Intuition of Reynold's Transport Theorem

Question

I have a textbook which states that for a function $H(t) = \int_{a(b)}^{b(t)} f(x, t) \mathop{}\!dx$, $$ \frac{\mathop{}\!dH}{\mathop{}\!dt}(0) = \left. \frac{\mathop{}\!d}{\mathop{}\!dt} \int_{a(0)}^{b(0)} f(x, t) \mathop{}\!dx \right\rvert_{t=0} + f(b(0), 0) \frac{\partial b}{\partial t}(0) - f(a(0), 0) \frac{\partial a}{\partial t}(0), $$ and I simply cannot understand this result. I am especially confused by the fact that to show this, the textbook rewrites $H$ to $$ H(t) = F\left(b(t), t\right) - F\left(a(t), t\right) $$ were $F$ is the anti-derivative of $f$. It then says it applies the chain rule to arrive at the above result. However, to me, it seems like we should get $$ \frac{\mathop{}\!dH}{\mathop{}\!dt} = f(b(t), t) \frac{\partial b}{\partial t} (t) - f(a(t), t) \frac{\partial a}{\partial t} (t). $$ I would appreciate all help in clearing up where I am getting it wrong.

Answer 1

Using Leibniz's Rule for differentiating under the integral, we see that

$$ H'(t)=f(b(t),t)b'(t)-f(a(t),t)a'(t)+\int_{a(t)}^{b(t)} \frac{\partial f(x,t)}{\partial t}\,dx$$

Now set $t=0$. Can you finish now?

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NOTE:

The last expression in the OP ommitted the term

$$\left. \frac{\partial F(x,y)}{\partial y } \right|_{(x,y)=(a(t),t)}^{(x,y)=(b(t),t)} =\int_{a(t)}^{b(t)} \frac{\partial f(x,t)}{\partial t}\, dx$$