If $\int_{0}^{4} f(x) \text{dx} = 17, \text{ what is } \int_{1}^{5} f(x-1) \text{dx}$?

Question

I tried manipulating the integral into a sum of integrals, but that isn't taking me anywhere:

$\int_{1}^{5} f(x-1) \text{ dx} = \int_{0}^{1} f(x-1) \text{ dx} + \int_{1}^{4} f(x-1) \text{ dx} + \int_{4}^{5} f(x-1) \text{ dx}$

First of all, I don't even know how to establish a relationship between $\int f(x) \text{dx}$ and $\int f(x-1) \text{dx}$.

Answer 1

Let $t = x-1$ , $dx =dt$

At $x =1 \ ,\ t = 0$

At $x =5 \ , \ t =4$

So $I = \int^5_1 f(x-1)dx = \int^4_0f(t)dt = 17$ (variable doesn't matter in definite integral)

Answer 2

If you integrate the second expression by the substitution $y = x - 1$ you will see that it becomes the first integral. Hence both integrals have the same value.