Let $\lambda$ denote the Lebesgue measure on $\mathcal B(\mathbb)^{\otimes d}$, $I$ be an open interval, $d\in\mathbb N$, $\Omega\in\mathcal B(\mathbb R^d)$, $\varphi:I\to\Omega$ and $f\in\mathcal L^1(\Omega)$.
By the general result of integration with respect to a pushforward measure, we know that $$\int_If\circ\varphi\:{\rm d}\lambda=\int f\:{\rm d}\varphi_\ast(\lambda).\tag1$$
Are we able to find a formula for the right-hand side of $(1)$? I'm particularly interested inthe case $\gamma=\left.\gamma_{x,\:y}\right|_I$ for some $x,y\in\mathbb R^d$, where $$\gamma_{x,\:y}:\mathbb R\to\mathbb R^d\;,\;\;\;t\mapsto x+ty$$ for $x,y\in\mathbb R^d$, are we able to find a formula for the right-hand side of $(1)$?
EDIT:
I think we don't need to restrict to the special form $\gamma=\left.\gamma_{x,\:y}\right|_I$ for some $x,y\in\mathbb R^d$. Remember the following result from differential geometry:
Theorem: Let $k\in\{1,\ldots,d\}$ and $\phi$ be a $C^1$-diffeomorphism from $\Omega$ onto an open subset $U$ of $\mathbb R^k$ (i.e. $\Omega$ is a $k$-dimensional $C^1$-submanifold of $\mathbb R^d$ and $\phi$ is a global $C^1$-chart for $\Omega$).
Then the surface measure $\sigma_\Omega$ on $\Omega$ is given by $$\sigma_\Omega=\phi^{-1}_\ast\left(\det\left|{\rm D}\phi^{-1}\right|\left.\lambda^{\otimes k}\right|_U\right)\tag{a}.$$
Applying this result for $k=1$ and $U=I$ and assuming that $\gamma=\phi^{-1}$ for some $\phi$ as in the theorem above, we obtain $$\sigma_\Omega=\gamma_\ast\left(\left\|\gamma'\right\|\left.\lambda\right|_I\right)\tag{b}$$ and hence $$\int f\:{\rm d}\sigma_\Omega=\int_I\left\|\gamma'\right\|(f\circ\gamma)\:{\rm d}\lambda.\tag{c}$$ Can we somehow show that the left-hand side is actually equal $$\int_{\gamma(I)}f\:{\rm d}\lambda^{\otimes d}?\tag{d}$$
Deprecated content of the original post:
If $$\sigma_t:\mathbb R^d\to\mathbb R^d$$ for $t\in\mathbb R$ and $$\tau_x:\mathbb R^d\to\mathbb R^d\;,\;\;\;y\mapsto x+y$$ for $x,y\in\mathbb R^d$, we know that \begin{align}\forall t>0&:\sigma_t\left(\lambda^{\otimes d}\right)=\frac1{t^d}\lambda^{\otimes d}\tag2\\\forall x\in\mathbb R^d&:\tau_x\left(\lambda^{\otimes d}\right)=\lambda^{\otimes d}\tag3.\end{align} While we may clearly write $$\forall t\in\mathbb R:\gamma_{x,\:y}(t)=\tau_x\left(\sigma_t(y)\right)\tag4,$$ this doesn't seem to help. In order to determine $\varphi_{x,\:y}(\lambda)$, it is clearly enough to consider $a,b\in\mathbb R^d$ with $a\le b$ and $$\gamma_{x,\:y}(\lambda)((a,b))=\lambda\left(\left\{t\in\mathbb R:ty\in(a-x,b-x)\right\}\right)\tag5.$$
Maybe the desired claim is some generalization of the substitution rule $$\int_I(f\circ\varphi)\varphi'\:{\rm d}\lambda=\int_{\varphi(I)}f\:{\rm d}\lambda\tag6$$ which holds in the case $d=1$ whenever $f\in C(\Omega)$ and $\varphi$ is differentiable.