Good day! Got stuck again on a Calc-III question, on the subject of differential 1-forms. I'll present the question and my thought process in trying to solve it up to now.
Let $f: [a,b] \rightarrow \mathbb{R}$ be a continuous function of bounded variation such that $f(a) = f(b) = 0$. Let $\varphi:= (t, f(t))$ for all $t \in [a,b]$ Prove that: $\int_\varphi (2xy + y)dx + x^2dy = \int_a^b f(x) dx$
First let us change notations a little by introducing - \begin{gather*} F: \mathbb{R}^2 \rightarrow \mathbb{R}^2 \\ (x, y) \rightarrow (2xy + y, x^2) \end{gather*}
Then, $F \cdot d(x, y)$ is a differential 1-form. It is easily deducible that the we are to prove that $\int_\varphi F \cdot d(x, y) = \int_a^b f(x) dx$.
Now, $f$ is integrable over $[a,b]$ as it is continuous there.
In addition, $\varphi$ is continuous as a collection of continuous functions, and it is also a collection of functions of bounded variation.
Thus, $\varphi$ is rectifiable and therefore $F \cdot dx$ is integrable over $\varphi$ as $F$ is also continuous in $\mathbb{R}^2$ let alone in the image of $\varphi$.
As for the actual question at hand, the first thing that came to mind is to show that $F$ is a conservative field (specifically using the 1-form case of Poincare's lemma). But as it turns out, (or immediately seen if you're used to these kind of processes unlike myself) the matrix $JF_a$ is asymmetrical for pretty much every $a$: $\frac{\partial [F]_1}{\partial y}(x) = 2x + 1$, $\frac{\partial [F]_2}{\partial x}(x) = 2x$. As you can see it is almost symmetric but not quite. So in fact $F$ is not a conservative field. Even if we try to restrict its domain, I don't think this will yield anything useful. (I don't think any possibly helpful restriction is star-shaped anyhow)
The next thing one might try is to show somehow that $\varphi$ is continuously differentiable but $f$ might very well not be differentiable in it's domain.
At a last-ditch effort I tried computing the Rieman's sum of the left side and somehow get the right side, but I got no where with that except maybe a waste of a page.
Thanks for reading! Would be grateful if anyone could lend a hint towards the solution.
Have a great time for the rest of your day!
Hint: $$\int_\varphi (2xy+y)\:dx+x^2\:dy = \int_\varphi 2xy\:dx+x^2\:dy + \int_\varphi y\:dx$$ $$= \int_\varphi d(x^2y) + \int_\varphi y\:dx$$