Let 2 C = sin dx. As f(x) = sin(1/x) is continuous except at 0 and bounded, f is integrable on [-1,1]. Since the integrand is positive (never negative), C > 0. Put u = 1/x and make the change of varibles to u so that f(x) = sin(u)² and dx = -u-2du. When x = u = -1. Therefore, 1, и %3 = 1 and when x = -1, 2 sin(u)*(u-?)du ´sin(u) C = du < 0. Where does this argument go wrong?
Let 2 C = sin dx. As f(x) = sin(1/x) is continuous except at 0 and bounded, f is integrable on [-1,1]. Since the integrand is positive (never negative), C > 0. Put u = 1/x and make the change of varibles to u so that f(x) = sin(u)² and dx = -u-2du. When x = u = -1. Therefore, 1, и %3 = 1 and when x = -1, 2 sin(u)*(u-?)du ´sin(u) C = du < 0. Where does this argument go wrong?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Riemann Integral
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Question
![Let
C =
sin
dx.
As f(x) = sin(1/x) is continuous except at 0 and bounded, ƒ is integrable on [-1, 1]. Since
the integrand is positive (never negative), C > 0. Put u =
varibles to u so that f(x) = sin(u)² and dx = -u-2du. When x =
u = -1. Therefore,
1/x and make the change of
= 1, u = 1 and when x =
-1,
´sin(u)
= = -
sin(u) (u-2)du
C =
du < 0.
и
Where does this argument go wrong?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7fd1602c-4cb7-4ab7-b037-059fcc7920df%2Ffb9fadca-687d-4ed0-b45b-8a2a06ea89ae%2F858noa3_processed.png&w=3840&q=75)
Transcribed Image Text:Let
C =
sin
dx.
As f(x) = sin(1/x) is continuous except at 0 and bounded, ƒ is integrable on [-1, 1]. Since
the integrand is positive (never negative), C > 0. Put u =
varibles to u so that f(x) = sin(u)² and dx = -u-2du. When x =
u = -1. Therefore,
1/x and make the change of
= 1, u = 1 and when x =
-1,
´sin(u)
= = -
sin(u) (u-2)du
C =
du < 0.
и
Where does this argument go wrong?
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