Why does having a function within a function change the integral value? 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.
 A: 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.
A: $ \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 $
A: 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.)
