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

#### jumbo1985

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

#### Idea

is it given that $f$ and $g$ are non-negative functions?

3 people

#### jumbo1985

Sorry, I forgot to mentioned anything about that. Yes, both functions are non-negative.

1 person

#### IvanM

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!

1 person

#### jumbo1985

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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#### IvanM

Yeah that's a really good approach to the problem as well, thanks!

1 person