Answer
See explanation.
Work Step by Step
Using the rules of integrals:
(a) $\int_{0}^{4} f(x) d x=\int_{1}^{4} \frac{1}{x^{2}} d x+\int_{0}^{1} \sqrt{x} d x$
$=\left.\left(\frac{x^{-1}}{-1}\right)\right|_{1} ^{4}+\left.\left(\frac{2}{3} x^{3 / 2}\right)\right|_{0} ^{1}$
$=1-\frac{1}{4}+\frac{2}{3}=\frac{17}{12}$
(b) $F(x)=\frac{2 x^{3 / 2}}{3}$ if $0 \leq x \leq 1$ and $F(x)=\frac{5}{3}-\frac{1}{x}$ if $1 \leq x \leq 4$
$(\mathrm{c}) \int_{0}^{4} f(x) d x=-F(0)+F(4)=\frac{17}{12}$
