Example A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equal squares from each corner and then folding


 Candice Charles
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1 1 Exmple A rectngulr box without lid is to be mde from squre crdbord of sides 18 cm by cutting equl squres from ech corner nd then folding up the sides.
2 1 Exmple A rectngulr box without lid is to be mde from squre crdbord of sides 18 cm by cutting equl squres from ech corner nd then folding up the sides.
3 1 Exmple A rectngulr box without lid is to be mde from squre crdbord of sides 18 cm by cutting equl squres from ech corner nd then folding up the sides. Find the length of the side of the squre tht must be cut off
4 1 Exmple A rectngulr box without lid is to be mde from squre crdbord of sides 18 cm by cutting equl squres from ech corner nd then folding up the sides. Find the length of the side of the squre tht must be cut off if the volume of the box is to be mximized.
5 1 Exmple A rectngulr box without lid is to be mde from squre crdbord of sides 18 cm by cutting equl squres from ech corner nd then folding up the sides. Find the length of the side of the squre tht must be cut off if the volume of the box is to be mximized. Wht is the mximum volume?
6 1 Exmple A rectngulr box without lid is to be mde from squre crdbord of sides 18 cm by cutting equl squres from ech corner nd then folding up the sides. Find the length of the side of the squre tht must be cut off if the volume of the box is to be mximized. Wht is the mximum volume? Anlysis Suppose the side is x.
7 1 Exmple A rectngulr box without lid is to be mde from squre crdbord of sides 18 cm by cutting equl squres from ech corner nd then folding up the sides. Find the length of the side of the squre tht must be cut off if the volume of the box is to be mximized. Wht is the mximum volume? Anlysis Suppose the side is x. Cn express length, width nd height in terms of x.
8 1 Exmple A rectngulr box without lid is to be mde from squre crdbord of sides 18 cm by cutting equl squres from ech corner nd then folding up the sides. Find the length of the side of the squre tht must be cut off if the volume of the box is to be mximized. Wht is the mximum volume? Anlysis Suppose the side is x. Cn express length, width nd height in terms of x. Cn express volume V in terms of x.
9 1 Exmple A rectngulr box without lid is to be mde from squre crdbord of sides 18 cm by cutting equl squres from ech corner nd then folding up the sides. Find the length of the side of the squre tht must be cut off if the volume of the box is to be mximized. Wht is the mximum volume? Anlysis Suppose the side is x. Cn express length, width nd height in terms of x. Cn express volume V in terms of x. There re restrictions on x (domin of the function V)
10 1 Exmple A rectngulr box without lid is to be mde from squre crdbord of sides 18 cm by cutting equl squres from ech corner nd then folding up the sides. Find the length of the side of the squre tht must be cut off if the volume of the box is to be mximized. Wht is the mximum volume? Anlysis Suppose the side is x. Cn express length, width nd height in terms of x. Cn express volume V in terms of x. There re restrictions on x (domin of the function V) Mximize V(x) for x belonging to the domin.
11 18 2
12 2 Solution Let length of side of the squre to be cut off be x cm. x 18 x
13 2 Solution Let length of side of the squre to be cut off be x cm. x x 18 2x x
14 2 Solution Let length of side of the squre to be cut off be x cm. The volume V, in cm 3, of the open box is V(x) = x(18 2x) 2, 0 < x < 9 18 x 18 2x 18 2x x
15 2 Solution Let length of side of the squre to be cut off be x cm. The volume V, in cm 3, of the open box is V(x) = x(18 2x) 2, 0 < x < 9 18 x = x(324 72x + 4x 2 ) 18 2x 18 2x x
16 2 Solution Let length of side of the squre to be cut off be x cm. The volume V, in cm 3, of the open box is V(x) = x(18 2x) 2, 0 < x < 9 18 x = x(324 72x + 4x 2 ) 18 2x = 324x 72x 2 + 4x x x
17 2 Solution Let length of side of the squre to be cut off be x cm. The volume V, in cm 3, of the open box is V(x) = x(18 2x) 2, 0 < x < 9 18 x = x(324 72x + 4x 2 ) 18 2x = 324x 72x 2 + 4x x V (x) = x + 12x 2 x
18 2 Solution Let length of side of the squre to be cut off be x cm. The volume V, in cm 3, of the open box is V(x) = x(18 2x) 2, 0 < x < 9 18 x = x(324 72x + 4x 2 ) 18 2x = 324x 72x 2 + 4x x V (x) = x + 12x 2 = 12(x 2 12x + 27) = 12(x 3)(x 9) x
19 2 Solution Let length of side of the squre to be cut off be x cm. The volume V, in cm 3, of the open box is V(x) = x(18 2x) 2, 0 < x < 9 18 x = x(324 72x + 4x 2 ) 18 2x = 324x 72x 2 + 4x x V (x) = x + 12x 2 = 12(x 2 12x + 27) = 12(x 3)(x 9) There is only one criticl point in (0, 9) x = 3 (x = 9 is rejected; not belong to the domin of V) x
20 2 Solution Let length of side of the squre to be cut off be x cm. The volume V, in cm 3, of the open box is V(x) = x(18 2x) 2, 0 < x < 9 18 x = x(324 72x + 4x 2 ) 18 2x = 324x 72x 2 + 4x x V (x) = x + 12x 2 = 12(x 2 12x + 27) = 12(x 3)(x 9) There is only one criticl point in (0, 9) x = 3 (x = 9 is rejected; not belong to the domin of V) x V (x) = x,
21 2 Solution Let length of side of the squre to be cut off be x cm. The volume V, in cm 3, of the open box is V(x) = x(18 2x) 2, 0 < x < 9 18 x = x(324 72x + 4x 2 ) 18 2x = 324x 72x 2 + 4x x V (x) = x + 12x 2 = 12(x 2 12x + 27) = 12(x 3)(x 9) There is only one criticl point in (0, 9) x = 3 (x = 9 is rejected; not belong to the domin of V) x V (x) = x, V (3) = 72
22 2 Solution Let length of side of the squre to be cut off be x cm. The volume V, in cm 3, of the open box is V(x) = x(18 2x) 2, 0 < x < 9 18 x = x(324 72x + 4x 2 ) 18 2x = 324x 72x 2 + 4x x V (x) = x + 12x 2 = 12(x 2 12x + 27) = 12(x 3)(x 9) There is only one criticl point in (0, 9) x = 3 (x = 9 is rejected; not belong to the domin of V) x V (x) = x, V (3) = 72 < 0
23 2 Solution Let length of side of the squre to be cut off be x cm. The volume V, in cm 3, of the open box is V(x) = x(18 2x) 2, 0 < x < 9 18 x = x(324 72x + 4x 2 ) 18 2x = 324x 72x 2 + 4x x V (x) = x + 12x 2 = 12(x 2 12x + 27) = 12(x 3)(x 9) There is only one criticl point in (0, 9) x = 3 (x = 9 is rejected; not belong to the domin of V) x V (x) = x, V (3) = 72 < 0 By 2nd deriv. test (sp ver), V hs mx t x = 3.
24 2 Solution Let length of side of the squre to be cut off be x cm. The volume V, in cm 3, of the open box is V(x) = x(18 2x) 2, 0 < x < 9 18 x = x(324 72x + 4x 2 ) 18 2x = 324x 72x 2 + 4x x V (x) = x + 12x 2 = 12(x 2 12x + 27) = 12(x 3)(x 9) There is only one criticl point in (0, 9) x = 3 (x = 9 is rejected; not belong to the domin of V) x V (x) = x, V (3) = 72 < 0 By 2nd deriv. test (sp ver), V hs mx t x = 3. Mximum volume is V(3) = 432 cm 3.
25 2 Solution Let length of side of the squre to be cut off be x cm. The volume V, in cm 3, of the open box is V(x) = x(18 2x) 2, 0 < x < 9 18 x = x(324 72x + 4x 2 ) 18 2x = 324x 72x 2 + 4x x V (x) = x + 12x 2 = 12(x 2 12x + 27) = 12(x 3)(x 9) There is only one criticl point in (0, 9) x = 3 (x = 9 is rejected; not belong to the domin of V) V (x) = x, V (3) = 72 < 0 By 2nd deriv. test (sp ver), V hs mx t x = 3. Mximum volume is V(3) = 432 cm x
26 3 Exmple A frmer wnts to construct rectngulr pen next to brn 40 m long, using ll of the brn s prt of one side of the pen. Find the dimensions of the pen with the lrgest re tht the frmer cn build if 200 m of fencing mteril is vilble.
27 3 Exmple A frmer wnts to construct rectngulr pen next to brn 40 m long, using ll of the brn s prt of one side of the pen. Find the dimensions of the pen with the lrgest re tht the frmer cn build if 200 m of fencing mteril is vilble. Pen 40 Brn
28 3 Exmple A frmer wnts to construct rectngulr pen next to brn 40 m long, using ll of the brn s prt of one side of the pen. Find the dimensions of the pen with the lrgest re tht the frmer cn build if 200 m of fencing mteril is vilble. x Pen 40 Brn Let x be length of side (in meters) of pen tht is prllel to brn
29 3 Exmple A frmer wnts to construct rectngulr pen next to brn 40 m long, using ll of the brn s prt of one side of the pen. Find the dimensions of the pen with the lrgest re tht the frmer cn build if 200 m of fencing mteril is vilble. x Pen 40 Brn Let x be length of side (in meters) of pen tht is prllel to brn Note tht x 40
30 3 Exmple A frmer wnts to construct rectngulr pen next to brn 40 m long, using ll of the brn s prt of one side of the pen. Find the dimensions of the pen with the lrgest re tht the frmer cn build if 200 m of fencing mteril is vilble. x Pen y 40 Brn Let x be length of side (in meters) of pen tht is prllel to brn Note tht x 40 Let y be length of side to brn.
31 Denote the re of the pen by A 4
32 4 Denote the re of the pen by A Note tht A = xy
33 4 Denote the re of the pen by A Note tht A = xy Moreover, x + (x 40) + 2y = 200
34 4 Denote the re of the pen by A Note tht A = xy Moreover, x + (x 40) + 2y = 200 2x + 2y =
35 4 Denote the re of the pen by A Note tht A = xy Moreover, x + (x 40) + 2y = 200 2x + 2y = y = 120 x
36 4 Denote the re of the pen by A Note tht A = xy Moreover, x + (x 40) + 2y = 200 2x + 2y = y = 120 x Therefore, A = x(120 x)
37 4 Denote the re of the pen by A Note tht A = xy Moreover, x + (x 40) + 2y = 200 2x + 2y = y = 120 x Therefore, A = x(120 x) 40 x < 120
38 4 Denote the re of the pen by A Note tht A = xy Moreover, x + (x 40) + 2y = 200 2x + 2y = y = 120 x Therefore, A = x(120 x) 40 x < 120 = 120x x 2
39 4 Denote the re of the pen by A Note tht A = xy Moreover, x + (x 40) + 2y = 200 2x + 2y = y = 120 x Therefore, A = x(120 x) 40 x < 120 = 120x x 2 Differentiting A (x) = 120 2x
40 4 Denote the re of the pen by A Note tht A = xy Moreover, x + (x 40) + 2y = 200 2x + 2y = y = 120 x Therefore, A = x(120 x) 40 x < 120 = 120x x 2 Differentiting A (x) = 120 2x = 2(60 x)
41 4 Denote the re of the pen by A Note tht A = xy Moreover, x + (x 40) + 2y = 200 2x + 2y = y = 120 x Therefore, A = x(120 x) 40 x < 120 = 120x x 2 Differentiting A (x) = 120 2x = 2(60 x) Criticl point of A is x 1 = 60
42 4 Denote the re of the pen by A Note tht A = xy Moreover, x + (x 40) + 2y = 200 2x + 2y = y = 120 x Therefore, A = x(120 x) 40 x < 120 = 120x x 2 Differentiting A (x) = 120 2x = 2(60 x) Criticl point of A is x 1 = 60 Remrk Hve to check whether A hs mx t x 1
43 5 Method 1 x (40, 60) 60 (60, 120) A (x) 0 A
44 5 Method 1 x (40, 60) 60 (60, 120) A (x) + 0 A
45 5 Method 1 x (40, 60) 60 (60, 120) A (x) + 0 A
46 5 Method 1 x (40, 60) 60 (60, 120) A (x) + 0 A
47 5 Method 1 x (40, 60) 60 (60, 120) A (x) + 0 A
48 5 Method 1 x (40, 60) 60 (60, 120) A (x) + 0 A Method 2 A (x) = 2
49 5 Method 1 x (40, 60) 60 (60, 120) A (x) + 0 A Method 2 A (x) = 2 A (60) < 0 nd 60 is the only criticl pt
50 5 Method 1 x (40, 60) 60 (60, 120) A (x) + 0 A Method 2 A (x) = 2 A (60) < 0 nd 60 is the only criticl pt Therefore, A hs globl mximum when x = 60
51 5 Method 1 x (40, 60) 60 (60, 120) A (x) + 0 A Method 2 A (x) = 2 A (60) < 0 nd 60 is the only criticl pt Therefore, A hs globl mximum when x = 60 When x = 60, y = 60
52 5 Method 1 x (40, 60) 60 (60, 120) A (x) + 0 A Method 2 A (x) = 2 A (60) < 0 nd 60 is the only criticl pt Therefore, A hs globl mximum when x = 60 When x = 60, y = 60 The dimensions of the lrgest pen re 60 m 60 m
53 6 Chpter 6: Integrtion Definite Integrls Indefinite Integrls Applictions of Definite Integrls Objectives To evlute definite integrls using Fundmentl Theorem of Integrl Clculus. To find indefinite integrls by formul. To find re of simple regions using definite integrls.
54 7 Problem Find the re of the region bounded by the curve y = x 2, the xxis nd the verticl line x =
55 7 Problem Find the re of the region bounded by the curve y = x 2, the xxis nd the verticl line x = 1. 1 Method Divide [0, 1] into n equl subintervls. [0, 1 ], n [1, 2 1 ],,..., [n, 1] n n n 1
56 7 Problem Find the re of the region bounded by the curve y = x 2, the xxis nd the verticl line x = 1. 1 Method Divide [0, 1] into n equl subintervls. [0, 1 ], n [1, 2 1 ],,..., [n, 1] n n n On ech subintervl, consider the lrgest rectngulr regions inside the given region, height of the ith region is f ( i 1 n )
57 7 Problem Find the re of the region bounded by the curve y = x 2, the xxis nd the verticl line x = 1. 1 Method Divide [0, 1] into n equl subintervls. [0, 1 ], n [1, 2 1 ],,..., [n, 1] n n n On ech subintervl, consider the lrgest rectngulr regions inside the given region, height of the ith region is f ( i 1 n ). Sum of res of rectngulr regions is n + 1 6n 2 Are of given region is 1/
58 7 Problem Find the re of the region bounded by the curve y = x 2, the xxis nd the verticl line x = 1. 1 Method Divide [0, 1] into n equl subintervls. [0, 1 ], n [1, 2 1 ],,..., [n, 1] n n n On ech subintervl, consider the lrgest rectngulr regions inside the given region, height of the ith region is f ( i 1 n ). Sum of res of rectngulr regions is n + 1 6n 2 Are of given region is 1/ Approximtion from below 1
59 1 8 Question Cn we use pproximtion from bove? 1
60 1 8 Question Cn we use pproximtion from bove? 1 tke n rbitrry point t i in ech subintervl nd use f (t i ) s height? 1 1
61 1 8 Question Cn we use pproximtion from bove? 1 tke n rbitrry point t i in ech subintervl nd use f (t i ) s height? 1 use unequl subintervls? 1
62 1 8 Question Cn we use pproximtion from bove? 1 tke n rbitrry point t i in ech subintervl nd use f (t i ) s height? 1 use unequl subintervls? Remrk Cn replce f (x) = x 2 by ny continuous function 1 [0, 1] by ny closed intervl.
63 Definition Let f be defined nd continuous on closed intervl [, b]. 9
64 9 Definition Let f be defined nd continuous on closed intervl [, b]. Tke finitely mny points in [, b]: = x 0 < x 1 < < x n 1 < x n = b = x 0 x 1 x n = b
65 9 Definition Let f be defined nd continuous on closed intervl [, b]. Tke finitely mny points in [, b]: = x 0 < x 1 < < x n 1 < x n = b These points divide [, b] into subintervls [x 0, x 1 ], [x 1, x 2 ],, [x n 1, x n ] = x 0 x 1 x n = b
66 9 Definition Let f be defined nd continuous on closed intervl [, b]. Tke finitely mny points in [, b]: = x 0 < x 1 < < x n 1 < x n = b These points divide [, b] into subintervls [x 0, x 1 ], [x 1, x 2 ],, [x n 1, x n ] For ech i = 1,..., n, tke n rbitrry point t i in the subintervl [x i 1, x i ]. t 1 = x 0 x 1 x n = b
67 9 Definition Let f be defined nd continuous on closed intervl [, b]. Tke finitely mny points in [, b]: = x 0 < x 1 < < x n 1 < x n = b These points divide [, b] into subintervls [x 0, x 1 ], [x 1, x 2 ],, [x n 1, x n ] For ech i = 1,..., n, tke n f (t 1 ) rbitrry point t i in the subintervl [x i 1, x i ]. Consider the sum t 1 = x 0 x 1 n f (t i ) x i := f (t 1 ) x 1 + f (t 2 ) x f (t n ) x n i=1 where x i = x i x i 1 is the length of the ith subintervl. x n = b
68 10 Cn show tht there exists rel number I (which is unique) such tht n ( ) f (t i ) x i is close to I whenever x 1,..., x n re smll enough. i=1
69 10 Cn show tht there exists rel number I (which is unique) such tht n ( ) f (t i ) x i is close to I whenever x 1,..., x n re smll enough. i=1 The number I is clled the definite integrl of f from to b, denoted by f (x) dx
70 10 Cn show tht there exists rel number I (which is unique) such tht n ( ) f (t i ) x i is close to I whenever x 1,..., x n re smll enough. i=1 The number I is clled the definite integrl of f from to b, denoted by f (x) dx b = upper limit of integrtion = lower limit
71 10 Cn show tht there exists rel number I (which is unique) such tht n ( ) f (t i ) x i is close to I whenever x 1,..., x n re smll enough. i=1 The number I is clled the definite integrl of f from to b, denoted by f (x) dx b = upper limit of integrtion = lower limit Remrk If f is nonnegtive, f (x) dx is the re under grph of f for x b. t 1 = x 0 x 1 x n = b
72 10 Cn show tht there exists rel number I (which is unique) such tht n ( ) f (t i ) x i is close to I whenever x 1,..., x n re smll enough. i=1 The number I is clled the definite integrl of f from to b, denoted by f (x) dx b = upper limit of integrtion = lower limit Remrk If f is nonnegtive, f (x) dx is the re under grph of f for x b. If f is somewhere negtive, definite integrl does not men re. t 1 = x 0 x 1 x n = b
73 11 Exmple 1 0 x 2 dx = 1 3 (cn be found esily using Fundmentl Theorem)
74 11 Exmple 1 0 x 2 dx = 1 3 (cn be found esily using Fundmentl Theorem) Nottion Cn use other symbols, sy t, insted of x, eg. f (t) dt
75 11 Exmple 1 0 x 2 dx = 1 3 (cn be found esily using Fundmentl Theorem) Nottion Cn use other symbols, sy t, insted of x, eg. For convenience, we define f (x) dx = 0 f (t) dt
76 11 Exmple 1 0 x 2 dx = 1 3 (cn be found esily using Fundmentl Theorem) Nottion Cn use other symbols, sy t, insted of x, eg. For convenience, we define f (x) dx = 0 f (t) dt b f (x) dx = f (x) dx
77 11 Exmple 1 0 x 2 dx = 1 3 (cn be found esily using Fundmentl Theorem) Nottion Cn use other symbols, sy t, insted of x, eg. For convenience, we define f (x) dx = 0 f (t) dt b f (x) dx = f (x) dx Exmple x 2 dx = 0 x 2 dx
78 11 Exmple 1 0 x 2 dx = 1 3 (cn be found esily using Fundmentl Theorem) Nottion Cn use other symbols, sy t, insted of x, eg. For convenience, we define f (x) dx = 0 f (t) dt b f (x) dx = f (x) dx Exmple x 2 dx = 0 x 2 dx = 1 3
79 12 Properties Let f, g : [, b] R be functions such tht both (1) g(x) dx exist. Let α R nd < c < b. Then [ f (x) + g(x) ] dx = f (x) dx + g(x) dx f (x) dx nd
80 12 Properties Let f, g : [, b] R be functions such tht both (1) g(x) dx exist. Let α R nd < c < b. Then [ f (x) + g(x) ] dx = f (x) dx + g(x) dx Reson Follows from the corresponding result for limits. f (x) dx nd
81 12 Properties Let f, g : [, b] R be functions such tht both (1) g(x) dx exist. Let α R nd < c < b. Then [ f (x) + g(x) ] dx = f (x) dx + g(x) dx Reson Follows from the corresponding result for limits. f (x) dx nd (2) α f (x) dx = α f (x) dx
82 12 Properties Let f, g : [, b] R be functions such tht both (1) g(x) dx exist. Let α R nd < c < b. Then [ f (x) + g(x) ] dx = f (x) dx + g(x) dx Reson Follows from the corresponding result for limits. f (x) dx nd (2) α f (x) dx = α f (x) dx (3) f (x) dx = c f (x) dx + c f (x) dx
83 12 Properties Let f, g : [, b] R be functions such tht both (1) g(x) dx exist. Let α R nd < c < b. Then [ f (x) + g(x) ] dx = f (x) dx + g(x) dx Reson Follows from the corresponding result for limits. f (x) dx nd (2) α f (x) dx = α f (x) dx (3) f (x) dx = c f (x) dx + c f (x) dx Reson Use res.
84 To find definite integrls using definition is very tedious. 13
85 13 To find definite integrls using definition is very tedious. Fundmentl Theorem of Integrl Clculus A convenient wy to evlute definite integrls (provided tht primitives / ntiderivtives cn be found).
86 13 To find definite integrls using definition is very tedious. Fundmentl Theorem of Integrl Clculus A convenient wy to evlute definite integrls (provided tht primitives / ntiderivtives cn be found). It is quite surprising tht integrtion is relted to differentition.
87 13 To find definite integrls using definition is very tedious. Fundmentl Theorem of Integrl Clculus A convenient wy to evlute definite integrls (provided tht primitives / ntiderivtives cn be found). It is quite surprising tht integrtion is relted to differentition. To tckle the problem, to find f (x) dx introduce n uxiliry function F : [, b] R defined by F(x) = x f (t) dt
88 13 To find definite integrls using definition is very tedious. Fundmentl Theorem of Integrl Clculus A convenient wy to evlute definite integrls (provided tht primitives / ntiderivtives cn be found). It is quite surprising tht integrtion is relted to differentition. To tckle the problem, to find f (x) dx introduce n uxiliry function F : [, b] R defined by F(x) = x f (t) dt x b
89 13 To find definite integrls using definition is very tedious. Fundmentl Theorem of Integrl Clculus A convenient wy to evlute definite integrls (provided tht primitives / ntiderivtives cn be found). It is quite surprising tht integrtion is relted to differentition. To tckle the problem, to find f (x) dx introduce n uxiliry function F : [, b] R defined by F(x) = The required integrl x f (t) dt f (x) dx is F(b). x b
90 Show tht F (x) = f (x) (ide only) 14
91 14 Show tht F (x) = f (x) (ide only) Recll tht F (x) = lim h 0 F(x + h) F(x) h
92 14 Show tht F (x) = f (x) (ide only) Recll tht F (x) = lim h 0 F(x + h) F(x) h By construction F(x + h) F(x) = x+h f (t) dt x f (t) dt
93 14 Show tht F (x) = f (x) (ide only) Recll tht F (x) = lim h 0 F(x + h) F(x) h By construction F(x + h) F(x) = x+h f (t) dt x f (t) dt = x+h x f (t) dt
94 14 Show tht F (x) = f (x) (ide only) Recll tht F (x) = lim h 0 F(x + h) F(x) h By construction F(x + h) F(x) = x+h f (t) dt x f (t) dt = x+h x f (t) dt x x + h b
95 14 Show tht F (x) = f (x) (ide only) Recll tht F (x) = lim h 0 F(x + h) F(x) h By construction F(x + h) F(x) = = x+h x+h x f (t) dt f (t) dt x f (t) dt = re of R (ssume f 0) R x x + h b
96 14 Show tht F (x) = f (x) (ide only) Recll tht F (x) = lim h 0 F(x + h) F(x) h By construction F(x + h) F(x) = = x+h x+h x f (t) dt f (t) dt x f (t) dt = re of R (ssume f 0) R x x + h b f (x) h when h is smll positive
97 14 Show tht F (x) = f (x) (ide only) Recll tht F (x) = lim h 0 F(x + h) F(x) h By construction F(x + h) F(x) = = x+h x+h x f (t) dt f (t) dt x f (t) dt = re of R (ssume f 0) R x x + h b f (x) h when h is smll positive Therefore F(x + h) F(x) h f (x) if h is smll.
98 Summry To find f (x) dx Introduce F(x) = x Required integrl is F(b). f (t) dt 15 It is shown tht F (x) = f (x).
99 Summry To find f (x) dx Introduce F(x) = x Required integrl is F(b). f (t) dt 15 It is shown tht F (x) = f (x). Definition Let f be (continuous) function. Suppose G is function such tht G = f (on n open intervl I), we sy tht G is primitive (or ntiderivtive) for f (on I).
100 Summry To find f (x) dx Introduce F(x) = x Required integrl is F(b). f (t) dt 15 It is shown tht F (x) = f (x). Definition Let f be (continuous) function. Suppose G is function such tht G = f (on n open intervl I), we sy tht G is primitive (or ntiderivtive) for f (on I). Exmple Let f (x) = 3x 2.
101 Summry To find f (x) dx Introduce F(x) = x Required integrl is F(b). f (t) dt 15 It is shown tht F (x) = f (x). Definition Let f be (continuous) function. Suppose G is function such tht G = f (on n open intervl I), we sy tht G is primitive (or ntiderivtive) for f (on I). Exmple Let f (x) = 3x 2. By inspection G(x) = x 3 is primitive for f (on R).
102 Summry To find f (x) dx Introduce F(x) = x Required integrl is F(b). f (t) dt 15 It is shown tht F (x) = f (x). Definition Let f be (continuous) function. Suppose G is function such tht G = f (on n open intervl I), we sy tht G is primitive (or ntiderivtive) for f (on I). Exmple Let f (x) = 3x 2. By inspection G(x) = x 3 is primitive for f (on R). Remrk Primitive is not unique.
103 Summry To find f (x) dx Introduce F(x) = x Required integrl is F(b). f (t) dt 15 It is shown tht F (x) = f (x). Definition Let f be (continuous) function. Suppose G is function such tht G = f (on n open intervl I), we sy tht G is primitive (or ntiderivtive) for f (on I). Exmple Let f (x) = 3x 2. By inspection G(x) = x 3 is primitive for f (on R). Remrk Primitive is not unique. Above exmple, G 1 (x) = x lso primitive.
104 Summry To find f (x) dx Introduce F(x) = x Required integrl is F(b). f (t) dt 15 It is shown tht F (x) = f (x). Definition Let f be (continuous) function. Suppose G is function such tht G = f (on n open intervl I), we sy tht G is primitive (or ntiderivtive) for f (on I). Exmple Let f (x) = 3x 2. By inspection G(x) = x 3 is primitive for f (on R). Remrk Primitive is not unique. Above exmple, G 1 (x) = x lso primitive. In fct, infinitely mny primitives: G(x) + ny constnt tht is, x 3 + C
105 Summry To find f (x) dx Introduce F(x) = x Required integrl is F(b). f (t) dt 15 It is shown tht F (x) = f (x). Definition Let f be (continuous) function. Suppose G is function such tht G = f (on n open intervl I), we sy tht G is primitive (or ntiderivtive) for f (on I). Exmple Let f (x) = 3x 2. By inspection G(x) = x 3 is primitive for f (on R). Remrk Primitive is not unique. Above exmple, G 1 (x) = x lso primitive. In fct, infinitely mny primitives: G(x) + ny constnt tht is, x 3 + C Any more??? If F = f, must F be in the bove form?
106 Summry To find f (x) dx Introduce F(x) = x Required integrl is F(b). f (t) dt 15 It is shown tht F (x) = f (x). Definition Let f be (continuous) function. Suppose G is function such tht G = f (on n open intervl I), we sy tht G is primitive (or ntiderivtive) for f (on I). Exmple Let f (x) = 3x 2. By inspection G(x) = x 3 is primitive for f (on R). Remrk Primitive is not unique. Above exmple, G 1 (x) = x lso primitive. In fct, infinitely mny primitives: G(x) + ny constnt tht is, x 3 + C Any more??? If F = f, must F be in the bove form? NO MORE If F = f, then F(x) x 3 + C for some constnt C.
107 16 Theorem Let F, G be functions on n open intervl (, b) such tht F (x) = G (x) for ll x (, b). Then F nd G differ by constnt,
108 16 Theorem Let F, G be functions on n open intervl (, b) such tht F (x) = G (x) for ll x (, b). Then F nd G differ by constnt, tht is, there exists constnt C such tht F(x) G(x) = C for ll x (, b)
109 16 Theorem Let F, G be functions on n open intervl (, b) such tht F (x) = G (x) for ll x (, b). Then F nd G differ by constnt, tht is, there exists constnt C such tht F(x) G(x) = C for ll x (, b) Geometry Mening Hypothesis: t corresponding points, tngents to grphs of F nd G re prllel. y = F(x) y = G(x)
110 16 Theorem Let F, G be functions on n open intervl (, b) such tht F (x) = G (x) for ll x (, b). Then F nd G differ by constnt, tht is, there exists constnt C such tht F(x) G(x) = C for ll x (, b) Geometry Mening Hypothesis: t corresponding points, tngents to grphs of F nd G re prllel. y = F(x) Conclusion: grph of F cn be obtined by moving tht of G upwrd (C > 0) or downwrd (C < 0). y = G(x)
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