| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
I can integrate the following integral two ways. Both ways seem
to use a correct integrating method, but only one returns the correct answer.
Whats is wrong with one of the methods that I am using.
Problem:
integrate dx/2x.
Method 1.
rewrite dx/2x as
1
integral----- dx
2x
1 1
----- integral --- dx ; take constant through integral
2 x
integrate 1
--- = ln | x | + c
x
final answer this way 1
---- ln | x | + C.
2
Method 2.
rewrite dx/2x as
1
integral----- dx
2x
Use u-substitution
let u = 2x
du
-- = 2 , du = 2dx
dx
set-up u-substitution,
1 1
---integral ---- du
2 u
integrate 1
--- = ln | u | + c
u
1
---- ln | u | + C.
2
Put equation back in terms of x, (u = 2x)
final answer this way 1
---- ln | 2x | + C.
2
What is wrong with my reasoning on one of these methods.
Mike
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 456.1 | BEING::POSTPISCHIL | Always mount a scratch monkey. | Mon Mar 17 1986 20:23 | 27 | |
First answer:
1
---- ln | x | + C0.
2
Second answer:
1
---- ln | 2x | + C1.
2
Let's make the answers equal:
1/2 ln |x| + C0 = 1/2 ln |2x| + C1.
Next, ln |2x| = ln 2 + ln |x|, so let's substitute (and distribute):
1/2 ln |x| + C0 = 1/2 ln 2 + 1/2 ln |x| + C1.
Subtract 1/2 ln |x| from both sides:
C0 = 1/2 ln 2 + C1.
That's it. You had the same answer both times, but the constants
coming out of the integration were hidden in different places.
-- edp
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