Inner product and Integral

Question

I was reading a proof that claims that the length of a curve $\alpha:[a,b]\rightarrow\mathbb{R}^n$ which is denoted by $l(\alpha)$ satisfies $l(\alpha)\geq|\alpha(a)-\alpha(b)|$ . It starts by

$\alpha(b)-\alpha(a)=\int_a^b\alpha'(t)dt$ and then by applying the inner product with $\alpha(b)-\alpha(a)$ at this equality we have:

$$|\alpha(b)-\alpha(a)|^2=\int_a^b\bigl\langle\alpha'(t),\alpha(b)-\alpha(a)\bigr\rangle dt$$ and then it continues with the proof.

My doubt is with this implication. The step $$|\alpha(b)-\alpha(a)|^2=\left\langle\int_a^b\alpha'(t) dt,\alpha(b)-\alpha(a)\right\rangle$$ is missing, but why $$\left\langle\int_a^b\alpha'(t) dt,\alpha(b)-\alpha(a)\right\rangle=\int_a^b\bigl\langle\alpha'(t),\alpha(b)-\alpha(a)\bigr\rangle dt?$$ Thank you in advance.

Answer 1

It is because\begin{align}\bigl\|\alpha(b)-\alpha(a)\bigr\|^2&=\bigl\langle\alpha(b)-\alpha(a),\alpha(b)-\alpha(a)\bigr\rangle\\&=\left\langle\int_a^b\alpha'(t)\,\mathrm dt,\alpha(b)-\alpha(a)\right\rangle\\&=\int_a^b\bigl\langle\alpha'(t),\alpha(b)-\alpha(a)\bigr\rangle\,\mathrm dt.\end{align}