If $\int\limits_{-2}^{3} x^2 \, dx = k \int\limits_{0}^{2} x^2 \, dx + \int\limits_{2}^{3} x^2 \, dx$, then the value of $k$ is:

$2$

The correct answer is Option (1) → 2

$\int\limits_{-2}^{3} x^2 \, dx = k \int\limits_{0}^{2} x^2 \, dx + \int\limits_{2}^{3} x^2 \, dx$

$\Rightarrow \left[ \frac{x^3}{3} \right]_{-2}^{3} = k \left[ \frac{x^3}{3} \right]_{0}^{2} + \left[ \frac{x^3}{3} \right]_{2}^{3}$

$\Rightarrow \left[ 9 + \frac{8}{3} \right] = k \left[ \frac{8}{3} \right] + \left[ 9 - \frac{8}{3} \right]$

$\Rightarrow k = 2$