Answer
$$\int^{4}_{1}\frac{2^{\sqrt x}}{\sqrt x}dx=\frac{4}{\ln2}$$
Work Step by Step
$$A=\int^{4}_{1}\frac{2^{\sqrt x}}{\sqrt x}dx$$
We set $\sqrt x=u$, which means $$\frac{1}{2\sqrt x}dx=du$$ $$\frac{1}{\sqrt x}dx=2du$$
- For $x=4$, we have $u=2$ and for $x=1$, we have $u=1$
Therefore, $$A=2\int^{2}_{1}2^udu$$ $$A=2\times\frac{2^u}{\ln2}\Big]^{2}_{1}$$ $$A=\frac{2}{\ln2}(2^2-2^1)$$ $$A=\frac{2\times2}{\ln2}=\frac{4}{\ln2}$$
