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
    $F'(x) = -\sqrt{x+\sin x}$
No comments:
Post a Comment