4
$\begingroup$

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.

$\endgroup$
8
$\begingroup$

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 $dy=-dx$.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Jazz Feb 17 '15 at 12:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.