Here I’ll introduce a very useful technique called differentiation under the integral sign.
Let f and \dfrac {\partial f} {\partial t} be continuous [a,b] \times [c,d]. Using the Fundamental Theorem of Calculus and the Mean Value Theorem, prove that:
\displaystyle \frac {\mathrm d} {\mathrm dt} \int_a^b f(x,t) \mathrm dx = \int_a^b \frac {\partial f} {\partial t} (x, t) \mathrm dx
for t \in (c,d).
That is, if F(t) = \displaystyle \int_a^b f(x,t) \mathrm dx and G(t) = \displaystyle \int_a^b \frac {\partial f} {\partial t} (x, t) \mathrm dx for all t \in (c,d), then F' = G.
This is Leibniz’s integral rule and we say we are differentiating under the integral sign.
Hint: Continuous functions are uniformly continuous on compact sets.
Let f:[a,b] \times [c,d] \to \mathbb R be continuous. Introduce a suitable parameter and use the above to prove (a particular case of) Fubini’s theorem:
\displaystyle \int_a^b \left(\int_c^d f(x,y) \mathrm dy\right) \mathrm dx = \int_c^d \left(\int_a^b f(x,y) \mathrm dx\right) \mathrm dy
