Assuming two functions are the same

Question

If we know the following: $$\int_a^b f(x) \,dx= k$$ and: $$\int_a^b g(x) \,dx= k$$ Can we deduce that $f(x) = g(x)$? If not, could someone find me an example where the above is the case and $f(x)$ does not equal $g(x)$? Thanks in advance.

Answer 1

$$\int_0^{\pi} \cos x \, \mathrm dx = 0 \quad \text{and} \quad \int_0^\pi 0 \, \mathrm dx=0,$$ (so $a=0$, $b=\pi$, $k=0$, $f(x) = \cos x$ and $g(x) =0$), but $\cos x \ne 0$ so $$f(x) \ne g(x).$$ There are many, many other examples. There is no reason that because the area under the graph of two functions is the same between two points, that they should be the same function.

Edit: I assume here that $a$ and $b$ are fixed. Please see the other answers to this question if they aren't!

Answer 2

The problem doesn't say whether $a$ and $b$ are existentially or universally quantified. If we assume the latter, and something more about $f$ and $g$, we can assert:

Claim: If $f$ and $g$ are continuous functions and $$\int_a^b f(x)\,dx = \int_a^b g(x)\,dx$$ for all real numbers $a < b$, then $f(x) = g(x)$ for all $x$.

To prove this, it is enough to show: If $h$ is continuous and $\int_a^b h(x)\,dx = 0$ for all $a < b$, then $h(x) = 0$ for all $x$.

To prove this, it is enough to show the contrapositive: If $f$ is continuous, and there exists $x_0$ such that $f(x_0) \neq 0$, then there exist real numbers $a < b $ such that $\int_a^b f(x)\,dx \neq 0$.

And finally, it is enough to show that: If $f$ is continuous, and there exists $x_0$ such that $f(x_0) > 0$, then there exist real numbers $a< b$ such that $\int_a^b f(x)\,dx > 0$.

The gist of the argument is that: if $f$ is continuous, and positive at a point, it is positive on an open interval around that point. The integral of $f$ over that interval must be positive.

Here is the formal proof: Let $\epsilon = f(x_0)/2$. There exists $\delta$ such that $|f(x) - f(x_0)| < \epsilon$ when $|x-x_0| < \delta$. Let $a =x_0 - \delta/2$ and $b = x_0 + \delta/2$. Then $f(x) > f(x_0)/2$ for all $x$ in $[a,b]$. So $$\int_a^b f(x) > (f(x_0)/2) \cdot (b-a) = \frac{f(x_0)}{2}\cdot \delta$$

Notice how the proof breaks down without the continuity assumption. The function $f$ which is $1$ at $x=0$ and $0$ elsewhere has $f(0) > 0$, but for all $a < b$, $\int_a^b f(x)\,dx = 0$.

Answer 3

Counterexample:

\begin{align} 1=\int_{0}^{1}dx=\int_{0}^{1}(2x)dx \end{align}

and

\begin{align} f(x)=1\ne g(x)=2x \end{align}

Answer 4

No and here’s a way to intuit why (note: this is not a proof or mathematically rigorous argument, just an aid in understanding intuitively why the answer is what it is):

The mean of a continuous function over an interval is calculated as the definite integral over the interval divided by the length of that interval. So saying two functions have the same integral over a non-zero length interval is identical to saying they share the same mean over that interval. And the mean of a continuous function doesn’t uniquely identify it. For example consider how many different continuous functions you could construct that have mean zero over some non-zero length interval. For a real-life analogy courtesy of @AmateurDotCounter, imagine you have two mountain ranges that have the same average (mean) height. Does this mean the two mountain ranges have exactly the same shape?

So just because two functions have the same definite integral over the same interval doesn’t mean they’re the same function. It’s also the intuition behind the counter-examples that have been provided in other answers: each function in such a pair share the same mean and thus the same integral over some non-zero length interval.

Note: in my answer I assume that OP means that there exists an interval [a,b] such that the integrals of the two functions over that interval are equal.