The integral forms of Maxwell's equations are always to be used with some care to get the signs right. In that sense they are more complicated than the fundamental differential forms of the law. So we start from the simple differential form of Faraday's law of induction (in SI units)
$$\vec{\nabla} \times \vec{E} = -\partial_t \vec{B}.$$
You get the integral law by using Stokes's integral theorem, i.e., you integrate the equation over an oriented surface, ##A##, i.e., a surface with surface-normal vectors ##\mathrm{d}^2 \vec{f}## oriented arbitrarily in one of two possible directions. Then on the left-hand side you can transform the surface integral as a line integral along the boundary, ##\partial A##, of the surface. Its orientation must be chosen using the right-hand-rule relative to the orientation of the surface-normal vectors. Then you get
$$\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{E} = -\int_A \mathrm{d}^2 \vec{f} \cdot \partial_t \vec{B}.$$
Finally, if we assume for simplicity that the surface and its boundary are at rest (and ONLY then) you can take the time derivative out of the surface integral
$$\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{E} = -\frac{\mathrm{d}}{\mathrm{d} t} \int_A \mathrm{d}^2 \vec{f} \cdot \vec{B}.$$
