How to compute $\frac{d}{dt} \int_{a(t)}^{b(t)} f(x,t) dx$

Question

I want to differentiate $ \int_{a(t)}^{b(t)} f(x,t) dx$ this integral with respect to $t$ $i.e$,

\begin{align} \frac{d}{dt} \int_{a(t)}^{b(t)} f(x,t) dx \end{align}

I only know for the case of constant $a,b$. For this case how can i differentiatie?

Answer 1

Consider the function $$ F(t, A, B) := \int_A^B f(x,t)\,dx $$ Then the integral in question is $I(t) = F(t, a(t), b(t))$. The chain rule gives $$ I'(t) = \partial_t F + \partial_A F \cdot a'(t) + \partial_B F \cdot b'(t) $$ Therefore, $$ I'(t) = \int_a^b \partial_t f(x,t)\, dx - f(a(t),t)a'(t) + f(b(t),t)b'(t) $$

Answer 2

$$\frac{db(t)}{dt}\times f(b(t),t)-\frac{da(t)}{dt}\times f(a(t),t)+\int_{a(t)}^{b(t)}\frac{\partial f(t,x)}{\partial t} dx$$ For more details , see Leibniz integral rule.