Answer
False.
For example,
let $f(x)=x^{2}$. Then
$$\int_{a}^{b} \sqrt{f(x)} d x= \int_{0}^{1} \sqrt{x^{2}} d x=\int_{0}^{1} x d x=\frac{1}{2},
$$
but
$$
\sqrt{\int_{a}^{b} f(x) d x}= \sqrt{\int_{0}^{1} x^{2} d x}=\sqrt{\frac{1}{3}}=\frac{1}{\sqrt{3}}
$$
Thus,
$$\int_{a}^{b} \sqrt{f(x)} d x\ne\sqrt{\int_{a}^{b} f(x) d x}$$
Work Step by Step
False.
For example,
let $f(x)=x^{2}$. Then
$$\int_{a}^{b} \sqrt{f(x)} d x= \int_{0}^{1} \sqrt{x^{2}} d x=\int_{0}^{1} x d x=\frac{1}{2},
$$
but
$$
\sqrt{\int_{a}^{b} f(x) d x}= \sqrt{\int_{0}^{1} x^{2} d x}=\sqrt{\frac{1}{3}}=\frac{1}{\sqrt{3}}
$$
Thus,
$$\int_{a}^{b} \sqrt{f(x)} d x\ne\sqrt{\int_{a}^{b} f(x) d x}$$
