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Unformatted text preview: 1 preview : integrals (theory/applications), differential equations, series 1 Tues 9 / 7 1st main goal : prepare for FTC Appendix E. : sigma notation def : n X i =1 a i = a 1 + a 2 + a 3 + ··· + a n- 1 + a n : sum , series i : index , a i : terms , 1 , n : limits ex 5 X i =1 i = 1 + 2 + 3 + 4 + 5 = 15 5 X i =1 i 2 = 1 + 4 + 9 + 16 + 25 = 55 note : A series can be written in different ways. ex 5 X i =1 i = 4 X j =0 ( j + 1) = 1 + 2 + 3 + 4 + 5 = 15 ok set j = i- 1 thm 1. n X i =1 ca i = c · n X i =1 a i , where c is any constant 2. n X i =1 ( a i + b i ) = n X i =1 a i + n X i =1 b i 3. n X i =1 a i b i 6 = n X i =1 a i · n X i =1 b i pf 1. n X i =1 ca i = ca 1 + ca 2 + ··· + ca n = c ( a 1 + a 2 + ··· + a n ) = c · n X i =1 a i ok 2. hw 3. example ( n = 2) 2 X i =1 a i b i = a 1 b 1 + a 2 b 2 2 X i =1 a i · 2 X i =1 b i = ( a 1 + a 2 ) · ( b 1 + b 2 ) = a 1 b 1 + a 1 b 2 + a 2 b 1 + a 2 b 2 ok 2 thm n X i =1 ( a i +1- a i ) = a n +1- a 1 : telescoping series pf n X i =1 ( a i +1- a i ) = ( a 2- a 1 ) + ( a 3- a 2 ) + ( a 4- a 3 ) + ··· + ( a n- a n- 1 ) + ( a n +1- a n ) =- a 1 + a n +1 ok thm 1. n X i =1 1 = 1 + 1 + 1 + ··· + 1 = n 2. n X i =1 i = 1 + 2 + 3 + ··· + n = n ( n + 1) 2 ( n = 5 ⇒ S = 15) implies ↓ 3. n X i =1 i 2 = 1 + 4 + 9 + ··· + n 2 = n ( n + 1)(2 n + 1) 6 ( n = 5 ⇒ S = 55) pf 1. ok 2. ( i + 1) 2- i 2 = i 2 + 2 i + 1- i 2 = 2 i + 1 n X i =1 (( i + 1) 2- i 2 ) = n X i =1 (2 i + 1) = 2 n X i =1 i + n X i =1 1 = 2 S + n , where S = n X i =1 i n X i =1 (( i + 1) 2- i 2 ) = (2 2- 1 2 ) + (3 2- 2 2 ) + (4 2- 3 2 ) + ··· + (( n + 1) 2- n 2 ) = ( n + 1) 2- 1 2 = n 2 + 2 n ⇒ 2 S + n = n 2 + 2 n ⇒ 2 S = n 2 + n = n ( n + 1) ⇒ S = n ( n + 1) 2 ok 3. hw , ( i + 1) 3- i 3 = ··· 3 5.1 area 2 Wed 9 / 8 Given f ( x ) ≥ , a ≤ x ≤ b . R x ) x ( f n x b 2 x 1 x x a y R = region in the xy-plane between y = 0 and y = f ( x ) for a ≤ x ≤ b = { ( x, y ) : a ≤ x ≤ b, ≤ y ≤ f ( x ) } problem : find the area of R solution : approximate by rectangles choose n ≥ 1 , set Δ x = b- a n , x i = a + i Δ x , i = 0 , . . . , n x = a x 1 = a + Δ x x 2 = a + 2Δ x . . . x n = a + n Δ x = a + b- a = b area of i th rectangle = f ( x i )Δ x area of region R ≈ n X i =1 f ( x i )Δ x : Riemann sum ↑ approximately area of region R = lim n →∞ n X i =1 f ( x i )Δ x 4 ex f ( x ) = x , ≤ x ≤ 1 1 R x ) x ( f y R = { ( x, y ) : 0 ≤ x ≤ 1 , ≤ y ≤ x } of course , area = 1 2 · base · height = 1 2 a = 0 , b = 1 , Δ x = b- a n = 1 n x i = a + i Δ x = i n f ( x i ) = x i = i n n X i =1 f ( x i )Δ x = n X i =1 i n · 1 n = 1 n 2 n X i =1 i = 1 n 2 · n ( n + 1) 2 = n + 1 2 n area = lim n →∞ n X i =1 f ( x i )Δ x = lim n →∞ n + 1 2 n = lim n →∞ 1 2 + 1 2 n !...
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This note was uploaded on 01/28/2011 for the course CHEM 210 taught by Professor Kiste during the Spring '10 term at Michigan Flint.

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lecture_notes2 - 1 preview integrals(theory/applications...

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