Answer
$F(x)=1-\cos x$
$F(2) \approx 1.4161$
$F(5) \approx 0.7163$
$F(8) \approx 1.1455$
Work Step by Step
Apply The Second Fundamental Theorem of Calculus (Th.4.11 )
If $f$ is continuous on an open interval I containing $a$, then, for every $x$ in the
interval,
$\displaystyle \frac{d}{dx}\left[\int_{a}^{x}f(t)dt\right]=f(x)$.
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$F(x)=\displaystyle \int_{0}^{x}\sin\theta d\theta=[-\cos\theta]_{0}^{x}$
$=-\cos x+\cos 0$
$=1-\cos x$
$F(x)=1-\cos x$
$F(2)=1-\cos 2\approx 1.4161$
$F(5)=1-\cos 5\approx 0.7163$
$F(8)=1-\cos 8\approx 1.1455$
