A = (a + 1) 2 = a 2 + 2a + 1

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Ralf Ferguson
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1 A = (a + 1) 2 = a 2 + 2a + 1 1

2 A = ( (a + b) + 1 ) 2 = (a + b) 2 + 2(a + b) + 1 = a 2 + 2ab + b 2 + 2a + 2b + 1 A = ( (a + b) + 1 ) 2 = (a + b) 2 + 2(a + b) + 1 = a 2 + 2ab + b 2 + 2a + 2b A = ( (a + b) + 1 ) 2 = (a + b) 2 + 2(a + b) + 1 A = (a + 1) 2 = a 2 + 2a + 1 A = (a + 1) 2 = a 2 + 2a + 1

3 A = (a + 1) 2 = a 2 + 2a + 1 A = (a + 1) 2 = a 2 + 2a + 1 A = ( (a + b) + 1 ) 2 = (a + b) 2 + 2(a + b) + 1 = a 2 + 2ab + b 2 + 2a + 2b + 1 A = (a + 1) 2 = a 2 + 2a + 1

4 A = (a + 1) 2 = a 2 + 2a + 1 A = ( (a + b) + 1 ) 2 = (a + b) 2 + 2(a + b) + 1 = a 2 + 2ab + b 2 + 2a + 2b (x + 1)2 dx = 1 0 (x2 + 2x + 1)dx = 1 0 x2 dx xdx dx = = (x + 1) 2 dx = = (x 2 + 2x + 1)dx x 2 dx + 2 = xdx dx = 7 3

5 (cos x + sin x) 2 = cos 2 x + 2 cos x sin x + sin 2 x = cos 2 x + sin 2 x + 2 sin x cos x = 1 + sin(2x) n n (x i + 1) 2 = (x 2 i + 2x i + 1) i=1 = i=1 n n x 2 i + 2 x i + n i=1 i=1 f = ( x (x + 1) 2) = ( ) x x 2 + 2x + 1

6 A = (a + b) 2 = a 2 + 2ab + b 2 0 ( I = ln 1 + tan ( π 4 u)) ( du) = = = = = π 4 π 4 0 π 4 0 π 4 0 π 4 0 π 4 0 ( ln 1 + tan ( π 4 u)) du ( ln tan u ) du 1 + tan u ( 1 + tan u + 1 tan u ln 1 + tan u ( ) 2 ln du 1 + tan u ( ln 2 ln(1 + tan u) ) du = π 4 ln 2 π 4 = π 4 ln 2 I 0 ln(1 + tan u) du ) du r l

7 I = = = = = = 0 π 4 π 4 0 π 4 0 π 4 0 π 4 0 π 4 0 ( ln 1 + tan ( π 4 u)) ( du) ( ln 1 + tan ( π 4 u)) du ( ln tan u ) du 1 + tan u ( 1 + tan u + 1 tan u ln 1 + tan u ( ) 2 ln du 1 + tan u ( ln 2 ln(1 + tan u) ) du = π 4 ln 2 π 4 = π 4 ln 2 I 0 ln(1 + tan u) du ) du 2 i (a + b)(a + ib)(a b)(a ib) = (a + b)(a b) (a + ib)(a ib) = (a 2 b 2 )(a 2 + b 2 ) = a 4 b 4 (x y)(x + y) = x 2 y 2 2xy 3y = x 2x(K y 0 + Ky 0) 3Ky 0 = x 2xK y 0 + K(2xy 0 3y 0 ) = x 2xK y 0 = x 2xK x 3 2 = x 1 2 K = 1 2x 2 K = 1 2x y 0 x a i(a) f(a) a b a b i(a), f(a) i(b), f(b)

8 A = B = C + D = D = E + F + G + H + I = K + L + M = N = O A = (a + 1) 2 = a 2 + 2a + 1 F = 1 2 G = H K = K

9 φ(x, y) = 0 (x + y) 2 + (x + 2y) 2 = 0 { x + y = 0 x + 2y = 0 x y φ(x, y) = 0 (x + y) 2 + (x + 2y) 2 = 0 { x + y = 0 x + 2y = 0 x y A = (a + 1) 2 = a 2 + 2a + 1

10 { f(x) = 3x 3 + 2x 2 x + 4 g(x) = 5x 2 5x + 6 n n (x i + 1) 2 = (x 2 i + 2x i + 1) i=1 = i=1 n n x 2 i + 2 x i + n i=1 i=1 n n (x i + 1) 2 = (x 2 i + 2x i + 1) i=1 = i=1 n n x 2 i + 2 x i + n i=1 i=1 φ(x, y) = 0 (x + 2y) 2 + (2x + 4y) 2 = 0 { x + 2y = 0 2x + 4y = 0 { x + 2y = 0 x + 2y = 0 x + 2y = 0

11 φ(x, y) = 0 (x + 2y) 2 + (2x + 4y) 2 = 0 { x + 2y = 0 2x + 4y = 0 { x + 2y = 0 2 x + 2y = 0 x + 2y = 0 φ(x, y) = 0 (x + 2y) 2 + (2x + 4y) 2 = 0 { x + 2y = 0 2x + 4y = 0 { x + 2y = 0 x + 2y = 0 x + 2y = φ(x, y) = 0 (x + 2y) 2 + (2x + 4y) 2 = 0 { x + 2y = 0 2x + 4y = 0 { x + 2y = 0 2 x + 2y = 0 x + 2y = 0

12 φ(x, y) = 0 (x + 2y) 2 + (2x + 4y) 2 = 0 { x + 2y = 0 2x + 4y = 0 { x + 2y = 0 2 x + 2y = 0 x + 2y = 0 A = B = C = D = E = F

13 A B + B + B + B + B + B + B + B + B + B + B + B + B { C D E F G H + H + H + H + H + H + H { J K I L M { N O P Q 0 A B + B + B + B + B + B + B + B + B + B + B + B + B { C D E F G H + H + H + H + H + H + H { J K I L M { N O P Q

14 A = B = C A = B = C A = (a + 1) 2 = a 2 + 2a + 1 A = (a + 1) 2 = a 2 + 2a + 1

15 A = (a + 1) 2 = a 2 + 2a + 1 A = (a + 1) 2 = a 2 + 2a + 1

16 ( S n = 1 n 1 ) n R ( ) e i π k 2n k=0 ) n ( = 1 ( 1 e i π n R 2n 1 e i π 2n = 1 n R ( 1 i 1 e i π 2n ) ) e i 2π n

17 E 3(2x + 4) = 6 2x + 4 = 2 2x = 2 2x = A = (a + b) 2 = a 2 + b 2 + 2ab f(x) g(x) f(x) 2 g(x) 2 f(x) 2 g(x) 2 0

18 arccos(x) = arcsin arcsin 5 13 x = sin ( arcsin arcsin 5 13 x = cos arcsin x = ( ) ) 13 cos arcsin ( ) [ π 2, π 2 ] x [ 1, 1], cos(arcsin x) = 1 x 2 E 3(2x + 4) = 6 2x + 4 = 2 2x = 2 2x =

19 S n = 1 n = 1 n n 1 cos ( π 2 ) k n k=0 n 1 k=0 ( R ( = 1 n 1 n R k=0 e i kπ 2n e i kπ 2n ) ) ( = 1 n 1 ) n R ( ) e i π k 2n k=0 ) n ( = 1 ( 1 e i π n R 2n 1 e i π 2n = 1 n R ( 1 i 1 e i π 2n ) ) cos x = R(e ix ) R(z + z ) = R(z) + R(z ) exp + e i 2π n

20 E (x+4) 3 + 5x+3 5 = 7 5(x + 4) + 3(5x + 3) = x x + 9 = x + 29 = x = 76 x = a. f E b. f 0 c. f d. K > 0 x E f(x) K x e. f

21 a. f E b. f 0 c. f d. K > 0 x E f(x) K x e. f

24 x

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Section 5.5 More Integration Formula (The Substitution Method) 2 Lectures. Dr. Abdulla Eid. College of Science. MATHS 101: Calculus I

Section 5.5 More Integration Formula (The Substitution Method) 2 Lectures College of Science MATHS : Calculus I (University of Bahrain) Integrals / 7 The Substitution Method Idea: To replace a relatively