## Calculus: Early Transcendentals 8th Edition

Given:$\int_{a}^{b}\sqrt {f(x)}dx=\sqrt {\int_{a}^{b}f(x)dx}$ Consider $f(x)=1,, a=0, b=4$ Therefore, $\int_{0}^{4}\sqrt {1}dx=\sqrt {\int_{0}^{4}(1)dx}$ Thus, $4\ne\sqrt4$ Hence, $\int_{a}^{b}\sqrt {f(x)}dx\ne\sqrt {\int_{a}^{b}f(x)dx}$ The given statement is false.