Let $w$ be a continuous form of degree $1$ in the open $U\subset \mathbb{R^m}$ and $\lambda:[a,b]\to U$ a path of class $C^1$. We have $\lambda^*w = f\ dx$, where $f:[a,b]\to\mathbb{R}$ is continuous and $\{dx\}$ is the canonical basis of $\mathbb{R}^*$. Show that the curved integral $\int_{\lambda} w$, defined in chapter IV, is nothing more than $\int_a^b f(x)\ dx$.
Chapter IV defines the curvi integral like this:
$$\int_\lambda w = \lim_{|P|\to 0} \sum_{j=1}^k w(\lambda(\xi(j))\cdot [\lambda(t_j)-\lambda(t_{j-1)}] =^* \int_a^b w(\lambda(t))\cdot \lambda'(t)\ dt$$
*in the case the path $\lambda$ is $C^1$
First, I don't know what $\lambda^*w = f\ dx$ means. Is it a definition of a product or composition between $\lambda$ and $w$?
What do I need to show exactly? That the integral of a $1$-form along a path is the same as integrating $f$ along $dx$?
UPDATE:
I found that my book defined the $^*$ operator like this:
Let $f:M\to N$ of class $C^k$. Given the form $w$, of degree $r$ over $N$, we define the form $f^*w$, of degree $r$ over $M$, by doing for each $x\in M$ and each $r$-list of vectors $w_1,\cdots, w_r\in T_xM$,
$$[(f^*w)(x)](w_1,\cdots, w_r) = w(f(x))\cdot(f'(x)\cdot w_1, \cdots, f'(x)\cdot w_r)$$
So, in my case, we have $f=\lambda$, and $w$ simply as $w$, we'd get something like this:
$$[(\lambda^*w)(x)](w_1,\cdots, w_r) = w(\lambda(x))\cdot(\lambda'(x)\cdot w_1, \cdots, \lambda'(x)\cdot w_r)$$
$\lambda$ is defined from $[a,b]\to U$, and $w$ from $U\subset\mathbb{R}^m$ to the space of linear functionals on $\mathbb{R}^m$. According to the book's definition, $M$ is $[a,b]$ and $N$ is $U$. So the vectors $w_1,\cdots, w_r$ are going to be on $T_xM = T_x[a,b]$, and since $w$ is of degree $1$, there will be only $1$ vector, that is, $r=1$, so we end up with:
$$[(\lambda^*w)(t)](w_0) = w(\lambda(t))\cdot(\lambda'(t)\cdot w_0)$$
I've exchanged $x$ by $t$ to remember that it's one dimensional. But what is $T_x[a,b]$? Which vector(s) generates it?
I need to end up with $\lambda^*w = w(\lambda(t))\cdot(\lambda'(t))$ or $\lambda^*w = w(\lambda(t))\cdot(\lambda'(t))\ dt$? And then integrate both sides to get that $\int_a^b f\ dx = \int_\lambda w$ as in the definition. How, exactly?
In the case of a one-form $\omega\in \Omega^1(U)$ and e.g. canonical coordinates $y_1,...,y_m$ on $U$ one usually writes: $$ \omega = \sum_{i=1}^m c_i(y) \; dy_i, \; \; y\in U .$$ Acting upon a vector $V=\sum_j v_i \frac{\partial}{\partial y_j}$ at the point $y\in U$, we get: $\omega_y(V)=\sum_i c_i(y) V_i$ which in the citation is written as a scalar-product $c\cdot V$.
The pull back $\lambda^*$ takes the one-form $\omega \in \Omega^1([a,b])$ to a one form on the interval $[a,b]$, the tangent-space of which is simply ${\Bbb R}$. Using the coordinate $x$ on $[a,b]$ and writing $\lambda_i = y_i \circ \lambda$ we have: $$ \lambda^* \omega = \sum_i c_i(\lambda(x)) d\lambda_i = \sum_i c_i(\lambda(x)) \frac{d\lambda_i}{dx} dx = f(x) dx$$ where again we may write $f$ as a scalar-product: $f(x) = c(\lambda(x)) \cdot \lambda'(x)$. Integrating this pulled back form over $[a,b]$ is then the same as the integral of $f$ from $a$ to $b$ and thus the same as integrating the form $\omega$ along the curve $\lambda$ (according to the definition you cite).