Two definite integrals are equal, what about intervals of integration?

Jan 2016
15
2
USA
Suppose I have two simple, integrateable functions of variable t such that

\(\displaystyle \int_{0}^{t_1}f(t)dt = \int_{0}^{t_2}g(t)dt\)

I believe that if \(\displaystyle f(t) \leq g(t)\) for all t then \(\displaystyle t_2 \leq t_1\)

Any tips on how to prove this? Thanks
 
Jun 2013
1,110
590
Lebanon
is it given that $f$ and $g$ are non-negative functions?
 
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Jan 2016
15
2
USA
Sorry, I forgot to mentioned anything about that. Yes, both functions are non-negative.
 
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May 2019
18
3
New York
Here is my approach to it, I hope this helps:

So here you are starting off with the equation,

\(\displaystyle \int_{0}^{t_{1}}f(t)\,dt=\int_{0}^{t_{2}}g(t)\,dt \).

If \(\displaystyle g(t) ≥ f(t) \) for all \(\displaystyle t \), then:

\(\displaystyle g(t)=f(t)+m(t) \)

where \(\displaystyle m(t) \) is an arbitrary function which makes the two equations above true.

In case that \(\displaystyle g(t)=f(t) \), \(\displaystyle m(t)=0 \).

So now we can try substituting that into your original equation and we get,

\(\displaystyle \int_{0}^{t_{1}}f(t)\,dt=\int_{0}^{t_{2}}f(t)\,dt+\int_{0}^{t_{2}}m(t)\, dt \).

Move over the \(\displaystyle \int_{0}^{t_{2}}f(t)\, dt \) to the left side and what's left is:

\(\displaystyle \int_{0}^{t_{1}}f(t)\,dt\, -\int_{0}^{t_{2}}f(t)\,dt\,=\int_{0}^{t_{2}}m(t)\, dt \).

If \(\displaystyle f(t) \) and \(\displaystyle g(t) \) are non-negative functions and \(\displaystyle g(t)≥f(t) \) for all \(\displaystyle t \), \(\displaystyle m(t) \) has to be a non-negative function as well, and the previous equation is only true when \(\displaystyle t_{1} ≥ t_{2} \). If \(\displaystyle t_{2} > t_{1} \) (which would therefore make \(\displaystyle f(t)≥g(t) \)), we would get \(\displaystyle -\int_{0}^{t_{2}}m(t)\, dt \). However, \(\displaystyle m(x) \) is not a negative function in our case, so its integral cannot be negative if all previous derivations hold true.

And of course, if \(\displaystyle t_{1} = t_{2} \), then:

\(\displaystyle \int_{0}^{t_{1}}f(t)\,dt\, -\int_{0}^{t_{2}}f(t)\,dt\,=0 \).

Hope that helped, and feel free to ask any questions!
 
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Jan 2016
15
2
USA
Thank you Ivan. I like your approach.

Here's my version. I make an additional assumption that $t_1, t_2 > 0$ which is missing from the original post. If you guys see any flaw with this please let me know. Thanks


Let $g(t) = f(t) + m(t)$. Then
\begin{eqnarray}
\int_{0}^{t_2} g(t)dt = \int_{0}^{t_2}f(t)dt + \int_{0}^{t_2}m(t)dt\nonumber
\end{eqnarray}

If $m(t) = 0$ then $t_1 = t_2$.

Otherwise $m(t) > 0$. Let $t_1 = t_2 + \triangle t$ with $\triangle t \in R$.

The case $\triangle t = 0$ is not possible as it leads to a contradiction in the form of
\begin{eqnarray}
\int_{0}^{t_1}f(t)dt < \int_{0}^{t_1}f(t)dt + \int_{0}^{t_1}m(t)dt
\end{eqnarray}

Then $\triangle t < 0 \lor \triangle t > 0$.

Assume that $\triangle t < 0$. Then
\begin{eqnarray}
\int_{0}^{t_2} g(t)dt = \int_{0}^{t_1} g(t)dt + \int_{t_1}^{t_2} g(t)dt\nonumber\\
= \int_{0}^{t_1}f(t)dt + \int_{0}^{t_1}m(t)dt + \int_{t_1}^{t_2}f(t)dt + \int_{t1}^{t_2}m(t)dt\nonumber\\
= \int_{0}^{t_1}f(t)dt + M
\end{eqnarray}
This is not possible since $M > 0$ and therefore $\triangle t > 0$. $\square$
 
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May 2019
18
3
New York
Yeah that's a really good approach to the problem as well, thanks!
 
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