Answer
$$\int^4_1\frac{\sqrt y-y}{y^2}dy=1-\ln4$$
Work Step by Step
$$A=\int^4_1\frac{\sqrt y-y}{y^2}dy$$ $$A=\int^4_1\frac{y^{1/2}-y}{y^2}dy$$ $$A=\int^4_1(y^{-3/2}-\frac{1}{y})dy$$ According to Table 1, we have $$\int x^ndx=\frac{x^{n+1}}{n+1}+C (n\ne-1)$$ $$\int \frac{1}{x}dx=\ln|x|+C$$
Therefore, $$A=(\frac{y^{-1/2}}{-\frac{1}{2}}-\ln|y|)\Bigg]^4_1$$ $$A=(-\frac{2}{\sqrt y}-\ln|y|)\Bigg]^4_1$$ $$A=(-\frac{2}{\sqrt 4}-\ln|4|)-(-\frac{2}{\sqrt 1}-\ln|1|)$$ $$A=(-1-\ln4)+(2+0)$$ $$A=1-\ln4$$
