Answer
$$\sqrt{5}$$
Work Step by Step
Theorem of Pythagoras
$L=\sqrt{(4-2)^{2}+(2-1)^{2}}=\sqrt{4+1}=\sqrt{5}$
(a) $L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x=\int_{1}^{2} \sqrt{1+2^{2}} d x=\sqrt{5}$
Furmula5 with g(y)=$\frac{1}{2}y$
(b) $L=\int_{c}^{d} \sqrt{1+\left(g^{\prime}(y)\right)^{2}} d y=\int_{2}^{4} \sqrt{1+\left(\frac{1}{2}\right)^{2}} d x=$
$ (4-2)\cdot \frac{\sqrt{5}}{2}=\sqrt{5}$
