# Integration Proof with constant

1. Dec 4, 2015

### OmniNewton

1. The problem statement, all variables and given/known data
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

2. Relevant equations
Let C be a constant
Let D be a constant
3. 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

2. Dec 4, 2015

### Simon Bridge

You have not used the interval.

3. Dec 4, 2015

### OmniNewton

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.

4. Dec 4, 2015

### PeroK

Here's a suggestion to get you started. Let $h(x) = f(x) - g(x)$. What can you say about $h(x)$?

5. Dec 4, 2015

### OmniNewton

h(x) is constant on (a,b). Should I be using the area beneath the curve formula?

6. Dec 4, 2015

### lordianed

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?