Answer
$$\int^{4}_1\frac{dt}{t\sqrt t}=1$$
Work Step by Step
$$A=\int^{4}_1\frac{dt}{t\sqrt t}=\int^{4}_1\frac{dt}{t^{3/2}}$$ $$A=\int^{4}_1t^{-3/2}dt$$ $$A=\Big(\frac{t^{-1/2}}{-\frac{1}{2}}\Big)\Big]^{4}_1=\Big(-\frac{2}{\sqrt t}\Big)\Big]^{4}_1$$ $$A=-\frac{2}{\sqrt4}-\Big(-\frac{2}{\sqrt1}\Big)$$ $$A=-\frac{2}{2}+\frac{2}{1}=-1+2$$ $$A=1$$
