Integration Proof with constant

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
5 replies · 2K views
OmniNewton
Messages
105
Reaction score
5

Homework Statement


Prove that if f'(x) = g'(x) for all x in an interval (a,b) then f-g is constant on (a,b) then f-g is constant on (a,b) that is f(x) = g(x) + C

Homework Equations


Let C be a constant
Let D be a constant

The Attempt at a Solution


f(x) = antiderivative(f'(x)) = f(x) + C
g(x)= antiderivative(g'(x)) = g(x) + D

f-g = f(x) + C - (g(x) + D)
f-g = f(x) - g(x) + C - D

but since f'(x) = g'(x) then f(x) = g(x) the only difference is their constant.

then,

f-g = f(x) - f(x) + C - D
f-g = C - D

Since C and D are constants

then,

f-g = constant

if C = D

then f-g = 0

Note: I feel like I proved it but my notation is wrong since I cannot use f(x) = f(x) + C. I would like guidance for the proper notation to use. The possibility also exists my proof is completely wrong. I would like help

Thanks in Advanced!
 
Physics news on Phys.org
How would I got about approaching this problem? I'm sorry perhaps I am lacking knowledge but I have only received a 30 minute lesson on integrals in my class, which was the last class of the semester.
 
OmniNewton said:
How would I got about approaching this problem? I'm sorry perhaps I am lacking knowledge but I have only received a 30 minute lesson on integrals in my class, which was the last class of the semester.

Here's a suggestion to get you started. Let ##h(x) = f(x) - g(x)##. What can you say about ##h(x)##?
 
PeroK said:
Here's a suggestion to get you started. Let ##h(x) = f(x) - g(x)##. What can you say about ##h(x)##?
h(x) is constant on (a,b). Should I be using the area beneath the curve formula?
 
OmniNewton said:
h(x) is constant on (a,b). Should I be using the area beneath the curve formula?
How did you deduce that ##h(x)## is constant on ##(a,b)##? PeroK has a very good suggestion, under what conditions would ##h(x)## be constant over an interval? Could this have something to do with its derivative?
 
  • Like
Likes   Reactions: PeroK