If $f$ is continuous in $\Bbb{R}$, then $\int_a^bf(x+c)dx = \int_{a+c}^{b+c}f(x)dx$.

Question

I cant show that if $f$ is continuous in $\Bbb{R}$, $$\int_a^bf(x+c)dx = \int_{a+c}^{b+c}f(x)dx.$$

What rule should I use?

Answer 1

It is sufficient that $f$ is (Riemann) integrable over the interval $[a+c,b+c]$:

If $${\cal P}:\quad a=x_0<x_1<\ldots<x_N=b,\qquad \xi_k\in[x_{k-1},x_k]\quad(1\leq k\leq N)$$ is a tagged partition of $[a,b]$ then $${\cal P}':\quad x_k':=x_k+c,\quad \xi_k':=\xi_k+c\qquad(0\leq k\leq N)$$ is a tagged partition of $[a+c,b+c]$ (and similarly in the other direction). Then $$\int_a^b f(x+c)\>dx\approx\sum_{k=1}^Nf(\xi_k+c)(x_k-x_{k-1})=\sum_{k=1}^Nf(\xi_k')(x_k'-x_{k-1}')\approx\int_{a+c}^{b+c}f(x)\>dx\ ,$$ whereby the errors implied by $\approx$ can be made arbitrarily small. It follows that in fact LHS=RHS.