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

Question

I am having trouble with this problem. f be integrable on $[a,b]$. Suppose $c\in R$ and $g: [a+c,b+c] \to R$ such that $g(x)=f(x-c), x\in[a+c,b+c]$ $$\int_a^bf(x)dx=\int_{a+c}^{b+c}g(x)dx$$ I am thinking that this problem makes use of the change of variable formula to solve. Can you give me some idea? Thank you very much!

Answer 1

You simply take $y=x+c$ (or $x=y-c$) which gives $\frac{dy}{dx}=1$ or $dy=dx$ and the new boundaries are $y=a+c$ and $y=b+c$ (because the old boundaries were $x=a$ and $x=b$) and the integral becomes $$\int_a^bf(x)dx = \int_{a+c}^{b+c}f(y-c)dy = \int_{a+c}^{b+c} g(y) dy = \int_{a+c}^{b+c} g(x)dx$$ where in the last step I renamed the dummy variable $y$ to $x$.

Answer 2

Indeed, if you set $t=x+c$, the jacobian is $1$ and thus $$\int_a^b f(x)dx=\int_{a+c}^{b+c}f(t-c)dt=\int_{a+c}^{b+c}g(t)dt.$$