Answer
(b) Negative
(a) Positive
Work Step by Step
(a) $x^{4}>0$ on [-3,-1], $\sqrt{-x+3}>0$ on $[-3,-1],$ so $f(x)=\frac{x^{4}}{\sqrt{-x+3}}>0$ on $[-3,-1] .$ By theorem 5.5.6, we have that $\int_{-3}^{-1} f(x) d x>0$
(b) $-9+x^{3}<0$ on [-2,2] , $1+|x|>0$ on $[-2,2],$ so $f(x)=\frac{-9+x^{3}}{1+|x|}<0$
on $[-2,2] .$ By theorem 5.5.6, we get that $\int_{-2}^{2} f(x) d x<0$
