{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

lecture_notes2

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

This preview shows pages 1–6. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 ) 0 , a x b . R x ) x ( f n x b 2 x 1 x 0 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, 0 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 0 = 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 x area of region R n X i =1 f ( x i x : Riemann sum approximately area of region R = lim n →∞ n X i =1 f ( x i x

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
4 ex f ( x ) = x , 0 x 1 1 0 R x ) x ( f y R = { ( x, y ) : 0 x 1 , 0 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 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 x = lim n →∞ n + 1 2 n = lim n →∞ 1 2 + 1 2 n ! = 1 2 ok
5 5.2 definite integral As before, given f ( x ) , a x b , set Δ x = b - a n , x i = a + i Δ x . Let x * i be any point such that x i - 1 x * i x i . Then n X i =1 f ( x * i x is a Riemann sum . ex x * i = x i : right-hand RS x * i = x i - 1 : left-hand RS x * i = x i - 1 + x i 2 : midpoint RS def : Z b a f ( x ) dx = lim n →∞ n X i =1 f ( x * i x : definite integral = area work probability · · · ex Z 1 0 dx = lim n →∞ n X i =1 1 · 1 n = lim n →∞ 1 = 1 a = 0 , b = 1 , Δ x = b - a n = 1 n , x i = a + i Δ x = i n , f ( x ) = 1 Z 1 0 x dx = lim n →∞ n X i =1 i n · 1 n = lim n →∞ 1 n 2 · n ( n + 1) 2 = 1 2 f ( x ) = x , f ( x i ) = x i = i n Z 1 0 x 2 dx = lim n →∞ n X i =1 i n !

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}