Does $\int_0^x f(t)dt=\int_0^x f(x-t)dt$ hold always?

Question

Question

I have found out that for many functions ($f(x)=x$, $f(x)=e^x$, $f(x)=\frac{1}{x+1}$, etc.) the following expression is satisfied: $\int_0^x f(t)dt=\int_0^x f(x-t)dt$. I am wondering: does this hold always, or it is just a nice property of a particular kind of functions? My question arises from the fact that in general $\int_a^x f(t)dt\neq\int_a^x f(x-t)dt$, but when $a=0$ I have observed the equality to hold.

Attempt of formal proof

I have tried to prove the expression following the path of the proof of the fundamental theorem of calculus.

Given $f(x)$, we know that $F(x)=\int_a^x f(t)dt$ satisfies $F^\prime(x)=f(x)$, where $a$ is set arbitrarily to $0$. If we can prove that $G(x)=\int_0^x f(x-t)dt$ satisfies $G^\prime(x)=f(x)$, then $\int_0^x f(t)dt=\int_0^x f(x-t)dt$. I tried with the following steps.

$G(x_1)=\int_0^{x_1} f(x_1-t)dt$

$G(x_1+\Delta x)=\int_0^{x_1+\Delta x} f(x_1+\Delta x-t)dt$

$G(x_1+\Delta x)-G(x_1)=\int_0^{x_1+\Delta x} f(x_1+\Delta x-t)dt-\int_0^{x_1} f(x_1-t)dt$

At this point the functions under the two integrals are different, so I get stuck.

Related question

I really struggle in finding an interpretation to the difference between $\int_0^x f(t)dt$ and $\int_0^x f(x-t)dt$, if there is any at all. Can anyone suggest a practical explanation?

Answer 1

Let $x-t=u \implies dt=-du$ at $t=0,u=x$ at $t=x,u=0$ thus $\int_0^x f(x-t)=\int_x^0 f(u)(-du)=\int_0^x f(u)du=\int_0^x f(t)dt..\text{without loss of generality}$ this property is know as $\text{King rule}$ and the generalized form is $\int_a^b f(x)dx=\int_a^b f(a+b-x)dx$

Answer 2

Just set $u = x-t$, so $t=x-u$.

Then $du = -dt$ so that

$\int_0^x f(t) dt =\int_x^0 f(x-u)(-du) =\int_0^x f(x-u)du $.

This is a special case of

$\int_a^b f(t)dt =\int_{g(a)}^{g(b)} f(g^{-1}(t))g^{-1}(t) $ with $g(t) = x-t$. (I think I have that righr.)

Answer 3

Assuming that $f$ is Riemann integrable,

$$\begin{align}I&=\int_0^x[f(t)-f(x-t)]dt\\ &=\lim_{n\to\infty}\sum_{k=0}^n\frac1n\left[f\left(\frac {ktx}n\right)-f\left(x-\frac{ktx}n\right)\right]\\ &\stackrel *=\lim_{n\to\infty}\sum_{k=0}^n\frac1n\left[f\left(x-\frac {ktx}n\right)-f\left(\frac{ktx}n\right)\right]=-I \end{align}$$

Thus, $I=0$.

$(*)$ Just reverse the order of the terms of the sum.