Is it ok to do this change of variable in integration: let $x = x - 1$

Question

In integrals like $\int \sqrt{x-1}\,dx$, is it ok to make this change of variable in integration: "let $x = x - 1$"?

It looks sketchy — like saying, let 5 = 4.

Answer 1

No, but you can let $u=x-1$. $x$ is never equal to $x-1$.

Answer 2

Is it ok to do this change of variable in integration: let $x = x - 1$

Certainly you may, if you do so carefully. $$\begin{align} g(x)+\int_{a}^{b} f(x) \operatorname d x & \;=\; g(x)+\int_{a}^{b} f(x-1) \operatorname d (x-1) \\[1ex] & \;=\; g(x)+\int_{a+1}^{b+1} f(x-1) \operatorname d x \end{align}$$

Notice, when we do this we may only so change the variable bound within the integration's scope.

It is probably safer to substitute with a different variable name though. (Less confusing.)

$$\begin{align} g(x)+\int_{a}^{b} f(x) \operatorname d x & \;=\; g(u-1)+\int_{a}^{b} f(u-1) \operatorname d (u-1) \\[1ex] & \;=\; g(u-1)+\int_{a+1}^{b+1} f(u-1) \operatorname d u \end{align}$$

Notice: here we can change the variable outside the integration scope.

You might more often see written out as:

$$\begin{align} \text{ Let } x & = u-1 & \therefore \mathrm d x = \mathrm d u \\[2ex] g(x)+\int_{a}^{b} f(x) \operatorname d x & \;=\; g(u-1)+\int_{a+1}^{b+1} f(u-1) \operatorname d u \end{align}$$