Leibniz' Rule: Prove $\int_0^y u_{tt}(x,t) dt = u_y (x,y) - u_y (x,0)$

Solution 1:

If I asked you why

$$\tag 1 \int_a^b f''(t)\,dt = f'(b)-f'(a),$$

you would have an answer: It follows from the FTC, since $f'$ is an antiderivative of $f''.$ That's all that is going on in your situation.

Things might be clearer if we use $D_1, D_2$ for the partial derivatives with respect the first and second variables. With $x$ fixed, define $f(t) = u(x,t).$ Then by the definition of partial derivatives, we have $f'(t) = D_2u(x,t)$ and $f''(t) = D_2(D_2 u)(x,t).$ Thus by $(1),$

$$\int_a^b D_2(D_2 u)(x,t) dt = \int_a^b f''(t)\,dt$$ $$ = f'(b)-f'(a) = (D_2 u)(x,b) - (D_2 u)(x,a).$$

If you think of $a=0,b=y,$ we obtain exactly the expression you asked about.