False: top-form integration needs no orientation

Orientation reversal changes the integral sign: Let $-M$ have the opposite orientation on every component of an oriented smooth manifold $M$. For every compactly supported top form, ${\int }_{-M}\omega =-{\int }_M\omega$, in all dimensions.

Chart integral with its orientation sign: Let $M^n$ be oriented and $\omega$ a smooth top form with compact support contained in a connected chart $(U,\varphi )$. For $n\ge 1$ write $({\varphi }^{-1}{)}^{\ast }\omega =fd{x}^1\wedge \cdots \wedge d{x}^n\mathrm{.}$ Let ${\sigma }_{\varphi }\in \{1,-1\}$ be the sign of its coordinate frame relative to the chosen orientation. Define the chart integral by $I_{\varphi }(\omega )={\sigma }_{\varphi }{\int }_{{\mathbb{R}}^n}\tilde{f}(x)dx\mathrm{.}$ Here $\tilde{f}$ is the Riemann-integrable zero extension, including across a genuine half-space face, as in lem-chart-supported-coefficients-have-well-defined-riemann-integrable-half-space-extensions. For $n=0$, a connected chart is a point $p$, and set $I_p(\omega )=\epsilon (p)\omega (p)$ using its determinant-line sign. Empty support gives zero. Negative charts are allowed: the upper-half-line chart $u=b-t$ at the right endpoint of an increasing interval has sign $-1$.