Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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My question is, if they intersect at -1 and 1, why are the limits integration set up for 0 to 1? And not -1 to 1?

6.
= 3x2, or 3x2 (x² +1) = 6. This can be written as 3z4+3r² -6 = 0.
6.2.11 The curves intersect when
%3D
x2 +1
6.
or (x4 +x² – 2) = 0. Factoring gives (x² + 2)(x² – 1) = (x² + 2)(x – 1)(x + 1) = 0, so the curves intersect
at r = -1 and x = 1. Using symmetry, the area is given by
-
2[G 3) dr = 2 (6 tan z
2).
1)
6T
-1
x2 +1
3D2
4
Зп - 2.
0.
0.
expand button
Transcribed Image Text:6. = 3x2, or 3x2 (x² +1) = 6. This can be written as 3z4+3r² -6 = 0. 6.2.11 The curves intersect when %3D x2 +1 6. or (x4 +x² – 2) = 0. Factoring gives (x² + 2)(x² – 1) = (x² + 2)(x – 1)(x + 1) = 0, so the curves intersect at r = -1 and x = 1. Using symmetry, the area is given by - 2[G 3) dr = 2 (6 tan z 2). 1) 6T -1 x2 +1 3D2 4 Зп - 2. 0. 0.
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