#### Answer

$$\sqrt{2}(b-a)$$

#### Work Step by Step

Since $0 \leq f^{\prime}(x) \leq 1,$ then
$$
\sqrt{1+\left(f^{\prime}(x)\right)^{2}} \leq \sqrt{1+1}=\sqrt{2}
$$
By comparison, we have
$$
\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x \leq \int_{a}^{b} \sqrt{2} d x=\sqrt{2}(b-a)
$$
For $f(x)=x,$ we have
$$
f^{\prime}(x)=1
$$
Therefore, we get
$$
\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x=\int_{a}^{b} \sqrt{2} d x=\sqrt{2}(b-a)
$$