1
$\begingroup$

Why does this integral equal this? It doesn't make sense to me why there is a one-half in this integral. It was on a test today and I got it wrong, my answer was $F(6) - F(2)$ Could someone explain why when I declare a function that is not directly $f(x)$, but $f(g(x))$ or some other function something weird like this occurs?

$$\int_{1}^3 \mathrm{f(2x)= \frac{1}2(F(6) - F(2))}$$ Where: $$F'(x) = f(x)$$

I'm more interested in this as a special case as I thought that having a function within a function only changes the values that are plugged into that function and I find this to be a little bit weird, and I want to know why this has to happen.

|cite|improve this question
$\endgroup$
  • $\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – RE60K Apr 7 '15 at 20:07
  • $\begingroup$ Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. $\endgroup$ – RE60K Apr 7 '15 at 20:07
  • $\begingroup$ Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. $\endgroup$ – RE60K Apr 7 '15 at 20:08
  • $\begingroup$ $\frac{\mathrm{d}}{\mathrm{dx}} F(2x) = 2f(2x)$, but you want $\frac{\mathrm{d}}{\mathrm{dx}} \frac{1}{2} F(2x) = f(2x)$ $\endgroup$ – Jerry Apr 8 '15 at 3:47
6
$\begingroup$

The answer is the chain rule. You want an antiderivative for $f(2x)$, but when you take $\frac{d}{dx}(F(2x))$ you get $2f(2x)$, so a factor of $1/2$ must be introduced so that the derivative will be $f(2x)$, as desired.

|cite|improve this answer
$\endgroup$
  • 6
    $\begingroup$ An intuitive/informal explanation is that the "vertical" part of the area is the same, but the "horizontal" part has been compressed by a factor of $2$, since the domain of integration is only half as long. $\endgroup$ – Ian Apr 7 '15 at 20:42
4
$\begingroup$

$ \int_1^3 f(2x) dx \\ \text{ Let } u=2x \\ du=2 dx \\ \frac{1}{2} du=dx \\ \int_1^3 f(2x) dx=\int_{2(1)}^{2(3)} f(u) \frac{1}{2} du=F(u) \cdot \frac{1}{2} |_2^6 \\ \text{ Example: Choose } f(x)=\sin(x) \text{ so then } f(2x)=\sin(2x) \\ \int_0^\frac{\pi}{2} \sin(x) dx=- \cos(x)|_0^\frac{\pi}{2}=-(\cos(\frac{\pi}{2})-\cos(0))=-(0-1)=1 \\ \text{ and } \\ \int_0^\frac{\pi}{2} \sin(2x) dx= \int_0^\pi \frac{1}{2} \sin(u) du=-\frac{1}{2} \cos(u)|_0^\pi=-\frac{1}{2}(\cos(\pi)-\cos(0))=-\frac{1}{2}(-1-1)=1 $

|cite|improve this answer
$\endgroup$
  • $\begingroup$ I don't know if the example makes things a little clearer or not but I put it in there just in case. $\endgroup$ – randomgirl Apr 7 '15 at 20:24
  • 1
    $\begingroup$ $\frac{d}{dx} \frac{-1}{2} \cos(2x)=-\frac{1}{2}(2x)'(-\sin(2x))=\frac{1}{2}(2) \sin(2x)=\sin(2x)$ So that is why we need the 1/2 in our result so we cancel the derivative of the inside as that one guy stated chain rule is definitely involved here $\endgroup$ – randomgirl Apr 7 '15 at 20:28
0
$\begingroup$

To see why the chain rule is necessary, consider the simplest possible case: $f(x) = 1$, $F(x) = x$.

$$\int_1^3 f(2x) = \int_1^3 1 = 2$$

but

$$\int_2^6 f(x) = F(6) - F(2) = 6 - 2 = 4$$

(You have to compensate for $\int_2^6$ being 4 units long rather than just 2. This may become clearer if you draw yourself a picture.)

|cite|improve this answer
$\endgroup$

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.