Let X be the random variable with probability density function
$$ f_X (x) = \begin{cases} \frac{3}{32} (-x^2+2x+3), & x\in(-1,3) \\ 0, & \text{otherwise} \\ \end{cases}. $$
Let $M_X$ be the median of $X$ and $m_X$ be the mode. Determine $M_X+m_X$.
To calculate the mode of $f_X$, it is the value of x at which $f_X$ attains its maximum. In this case, it is $1$.
I know that generally to calculate the median,
$$ \int_{-\infty}^{M_X} f_X (x) = \frac{1}{2} $$
But as we have an interval on $x$, should the lower limit be $-1$?
Yes, you can integrate from $x = -1$ instead of $x = -\infty$. The relevant concept here is the support of the distribution. When working with probability densities, you can generally ignore the parts of the domain where $f_X(x) = 0$. In these portions of the domain, the CDF is unchanged, so they have no effect on probabilities.
In this particular problem, you can split up the integral used to calculate the median into two parts, only one of which is over the support of the distribution:
$$ \int_{-\infty}^{M(x)} {f_X(x) dx} = \int_{-\infty}^{-1} {f_X(x) dx} + \int_{-1}^{M(x)} {f_X(x) dx} . $$
Then, since $f_X(x) = 0$ for $x \in (-\infty, -1)$, we have
$$ \int_{-\infty}^{M(x)} {f_X(x) dx} $$ $$ = \int_{-\infty}^{-1} {0 dx} + \int_{-1}^{M(x)} {f_X(x) dx} $$ $$ = 0 + \int_{-1}^{M(x)} {f_X(x) dx} $$ $$ = \int_{-1}^{M(x)} {f_X(x) dx} $$