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