$$ f-g = (x-2)(x+2)(k^2-k) $$
$$ \int_{-2}^2 f-g \,dx = 8 $$
$$ (k^2-k) \int_{-2}^2 x^2-4 \,dx = 8 $$
$$ (k^2-k) \left[ {1\over3}x^3-4x \right]_{-2}^2 = 8 $$
$$ (k^2-k) \left| -{32\over3} \right| = 8 $$
$$ 4k^2-4k-3 = 0 $$
$$ (2k+1)(2k-3) = 0 $$
$$ k_1 = -{1\over2} $$
$$ k_2 = {3\over2} $$
Grüße,
M.B.