A step involving integration by part in a proof

Question

This is lemma 4 from Gerfand and Fomin.

I want to ask how to get $\int^b_a[\alpha(x)h(x)dx=-\int^b_aA(x)h'(x)dx\ $ by applying integration by part? I try it for myself and can't get the exact expression. In particular, $A'(x)=\alpha(x)-0+\underbrace{\int^x_a\frac{\partial\alpha (t)}{\partial x}dt}_{=0}=\alpha(x)$ by the Leibniz integral rule.

Answer 1

First note that $A'(x)=\alpha(x)$. Then, by integration by parts $$\int^b_a \alpha(x)h(x)dx=A(b)h(b)-A(a)h(a)-\int^b_a A(x)h'(x)dx$$ Now we know that $h(a)=h(b)=0$ and we may conclude that $$\int^b_a \alpha(x)h(x)dx=-\int^b_a A(x)h'(x)dx$$ that is $$\int^b_a (\alpha(x)h(x)+A(x)h'(x))dx=0.$$

Answer 2

Note $$\left[h(x)A(x)\right]^{b}_{a} = 0 $$ since $h(b) = h(a) = 0$