1 (a - b) In K2 = | (a − x) (b - x) After solving this expression we get, 1 t(a - b) 31 = K₂ ti t] ln b(a - x) a(b − x) (201

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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1
(a - b)
X
(a − X
-[n | (8 - x)] = [K₂ t)
In
t]
(b-x)
After solving this expression we get,
K2
=
1
t(a - b)
b(a - x)
a(b − x)
-
ln
(2)
Transcribed Image Text:1 (a - b) X (a − X -[n | (8 - x)] = [K₂ t) In t] (b-x) After solving this expression we get, K2 = 1 t(a - b) b(a - x) a(b − x) - ln (2)
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