Determine the derivative of the function $\displaystyle F(x) = \int^1_x \sqrt{t + \sin t} dt$ using the properties of integral.

Using the properties of integral

$\displaystyle \int^a_b f(x) dx = - \int^b_a f(x) dx$

Then,

$\displaystyle F(x) = \int^1_x \sqrt{t + \sin t} d t = - \int^x_1 \sqrt{1 + \sin t} dt$

Since $F(t) = - \sqrt{t + \sin t}$, using the first fundamental theorem of calculus

$\displaystyle g(x) = \int^x_a f(t) dt$, then

$F'(x) = -\sqrt{x+\sin x}$

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