# How to derive this formula: $\int_a^bf(c-x)dx = \int_{c-b}^{c-a}f(x)dx$?

I'm stuck in this exercise:

$$\int_a^bf(c-x)dx = \int_{c-b}^{c-a}f(x)dx$$

My attempt is this:

 \begin{align*} \int_a^bf(c-x)dx &= - \int_{-a}^{-b}f(x-c)dx\\ &= \int_{-b}^{-a}f(x-c)dx \end{align*} 

But at this point I'm not sure what to do. To my understanding if one wants to integrate $f(x-c)$, which is shifted to the right, it would be the same as integrating $f(x)$ with the interval of integration shifted to the left:

$$= \int_{-b-c}^{-a-c}f(x)dx$$

But that does not seem to be the right answer.

## 2 Answers

Make the Substitution $y=c-x$. The upper bound of your integral $x=b$ is changing to $y=c-b$. Moreover the lower bound $x=a$ transforms to $y=c-b$. Also the differential $dx$ is changed; you derive $y$ by $x$ for constant $c$ and you obtain $dy=-dx$.

Let's say what you need to do, how it follows naturally and then why the claims are true. First, as the second equation contains only $x$ in the $f(x)$, and a definite integral doesn't really depend on the variable with respect to which you are integrating (the variable is really a label for the integral), you may think of an almost natural substitution of $y = c-x$. Let me just elaborate what happens with this; the crucial point is that nothing has changed with the integral, as in our old integral, $x$ traverses values from $a$ to $b$, the new substituted variable $y = c-x$ traverses values from $c-b$ to $c-a$, however, the direction of traversal of values has changed due to the negative sign in front of the $x$ in the expression $f(c-x)$. Hence, the integral on the left hand side is really the same as on the right hand side, written in a different way. It has not been simplified or operated, or anything with the series of operations of substitution. It's better to have such a picture in your mind, at least in real valued functions such as these. Although the same picture is of much help in the multivariate integration as well.

• Thank you. Although making a substitution seemed the (obvious?) way to derive the formula, I didn't understand what was happening behind it. Your comment clears things up a lot. – Jazz Feb 17 '15 at 12:42