Prove that $\int_a^{b} f(x) dx$=$\int_{a+c}^{b+c} f(x-c)dx$

I want to prove that if $f$ is continuous, then $\int_a^{b} f(x) dx$=$\int_{a+c}^{b+c} f(x-c)dx$. I have observed in some places that the following is done:

Letting $t=x+c$, then we have the corresponding differentials $dt=dx$, and so $\int_a^bf(x)\,dx=\int_{t(a)}^{t(b)}f(t-c)\,dt=\int_{a+c}^{b+c}f(t-c)\,dt$

However, I do not understand why they say that t=x+c and if the demonstration is correct.

Any help? thanks

Solution 1:

It is a rephrasing of the change of variables formula, which says that $$\int_{\varphi(\alpha)}^{\varphi(\beta)}f(x)dx=\int_{\alpha}^\beta f(\varphi(t))\varphi'(t)dt$$ for $\varphi\in C^1(\mathbb{R})$ and $f$ continuous. We like to think about this formula as "we replace the variable $x$ by the variable $\varphi(t)$", so here writing "let $t=x+c$" is the short way for writing "we apply the change of variables formula for $\varphi(t)=x=t-c$, $\alpha=a+c$ and $\beta=b+c$".