Integral oben 2 unten -1 (pq) dp

Integral oben 2 unten -1 (pq) dp

$\boxed{\int^{2}_{-1}pq\;\mathtt{dp}\\=q([\dfrac{p^2}{2}]^{2}_{-1})\leftarrow\boxed{\int p \mathtt{dp}=\dfrac{p^2}{2}}\\=q(\dfrac{2^2}{2}-\dfrac{(-1)^2}{2})\\=\dfrac{3q}{2}}$