Let $$I:=\frac{\partial}{\partial \epsilon} \left[\epsilon \int^{b(\epsilon)}_{a(\epsilon)} x f(x) dx \right]_{\epsilon = 0}.$$ My textbook claims that, as a consequence of the Leibniz integral rule:
$$I= \int^{b(0)}_{a(0)} x f(x)dx + f(b(0)) \ b(0) \left( \frac{\partial}{\partial \epsilon} b(\epsilon) \right)_{\epsilon=0} - f(a(0)) \ a(0) \left( \frac{\partial}{\partial \epsilon} a(\epsilon) \right)_{\epsilon =0} .$$
I found that (by an application of the product rule):
$$I= \int^{b(0)}_{a(0)} x f(x)dx + \left[ \epsilon \left(\frac{\partial}{\partial \epsilon} \int^{b(\epsilon)}_{a(\epsilon)} x f(x) dx \right) \right]_{\epsilon = 0} = \int^{b(0)}_{a(0)} x f(x)dx + 0$$
which appears to be in contradiction with the first claim...
Am I missing something here ?
What you are given is suitable for Product Rule & you get the Correct $I$.
It is not suitable for Leibniz Integral Rule due to Extra Product term , which the text book is wronging ignoring.
Let us rewrite it to make it suitable for Leibniz Integral Rule :
$$ I = \frac{\partial}{\partial \epsilon} \left[\int^{b(\epsilon)}_{a(\epsilon)} \epsilon x f(x) dx \right]_{\epsilon = 0} $$
Consequence of the Leibniz Integral Rule is now :
$$ \begin{align} I &= \int^{b(0)}_{a(0)} x f(x)dx + \left( \epsilon f(b(0)) \ b(0) \frac{\partial}{\partial \epsilon} b(\epsilon) \right)_{\epsilon=0} - \left( \epsilon f(a(0)) \ a(0) \frac{\partial}{\partial \epsilon} a(\epsilon) \right)_{\epsilon =0} \\ &= \int^{b(0)}_{a(0)} x f(x)dx + 0 \end{align} $$
It will match with Product Rule ...