$\int_a^{b} f(x) dx$ exists then so does $\int_{a+c}^{b+c} f(x-c)dx$ I am with a friend doing some late night math at math club and we came across this problem: 
$f : [a,b] \rightarrow \mathbb R$ and let $c \in \mathbb R$. I am trying to show that 
$\int_a^{b}  f(x) dx$ exists then so does $\int_{a+c}^{b+c} f(x-c)dx$ and these two integrals are equal. 
This seems almost trivial to us as we quickly computed some integrals and this was obvious, however we are having some trouble proving this. Is there some rule where you can subtract a constant in the integral and add it in the integral bounds. I have never used this before. Help would be great! Just want to understand how this is actually proved. Thanks! 
 A: If $f$ is continuous, one can use substitution.  Letting $u=x+c$ be a function of $x$, then we have the corresponding differentials $du=dx$, and so $\int_a^bf(x)\,dx=\int_{u(a)}^{u(b)}f(u-c)\,du=\int_{a+c}^{b+c}f(u-c)\,du$.
However, $f$ is only specified to be a function, and that the integral exists.  For this, we may use the Riemann integral $\int_a^bf(x)\,dx=\lim_{n\to\infty}\sum_{i=1}^n\frac{b-a}{n}f(a+\frac{b-a}{n}i)$, and by some algebra get $\lim_{n\to\infty}\sum_{i=1}^n\frac{(b+c)-(a+c)}{n}f(a+c+\frac{(b+c)-(a+c)}{n}i - c)=\int_{a+c}^{b+c}f(x-c)\,dx$.
A: It is straightforward to prove from the definition.
Let $f'(x) = f(x-c)$.
Suppose $P=(a=x_0,x_1,...,x_{n-1},x_n = b)$ is a partition of $[a,b]$ and let
$P' = (x_0+c,x_1+c,...,x_{n-1}+c,x_n+c)$ (which is a partition of $[a+c,b+c]$).
Then $U(f,P) = U(f',P')$ and $L(f,P) = L(f',P')$. 
Similarly, if $P'$ is a partition of $[a+c,b+c]$, we can construct an corresponding partition of $[a,b]$.
It follows that $\inf_P U(f,P) = \inf_{P'} U(f',P')$, and similarly $\sup_P L(f,P) = \sup_{P'} L(f',P')$. Hence $f'$ is integrable and the integrals are the same.
A: Consider $\int_a^b f(x) \ dx,$
Let $t = x + c$, we immediately learn a few things:   


*

*$f(t - c) = f(x)$ 

*$dx = dt$

*$x=a\implies t = a + c$

*$x = b\implies t = b + c$


It follows from substitution that 
$$\int_a^b f(x) \ dx = \int_{a+c}^{b+c}f(t-c)\ dt$$
Further, since these are both definite integrals, the use of $t$ and $x$ is interchangeable. So we have
$$\int_a^b f(x) \ dx = \int_{a+c}^{b+c}f(x-c)\ dx$$
A: Sketch the graphs of $y=f(x)$ and $y=f(x-c)$ over the intervals $(a,b)$ and $(a+c, b+c)$, respectively. You will find they are identical except for a horizontal translation. The integrals represent the (net) areas under these curves over the respective intervals, which will of course be the same.
