Answer
$$\int\sqrt{3-2s}ds=-\frac{(3-2s)^{3/2}}{3}+C$$
Work Step by Step
$$A=\int\sqrt{3-2s}ds$$
We set $u=3-2s$.
Then $$du=-2ds$$
That means, $$ds=-\frac{1}{2}du$$
Therefore, $$A=-\frac{1}{2}\int\sqrt udu=-\frac{1}{2}\int u^{1/2}du$$ $$A=-\frac{1}{2}\times\frac{u^{3/2}}{\frac{3}{2}}+C$$ $$A=-\frac{u^{3/2}}{3}+C$$ $$A=-\frac{(3-2s)^{3/2}}{3}+C$$
