Answer
$\dfrac{625}{12}$
Work Step by Step
Consider $V= \int_{-4}^{1} \int_{3x}^{4-x^2} (x+4) dy dx$
or, $= \int_{-4}^{1} [x y+4y]_{3x}^{4-x^2} dx$
or, $= \int_{-4}^{1} [4x-x^3+16-4x^2]-[3x^2+12x]dx$
or, $= \int_{-4}^{1} [-8x-x^3+16-7x^2]dx$
or, $= [-4x^2-\dfrac{x^4}{4}+16x-\dfrac{7x^3}{3}]_{-4}^{1}$
Thus, we have $I= [-4- \dfrac{1}{4}+16-\dfrac{7}{3}]- [ -64-64-64+\dfrac{448}{3}]=\dfrac{625}{12}$
