## Calculus (3rd Edition)

$$\sqrt{2}(b-a)$$
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 $$\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x=\int_{a}^{b} \sqrt{2} d x=\sqrt{2}(b-a)$$