If $\int_{-\infty}^{\infty}f(x)\ \mathrm dx=100$ then $\int_{-\infty}^{\infty}f(100x+9)\ \mathrm dx =?$ 
Given $\displaystyle\int_{-\infty}^{\infty}f(x)\ dx=100$, evaluate $\displaystyle\int_{-\infty}^{\infty}f(100x+9)\ dx.$

Question is as above. I'm not sure how to even start. Is the answer $100$? Seems like if the function is bounded from negative infinity to infinity, any transformation just changes the shape. Not sure how to explain this properly though.

We know that
$$\int_{-\infty}^\infty f(x)dx=\lim_{x\to \infty}(F(x)-F(-x))=100\tag 1$$
Considering $dF(ax+b)\over dx$ we get $af(ax+b)$, so we have
$$\int_{-\infty}^\infty f(ax+b)dx=\lim_{x\to \infty}\frac{F(ax+b)-F(-ax+b)}a=\frac {100}a\tag 2$$
With $a=100$ we get $1$ as the value of the right-hand side of $(2)$.
