To determine The value of ∫ x = 1 x = 4 y d x when y 2 = 4 x ( y > 0 ) . Explanation Formula used: The definite integral of a function f ( x ) as, ∫ a b f ( x ) d x = F ( b ) − F ( a ) , where F ′ ( x ) = f ( x ) . The numbers a and b are called the lower and upper integral. Calculation: Obtain the value of ∫ x = 1 x = 4 y d x when y 2 = 4 x ( y > 0 ) as follows. Rewrite the ∫ x = 1 x = 4 y d x as ∫ 1 4 2 x d x since y 2 = 4 x ⇒ y = ± 2 x and also y > 0

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ISBN: 9780134437705

Author: Washington

Publisher: PEARSON

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Chapter 25.4, Problem 37E

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The value of ∫x=1x=4ydx when y2=4x(y>0).

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Chapter 25.4, Problem 36E

Chapter 25.4, Problem 38E

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- Let n = 7, let p = 23 and let S be the set of least positive residues mod p of the first (p − 1)/2 multiple of n, i.e. n mod p, 2n mod p, ..., p-1 2 -n mod p. Let T be the subset of S consisting of those residues which exceed p/2. Find the set T, and hence compute the Legendre symbol (7|23). 23 32 how come? The first 11 multiples of 7 reduced mod 23 are 7, 14, 21, 5, 12, 19, 3, 10, 17, 1, 8. The set T is the subset of these residues exceeding So T = {12, 14, 17, 19, 21}. By Gauss' lemma (Apostol Theorem 9.6), (7|23) = (−1)|T| = (−1)5 = −1.

Let n = 7, let p = 23 and let S be the set of least positive residues mod p of the first (p-1)/2 multiple of n, i.e. n mod p, 2n mod p, ..., 2 p-1 -n mod p. Let T be the subset of S consisting of those residues which exceed p/2. Find the set T, and hence compute the Legendre symbol (7|23). The first 11 multiples of 7 reduced mod 23 are 7, 14, 21, 5, 12, 19, 3, 10, 17, 1, 8. 23 The set T is the subset of these residues exceeding 2° So T = {12, 14, 17, 19, 21}. By Gauss' lemma (Apostol Theorem 9.6), (7|23) = (−1)|T| = (−1)5 = −1. how come?

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The value of ∫ x = 1 x = 4 y d x when y 2 = 4 x ( y &gt; 0 ) .

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The value of ∫x=1x=4ydx when y2=4x(y>0).

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The value of ∫ x = 1 x = 4 y d x when y 2 = 4 x ( y > 0 ) .