Answer
a. $y=\sqrt x+C$.
b. $y=2\sqrt x+C$.
c. $y=2x^2-2\sqrt x+C$.
Work Step by Step
a. Given $y'=\frac{1}{2\sqrt x}$, we can find a function $f(x)=\sqrt x$ such that $f'(x)=\frac{1}{2\sqrt x}$. Based on Corollary 2, we have $y=f(x)+C=\sqrt x+C$ where $C$ is a constant.
b. Given $y'=\frac{1}{\sqrt x}$, we can find a function $f(x)=2\sqrt x$ such that $f'(x)=\frac{1}{\sqrt x}$. Based on Corollary 2, we have $y=f(x)+C=2\sqrt x+C$ where $C$ is a constant.
c. Given $y'=4x-\frac{1}{\sqrt x}$, we can find a function $f(x)=2x^2-2\sqrt x$ such that $f'(x)=4x-\frac{1}{\sqrt x}$. Based on Corollary 2, we have $y=f(x)+C=2x^2-2\sqrt x+C$ where $C$ is a constant.
