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Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK

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Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK (KSC) ** School Info Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK (KSC) has 3 departments in Course Hero with 88 documents.

Karak, Khyber Pakhtunkhwa
http://www.eduvision.edu.pk/institutions-detail.php?city=927k&institute=1312606888_khyber-scince-college%5btakht-e-nasrati%5d&submit=search

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Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK Documents (88)

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2 pages
- Campaign Finance and Mass Media Quiz Davon Etienne
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- vibration
- MECHANICAL 10 - Fall 2016
- Campaign Finance & Mass Media Quiz 1. Which of the following most accurately describes media coverage of elections? A) Coverage of presidential primaries gives relatively equal power to states regardless of when they hold their primaries. B) Coverage tend
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Campaign Finance and Mass Media Quiz Davon Etienne
5 pages
- htmllol
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- vibration
- MECHANICAL 10 - Fall 2016
- <html> <head> <title>Web Design Final</title> </head> <body> <img src="h:/Web Design/webdesignbanner.png" height="200" width="1024"/></br></br> <header class="banner"> <h1>Web Design Final Exam Study Guide Site</h1> <nav> <ul> <li><a href="careers.html">C
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htmllol
1 pages
- Good Vs. Bad Website Assignment
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- vibration
- MECHANICAL 10 - Fall 2016
- Class: Web Design Unit: Web Design Teacher: Mrs. Swaby Browse the web for the following: - 3 high quality sites - 3 poorly designed sites Provide the following information for the sites that you select. Rating Good/Bad Website Name URL Explain Design Pros
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Good Vs. Bad Website Assignment
1 pages
- Aztec Religion
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- SD
- SA DX - Winter 2017
- Aztec Religion The Aztecs believed in many gods. Some of their gods controlled nature. The Aztecs believed that different gods watched over their crops. Numerous gods were in charge of the rain, water and wind. They also believed that gods needed rest, ju
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Aztec Religion
2 pages
- Notes-73
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- maths
- MATH 26 - Summer 2016
- 7.7. CHAPTER 4 57 Suppose that g is integrable on [a, b]. If so, then there is a Cousin cover with the property that for any partition of [a, b] from Let Define 00 = 00 = cfw_([ai , bi ], i ) : i = 1, 2, . . . , n , 0 \ n X g(i )L([ai , bi ]) Z b g(x) dx
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Notes-73
2 pages
- Notes-109
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- maths
- MATH 26 - Summer 2016
- 8.3. CHAPTER 3 93 so that k (k = 1, 2, 3, . . . ) just contains elements ([ai , bi ], i ) from that happen also to belong to k . Most of these new subpartitions are empty: in fact since contains only n elements, there are at most n of the k that are nonem
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Notes-109
2 pages
- Notes-57
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- maths
- MATH 26 - Summer 2016
- 6.9. DESCRIPTIVE CHARACTERIZATION OF THE INTEGRAL 41 [Fine version] A set E is a null set if for every > 0 there is a fine cover of E so that n X (bi ai ) < i=1 whenever = cfw_([ai , bi ], xi ) : i = 1, 2, . . . , n is a subpartition chosen from . L(E)
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Notes-57
2 pages
- Notes-53
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- maths
- MATH 26 - Summer 2016
- CHAPTER 6 Measure Theory Warning: This is not an elementary chapter! The measure theory in this chapter should not be considered in the same pedagogical light as the earlier four chapters. Up to now no techniques beyond simple compactness arguments are in
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Notes-53
2 pages
- Notes-85
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- maths
- MATH 26 - Summer 2016
- 7.7. CHAPTER 4 69 now a partition of the larger interval [a, b]. We note that t = a1 and 1 = a so that f (1 ) = f (a) = 0. In particular then X X f (x)L(I) = f (x)L(I). (I,x)21 (I,x)22 Now we recall, since 2 is a partition of [a, b] from , that F (b) F (a
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Notes-85
2 pages
- Notes-55
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- maths
- MATH 26 - Summer 2016
- 6.3. LEBESGUES MEASURE 39 6.3. Lebesgues measure The measure L is known as Lebesgues measure and assigns a natural length to every set. There are numerous presentations of Lebesgues measure and any one of these can be consulted. We oer no proofs of the st
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Notes-55
2 pages
- Notes-67
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- maths
- MATH 26 - Summer 2016
- 7.6. CHAPTER 3 from have 51 chosen in such a way that the intervals cfw_[ai , bi ] do not overlap, we must n X (bi ai ) < /2. i=1 It is easy to check that partition is a Cousin cover of the interval [c, d]. Thus there is a cfw_([ai , bi ], xi ) : i = 1, 2
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Notes-67
3 pages
- Notes-04
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- maths
- MATH 26 - Summer 2016
- iv PREFACE function is adequate justification in the correct theory of integration, but not at all for the Riemann integral. More distressingly would be their solution to this problem: Verify that Z 1 1 xp dx = ( 1 < p < 0). 1 + p 0 My students would use
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Notes-04
2 pages
- Notes-71
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- maths
- MATH 26 - Summer 2016
- 7.7. CHAPTER 4 55 many steps, to handle all the extra points where the derivative might fail. This suggests that we split into infinitely many parts: = + + + + n. 2 4 8 2 Proof. /8: Let > 0. Construct 1 and 2 as before but using the choices /4) and 1 = c
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Notes-71
2 pages
- Notes-45
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- maths
- MATH 26 - Summer 2016
- 5.10. COMPARING RIEMANN SUMS 29 Some rewriting will help in the comparison. First of all use L([a, b]) = b a to denote the length of the interval [a, b]. We can even allow a = b so that [a, a] is a degenerate interval of length L([a, a]) = 0. The length o
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Notes-45
2 pages
- Notes-63
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- maths
- MATH 26 - Summer 2016
- CHAPTER 7 PROOFS The proofs assume the usual calculus background for definitions of continuity and derivatives. We need the mean-value theorem of the calculus but little more than that. It would be useful if the student knew that continuous functions on c
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Notes-63
2 pages
- Notes-51
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- maths
- MATH 26 - Summer 2016
- 5.18. INFINITE INTEGRALS 35 5.18.1. Summing inside the integral. We now can extend the formula we have used before: ! ! Z b X Z b 1 1 X fn (x) dx = fn (x) dx . a n=1 n=1 a Our assumptions previously (Theorem 5.22) required the integrals to exist as finite
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Notes-51
2 pages
- Notes-69
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- maths
- MATH 26 - Summer 2016
- 7.7. CHAPTER 4 53 At the remaining points we take advantage of the derivative formula F 0 (x) = f (x) to define 3 where = cfw_([c, d], x) : c x d and |F (d) = F (c) 2(b a) f (x)(d c)| < (d c) . Combine the covering relations = 1 [ [ 2 3. Simply check (poi
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Notes-69
2 pages
- Notes-47
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- maths
- MATH 26 - Summer 2016
- 5.14. THE INDEFINITE INTEGRAL 31 5.14. The indefinite integral The essential ingredient in the Newton definition of the integral of a function f on a compact interval [a, b] is that there must be an indefinite integral, i.e., a function F defined on [a, b
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Notes-47
2 pages
- Notes-105
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- maths
- MATH 26 - Summer 2016
- CHAPTER 8 SOLUTIONS FOR SELECTED EXERCISES 8.1. Chapter 1 8.1.1. Solution for Exercise 5. To evaluate Z 1 1 p dx |x| 0 it is enough to find explicitly an indefinite integral. (Not always possible but here it is.) p 1 The function F (x) = 2 x = 2x 2 has f
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Notes-105
2 pages
- Notes-65
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- maths
- MATH 26 - Summer 2016
- 7.5. CHAPTER 2 49 7.4.2. Proof of Lemma 1.3. If F and G are continuous functions on an interval [a, b] and if F 0 (x) = G0 (x) for all a < x < b with at most finitely many exceptions then F (b) Proof. F (a) = G(b) G(a). Let us suppose that the exceptions
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Notes-65
2 pages
- Notes-83
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- maths
- MATH 26 - Summer 2016
- 7.7. CHAPTER 4 67 case ([ai , bi ], i ) is from 2 or j0 belongs to N and so the corresponding element ([a0j , b0j ], j0 ) is from 2 . That sum is smaller than X f (i ) f (j0 ) L([ai , bi ] \ [a0j , b0j ]) < 2M [/8M ] + 2M [/8M ] = /2. 1 The second part co
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Notes-83
2 pages
- Notes-75
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- maths
- MATH 26 - Summer 2016
- 7.7. CHAPTER 4 But is also a subset of n X i=1 59 , so we know that Z b g(i )L([ai , bi ]) g(x) dx < /2. 00 a Putting these two inequalities together we deduce that Z b n X f (i )L([ai , bi ]) g(x) dx a i=1 n X n X f (i )L([ai , bi ]) i=1 g(i )L([ai , bi
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Notes-75
2 pages
- Notes-87
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- maths
- MATH 26 - Summer 2016
- 7.7. CHAPTER 4 71 Fix an integer n and let > 0. There exists a Cousin cover that X (7.1) | F (I) f (x)L(I)| < /2, 1 of [a, b] such (I,x)2 for every partition of [a, b] contained in 1 . Since Nn is a null set there is a Cousin cover 2 so that X (7.2) L(I)
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Notes-87
2 pages
- Notes-59
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- maths
- MATH 26 - Summer 2016
- 6.13. LEBESGUES PROGRAM: CONSTRUCTION OF THE INTEGRAL 43 A function f : [a, b] ! R would be measurable if the corresponding set E = cfw_x 2 [a, b] : f (x) < r is a measurable set. An equally simple way to think of this is that a function f defined on an i
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Notes-59
2 pages
- Notes-43
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- maths
- MATH 26 - Summer 2016
- 5.6. IGNORING NULL SETS 27 . PROOF IN SECTION 7.7.5 Lemma 5.4. Let e1 , e2 , e3 , . . . be a sequence of points from the interval [a, b] and suppose that f and g are functions defined on [a, b] and for which f (x) = g(x) for every point x not belonging to
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Notes-43
2 pages
- Notes-93
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- maths
- MATH 26 - Summer 2016
- 7.8. CHAPTER 5 77 7.8.3. Step 1: Vitali covering theorem. For the first step in the proof of the Vitali covering theorem we prove a preliminary covering theorem that is of interest on its own. Lemma 7.1 (Rad o Covering Theorem). Let E be a set of finite L
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Notes-93
2 pages
- Notes-107
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- maths
- MATH 26 - Summer 2016
- 8.3. CHAPTER 3 91 8.2.4. Solution for Exercise 13. Let F be dierentiable at a point x0 , let > 0. Then we can choose > 0 so that |F (x) F (x0 ) F 0 (x0 )(x x0 )| |x x0 |/2 for all |x x0 | < . Suppose that [c, d] is an interval containing x0 and with leng
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Notes-107
2 pages
- Notes-77
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- maths
- MATH 26 - Summer 2016
- 7.7. CHAPTER 4 61 7.7.11. Proof of Theorem 5.9. Our goal is to establish the useful formula Z b Z c Z b f (x) dx = f (x) dx + f (x) dx a a c for integrating over two adjacent intervals [a, c] and [c, b]. Proof. We begin by invoking our right to change the
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Notes-77
2 pages
- Notes-89
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- maths
- MATH 26 - Summer 2016
- 7.7. CHAPTER 4 73 To prove this formula is not sophisticated. It merely uses the definition of the integral. Given an > 0 can we find a Cousin cover so that the Riemann sums for the cover are within of the value we require for the integral? The computatio
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Notes-89
2 pages
- Notes-91
- Khyber Science Degree College, Takht-e-Nasrati, Karak, KPK
- maths
- MATH 26 - Summer 2016
- 7.8. CHAPTER 5 Z N X b fj (x) dx + 2 j a j=1 ! 75 Z 1 X j=1 b ! fj (x) dx a + . We have established this inequality for all t < 1: ! ! Z b 1 X X t tf (x)L(I) fj (x) dx + . a j=1 This gives the upper bound for all Riemann sums from 00 that we wanted. 7.7.
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Notes-91

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