + 2 = (7,1)
- 39. Theorem For any vectors a,b,c and scalars k ,l, we have (a) a+ b =b +a Commutative Law (b) a+( b+c) = (a +b)+c Associative Law (c) a+0=0+a Additive Identity (d) a+(-a)=0 Additive Inverse (e) k(la) = (kl)a Associative Law (f) k(a+b) = ka+ kb Distributive Law (g) (k + l)a = ka+ l b Distributive Law Example Let v = (2,1,−1) and w = (3,−4,2) in 𝑅3. (a) Find v−w. Solution: v−w = (2−3,1−(−4),−1−2) = (−1,5,−3) (b) Find 3v+2w. Solution: 3v+2w = (6,3,−3)+(6,−8,4) = (12,−5,1)
- 40. The unit vectors: A vector with length equal 1 is called unit vector .The standard unit vectors is i= (1,0) ,j= (0,1) in two dimension and in 3D is i=(1,0,0) ,j=(0,1,0) ,k=(0,0,1) Component of Vector in Term of Unit Vectors:
- 42. Vector in The Space :
- 43. EX: a= -i+3j+k and b= 4i+7j .find 1. 2a+3b 2(-i+3j+k) +3(4i+7j) = (-2i+6j+2k)+(12i+21j) = 10i+27j+2k 1. a-b (-i+3j+k) –( 4i+7j)= -5i- 4j+k 2. 1 2 𝑎 = (− 1 2 )2 + ( 3 2 )2 + ( 1 2 )2 = 1 4 + 9 4 + 1 4 = 11 4 = 11 2 The Dot Product:
- 44. Let a= (𝑎1, 𝑎2, 𝑎3) 𝑎𝑛𝑑 𝑏 = 𝑏1, 𝑏2, 𝑏3 in 3D then, the dot product denoted by a.b is 𝑎1𝑏1 + 𝑎2𝑏2 + 𝑎3𝑏3 Similarly if a= 𝑎1𝑖 + 𝑎2𝑗 + 𝑎3𝑘 𝑎𝑛𝑑 𝑏 = 𝑏1𝑖 + 𝑏2𝑗 + 𝑏3𝑘 a.b= 𝑎1𝑏1 + 𝑎2𝑏2 + 𝑎3𝑏3 EX: If p=3i-2j+k and q=2i+3j-4k
- 45. The Angle between Two Vectors : The angle between two non zeros vectors is the smallest angle between them . Theorem: Let v ,w are two non zeros vectors and 𝜃 𝑖𝑠the angle between a and b then cos 𝜃 = 𝑣. 𝑤 𝑣 𝑤
- 46. EX: Is a= -i+5j+2k perpendicular to b=3i+j-k ? a.b = (-1.3+5.1+2.-1)=0 a is perpendicular to b Theorem : For any vectors u,v,w and k be a scalar :
- 47. The cross product: Let a= 𝑎1𝑖 + 𝑎2𝑗 + 𝑎3𝑘 𝑎𝑛𝑑 𝑏 = 𝑏1𝑖 + 𝑏2𝑗 + 𝑏3𝑘 in 3D then, the cross product denoted by a×b is 𝑡he vector :
- 48. a×b= 𝑖 𝑗 𝑘 𝑎1 𝑎2 𝑎3 𝑏1 𝑏2 𝑏3 = 𝑎2 𝑎3 𝑏2 𝑏3 𝑖 − 𝑎1 𝑎3 𝑏1 𝑏3 𝑗 + 𝑎1 𝑎2 𝑏1 𝑏2 𝑘= (𝑎2𝑏3 − 𝑎3𝑏2)𝑖 − (𝑎1𝑏3 − 𝑎3𝑏1)𝑗 +(𝑎1𝑏2 − 𝑎2𝑏1)𝑘 EX: let p=3i-4j+2k and q= 2i+5j-k ,find p×q?
- 50. Function Numbers: 1. N={1,2,3,…….} the set of Natural numbers 2-Z={-1,-2,…..,0,1,2,……} the set of integer numbers 3-Q={ N,Z, 𝑝 𝑞 , q≠ 0 and p,q 𝜖𝑍} the set of rational numbers 4- H= { 𝑥} the set of irrational numbers 5- R=Q+H = the set of real numbers Interval
- 52. Ex: if f(x)=𝑥3 then f(2) =23 = 8 Ex: verify the domains and ranges of these functions : y= 𝑥2 domain(x)= (-∞, ∞) range(y)=[0,∞) y=1 𝑥 domain(x)= (-∞, 0) ∪ (0, ∞) range(y)= (-∞, 0) ∪ (0, ∞) y= 𝑥 domain(x)= [0, ∞) range(y)= [0, ∞) y= 4 − 𝑥 domain(x)= (−∞, 4) range(y)= [0, ∞)
- 53. The gargh of function
- 54. Ex:
- 58. Trigonometric function : 1. sin 𝜃 = 𝑎 𝑐
- 63. Y=Sin x ,y= cos x, y= tan x, y=cot x , y= sec x, y=csc x, 1. y=sin x : domain (x) −∞ ≤ 𝑥 ≤ ∞ range(y) -1≤ 𝑦 ≤ 1 2. y= cos x : domain (x) −∞ ≤ 𝑥 ≤ ∞ range(y) -1≤ 𝑦 ≤ 1 3. y= tan x : domain (x)={x≠ 𝜋/2, 𝑥 ≠ 3𝜋/2}
- 73. The Continuity :
- 75. The Derivative:
- 84. The Integration : Integration is considered as the anti derivative operation. The formula 𝑑 𝑑𝑥 F(x)=f(x) is the same as 𝑓 𝑥 𝑑𝑥 = 𝐹 𝑥 + 𝑐 ex: 𝑑 𝑑𝑥 sin 𝑥 = cos 𝑥 cos 𝑥 = sin 𝑥 + 𝑐 Rules: 1. 𝑑𝑥 = 𝑥 + 𝑐 2. 𝑘𝑑𝑥 = 𝑘𝑥 + 𝑐 ,k is constant 3. 𝑥𝑛 𝑑𝑥 = 𝑥𝑛+1 𝑛+1 + 𝑐 n≠ −1 4. 𝑓 𝑥 + 𝑔 𝑥 𝑑𝑥 = 𝑓 𝑥 𝑑𝑥 + 𝑔 𝑥 𝑑𝑥 5. 𝑘𝑓 𝑥 = 𝑘 𝑓 𝑥 𝑑𝑥 , k is constant 6. 𝑓 𝑥 𝑛 𝑓′ (𝑥)𝑑𝑥 = 𝑓(𝑥)𝑛+1 𝑛+1 + 𝑐
- 85. Ex: 1. 2𝑑𝑥 = 2𝑥 + 𝑐 2. 𝑥3 𝑑𝑥 = 𝑥4 4 + 𝑐 3. −2𝑥2 𝑑𝑥 = −2𝑥3 3 + 𝑐 4. (𝑥2 + 2)2. 2𝑥𝑑𝑥 = (𝑥2+2)3 3 + 𝑐 5. (𝑥3 − 1)3. 𝑥2𝑑𝑥 = 1 3 (𝑥3 − 1)3. 3𝑥2𝑑𝑥 = (𝑥3−1)4 12 + 𝑐 The Trigonometric functions 1. sin 𝑥 𝑑𝑥 = − cos 𝑥 + 𝑐 2. cos 𝑥 𝑑𝑥 = sin 𝑥 + 𝑐 3. 𝑠𝑒𝑐2𝑥𝑑𝑥 = tan 𝑥 + 𝑐 4. 𝑐𝑠𝑐2𝑥𝑑𝑥 = −𝑐𝑜𝑡𝑥 + 𝑐 5. 𝑠𝑒𝑐 𝑥𝑡𝑎𝑛𝑥𝑑𝑥 = sec 𝑥 + 𝑐 6. 𝑐𝑠𝑐 𝑥𝑐𝑜𝑡𝑥𝑑𝑥 = −csc 𝑥 + 𝑐
- 86. EX: 1. 𝑥𝑐𝑜𝑠(𝑥2 + 1)𝑑𝑥 = 1 2 2𝑥𝑐𝑜𝑠 𝑥2 + 1 𝑑𝑥 = 1 2 sin 𝑥2 + 1 + 𝑐 2. 𝑠𝑒𝑐2 (4𝑥) 𝑑𝑥 = tan(4𝑥) 4 + 𝑐 3. cot 𝑥 𝑠𝑖𝑛3𝑥 𝑑𝑥 = cos 𝑥 sin 𝑥 𝑠𝑖𝑛3𝑥𝑑𝑥 = cos 𝑥 𝑠𝑖𝑛2 𝑥 𝑑𝑥 = 𝑠𝑖𝑛3𝑥 3 +c األتي فنتبع المشتقة توفر عدم حالة في أما : 1.if power is even : 𝑐𝑜𝑠2 𝑥 = (1 + cos 2𝑥) 2 𝑠𝑖𝑛2 𝑥 = (1 − cos 2𝑥) 2 2.if power is odd: 𝑠𝑖𝑛2 𝑥 = 1 − 𝑐𝑜𝑠2 𝑥 𝑐𝑜𝑠2 𝑥 = 1 − 𝑠𝑖𝑛2 𝑥 𝑠𝑒𝑐2 𝑥 = 1 + 𝑡𝑎𝑛2 𝑥 EX: 𝑐𝑜𝑠3 𝑥 𝑑𝑥 = cos 𝑥 𝑐𝑜𝑠2 𝑥 𝑑𝑥 = cos 𝑥(1
- 87. Integration of multiply sin and cos with diff. angles 1. 𝑠𝑖𝑛 𝑎𝑥𝑐𝑜𝑠 𝑏𝑥𝑑𝑥 = 1 2 (sin 𝑎 − 𝑏 𝑥 + sin 𝑎 + 𝑏 𝑥)𝑑𝑥 2. 𝑠𝑖𝑛 𝑎𝑥𝑠𝑖𝑛 𝑏𝑥𝑑𝑥 = 1 2 (cos 𝑎 − 𝑏 𝑥 − cos 𝑎 + 𝑏 𝑥)𝑑𝑥 3. 𝑐𝑜𝑠 𝑎𝑥𝑐𝑜𝑠 𝑏𝑥𝑑𝑥 = 1 2 (cos 𝑎 − 𝑏 𝑥 + cos 𝑎 + 𝑏 𝑥)𝑑𝑥 EX 1. 𝑠𝑖𝑛 4𝑥𝑐𝑜𝑠2𝑥𝑑𝑥 = 1 2 (sin 4 − 2 𝑥 + sin 4 + 2 𝑥)𝑑𝑥 = 1 2 (sin 2 𝑥 + sin 6 𝑥)𝑑𝑥 = 1 2 − 1 2 𝑐𝑜𝑠2𝑥 − 1 6 𝑐𝑜𝑠6𝑥 + 𝑐 The Logarithmic functions 1. Natural logarithm of x 𝑑 𝑑𝑥 𝑙𝑛𝑥 = 1 𝑥 ∴ 1 𝑥 𝑑𝑥 = 𝑙𝑛 𝑥 + 𝑐
- 88. 2.𝑒𝑥 function 𝑑 𝑑𝑥 𝑒𝑥 = 𝑒𝑥 ∴ 𝑒𝑥 𝑑𝑥 = 𝑒𝑥 + 𝑐 3. 𝑎𝑥 function 𝑑 𝑑𝑥 𝑎𝑥 = 𝑎𝑥𝑙𝑛𝑎 ∴ 𝑎𝑥 𝑑𝑥 = 𝑎 𝑙𝑛𝑎 𝑥 + 𝑐 Ex: 1. 2𝑥 𝑥2 𝑑𝑥 = 𝑙𝑛 𝑥2 + 𝑐 2. 𝑒3𝑥 𝑑𝑥 = 1 3 3𝑒3𝑥 𝑑𝑥 = 1 3 𝑒3𝑥 + 𝑐 3. 24𝑥 𝑑𝑥 = 1 4 2 𝑙𝑛2 𝑥 + 𝑐 ∴. tan 𝑥 𝑑𝑥 = −𝑙𝑛 cos 𝑥 + 𝑐 cot 𝑥 𝑑𝑥 = 𝑙𝑛 sin 𝑥 + 𝑐
- 89. The Hyperbolic Functions 1. sinℎ 𝑥 𝑑𝑥 = cos ℎ𝑥 + 𝑐 2. cosh 𝑥 𝑑𝑥 = sin ℎ𝑥 + 𝑐 3. 𝑠𝑒𝑐ℎ2𝑥𝑑𝑥 = tanh 𝑥 + 𝑐 4. 𝑐𝑠𝑐ℎ2𝑥𝑑𝑥 = −𝑐𝑜𝑡ℎ𝑥 + 𝑐 5. 𝑠𝑒𝑐 ℎ𝑥𝑡𝑎𝑛ℎ𝑥𝑑𝑥 = − sec ℎ𝑥 + 𝑐 6. 𝑐𝑠𝑐ℎ 𝑥𝑐𝑜𝑡ℎ𝑥𝑑𝑥 = −csc ℎ𝑥 + 𝑐 7. tanh 𝑥 𝑑𝑥 = 𝑙𝑛 cosh 𝑥 + 𝑐 8. coth 𝑥 𝑑𝑥 = 𝑙𝑛 sinh 𝑥 + 𝑐 Ex: 1. 𝑑𝑥 𝑠𝑖𝑛ℎ𝑥+𝑐𝑜𝑠ℎ𝑥 = 𝑑𝑥 𝑒𝑥 = 𝑒−𝑥𝑑𝑥 = −𝑒−𝑥 + 𝑐 2. 𝑒2𝑥 𝑠𝑖𝑛ℎ3𝑥𝑑𝑥 = 𝑒2𝑥 ( 𝑒3𝑥−𝑒−3𝑥 2 )𝑑𝑥 = 1 2 ( 𝑒5𝑥 − 𝑒−𝑥 𝑑𝑥 = 1 10 𝑒5𝑥 + 1 2 𝑒−𝑥 + 𝑐
- 92. Method of Integral : 1-. Integration by Substitution: EX: 1. 𝑥(𝑥2 +1)20 𝑑𝑥 Let u=(𝑥2 +1) → 𝑥2 = 𝑢 − 1 → x= 𝑢 − 1 du=2xdx dx= 𝑑𝑢 2𝑥 dx= 𝑑𝑢 2 𝑢−1 𝑢 − 1 (𝑢)20 𝑑𝑢 2 𝑢−1 = 1 2 (𝑢)20 𝑑𝑢 = (𝑢)21 42 + 𝑐 2. 𝑥 𝑥 − 3 𝑑𝑥 = u=x-3 x=u+3 du=dx (𝑢 + 3) 𝑢 𝑑𝑢 = ( 𝑢 3 2 + 3 𝑢)1 2 𝑑𝑢 = (𝑢)5 2 5 2 + 3 (𝑢)3 2 3 2 + 𝑐 = 2 5 (𝑢)5 2+2(𝑢)3 2+𝑐
- 93. 2-.Integration by parts( udv) 𝑢𝑑𝑣 = 𝑢𝑣 − 𝑣𝑑𝑢 1. 𝑥𝑒3𝑥 𝑑𝑥 = u=x du=dx dv=𝑒3𝑥 𝑑𝑥 𝑣 = 1 3 𝑒3𝑥 ∴ 𝑢𝑑𝑣 = 1 3 𝑥𝑒3𝑥 − 1 3 𝑒3𝑥 𝑑𝑥 = 1 3 𝑥𝑒3𝑥 − 1 9 𝑒3𝑥 + 𝑐 = 1 3 𝑒3𝑥(𝑥 − 1 3 ) + 𝑐 EX:
- 95. The Definite Integral : Theorem: (Fundamental Theorem I): 𝑎 𝑏 𝑓 𝑥 𝑑𝑥 = 𝐹 𝑥 ]𝑎 𝑏 = 𝐹 𝑏 − 𝐹 𝑎 , 𝑑 𝑑𝑥 𝐹 𝑥 = 𝑓(𝑥) EX: 1. Find 2 3 𝑥2 𝑑𝑥 = 𝑥3 3 ]2 3 = 33 3 − 23 3 = 27 3 − 8 3 = 19 3 2. 0 𝜋 sin2 𝑥 𝑑𝑥 = 0 𝜋 1 2 (1 − cos 2𝑥)𝑑𝑥 = 1 2 𝑥 − sin 2𝑥 2 ]0 𝜋 = 1 2 𝜋 − sin 2𝜋 2 − 1 2 0
- 98. The applications of definite integral 1-Area under the curve 2.Area between two curves 3.Volumes 4 The length of the curves 5. Surface area 1. Area under the curve The integral 𝑎 𝑏 𝑓 𝑥 𝑑𝑥 is defined the area under the curve y=f(x) from x=a to x=b A= 𝑎 𝑏 𝑦 𝑑𝑥
- 100. 2. Area between two curves : Case 1(No intersecting of two curves in [a,b]) If 𝑦1 = 𝑓 𝑥 , 𝑦2 = 𝑔 𝑥 then the area is defined as 𝐴𝑎 𝑏 = 𝑎 𝑏 𝑦1 − 𝑦2 𝑑𝑥 Case 2(An intersecting point exists in [a,b]) The area between two curves 𝑦1, 𝑦2 from x=a to x=b is defined as 𝐴𝑎 𝑏 = 𝑎 𝑐 𝑦2 − 𝑦1 𝑑𝑥 + 𝑐 𝑏 (𝑦1 − 𝑦2) 𝑑𝑥
- 104. Inverse of Trigonometric Functions:
- 108. EX: 1. 𝑥2 1+𝑥6 𝑑𝑥 = 𝑥2 1+(𝑥3)2 𝑑𝑥 = 1 3 3𝑥2 1+(𝑥3)2 𝑑𝑥 = 1 3 𝑡𝑎𝑛−1 𝑥3 + 𝑐 2. 𝑥 1−𝑥4 𝑑𝑥 = 1 2 2𝑥 1−(𝑥2)2 𝑑𝑥 = 1 2 𝑠𝑖𝑛−1𝑥2 + 𝑐 3- 𝑥𝑠𝑒𝑐−1𝑥 𝑑𝑥 = u = sec−1 𝑥 𝑑𝑢 = 𝑑𝑥 𝑥 𝑥2−1 dv = xdx v = 𝑥2 2 ∴ 𝑥𝑠𝑒𝑐−1𝑥 𝑑𝑥 = sec−1 𝑥 𝑥2 2 − 𝑥2 2 𝑑𝑥 𝑥 𝑥2−1 = sec−1 𝑥 𝑥2 2 − 1 2 𝑥𝑑𝑥 𝑥 𝑥2−1 = sec−1 𝑥 𝑥2 2 − 1 2 ∗ 1 2 ∗ (𝑥2−1) 1 2 1 2 +c=sec−1 𝑥 𝑥2 2 − 1 2 ∗ 𝑥2 − 1 + 𝑐
- 109. 4-. tan−1 𝑥 𝑑𝑥 = u= tan−1 𝑥 du= 𝑑𝑥 1+𝑥2 dv= dx v= 𝑥 tan−1 𝑥 𝑑𝑥 = tan−1 𝑥 ∗ 𝑥 − 𝑥 𝑑𝑥 1 + 𝑥2 = tan−1 𝑥 ∗ 𝑥 − 1 2 𝑙𝑛 1 + 𝑥2 + 𝑐