Integral of Pointwise Multiplication of Mollifier and Function?

Question

Say I have a mollifier defined by $$ \Phi\rho(x)=\int\rho(\tau)\mathbb{1}_{-1-\epsilon,1+\epsilon}(x-\tau) d\tau=\varphi(x) $$ where $$ \rho(x)=\begin{cases} e^{-1/(1-\lvert x \rvert^{2})}/K & \text{ if } x \in (-1,1) \\ 0 & \text{otherwise}\end{cases} $$ where $K=\int_{-1}^{1} e^{-1/(1-\lvert x \rvert^{2})} \, dx$ and then for some function $f \in L^{1}(\mathbb{R})$ I want to evaluate $$ \int_{\mathbb{R}}f(x)\varphi(x) \, dx $$ So I'm (almost) sure that I can simplify this to be $$ \int_{\mathbb{R}}f(x)\varphi(x) \, dx=\int_{-1}^{1} f(x)\varphi(x) \, dx $$ but here I reach a problem. On one hand, this seems like it should be like multiplying by the indicator function of a set so we'd have $$ \int_{-1}^{1} f(x)\varphi(x) \, dx=\int_{-1}^{1} f(x) \, dx $$ since the mollifier and the indicator function are a.e. equal pointwise on $(-1,1)$ (or perhaps I'm mistaken here?). Basically my question is, how does one evaluate the integral of the pointwise product of a mollifier and some $f \in L^{1}$? I apologize if this question is too elementary but I'd appreciate any insight anyone can offer.