Answer
a. $-\sqrt x+C$
b. $x+C$
c. $\sqrt x+C$
d. $-x+C$
e. $x-\sqrt x+C$
f. $-x-\sqrt x+C$ where $C$
where $C$ represents a general constant.
Work Step by Step
Given $f(x)=\frac{d}{dx}(1-\sqrt x)$ and $g(x)=\frac{d}{dx}(x+2)$, we have:
a. $\int f(x)dx=\int (\frac{d}{dx}(1-\sqrt x))dx=\int d(1-\sqrt x)=1-\sqrt x+C=-\sqrt x+D$
where $C, D$ are constants.
b. $\int g(x)dx=\int (\frac{d}{dx}(x+2))dx=\int d(x+2)=x+2+C=x+D$
where $C, D$ are constants.
c. $\int[- f(x)]dx=\int (-\frac{d}{dx}(1-\sqrt x))dx=-\int d(1-\sqrt x)=-1+\sqrt x+C=\sqrt x+D$
where $C, D$ are constants.
d. $\int [-g(x)]dx=\int (-\frac{d}{dx}(x+2))dx=-\int d(x+2)=-x-2+C=-x+D$
where $C, D$ are constants.
e. $\int[f(x)+g(x)]dx=\int f(x)dx+\int g(x)dx=x-\sqrt x+C$
where $C$ is a constant.
f. $\int[f(x)-g(x)]dx=\int f(x)dx-\int g(x)dx=-x-\sqrt x+C$
where $C$ is a constant.
