Answer
$\int_0^\frac{\pi}{2}sin2xdx=1$
Work Step by Step
$y=sin2x$, find the area of the region on the interval $(0,\frac{\pi}{2})$
$\int_0^\frac{\pi}{2}sin2xdx$
Let $u=2x$
$\frac{du}{dx}=2$
$dx=\frac{1}{2}du$
$=\frac{1}{2}\int_0^\frac{\pi}{2}sinudu$
$=\frac{1}{2}(-cosu)|_0^\frac{\pi}{2}$
$=-\frac{1}{2}(cos2x)|_0^\frac{\pi}{2}$
$=-\frac{1}{2}[(cos2(\frac{\pi}{2}))-(cos2(0))]$
$=-\frac{1}{2}[(cos\pi)-(cos0)]$
$=-\frac{1}{2}(-1-1)$
$=(-\frac{1}{2})(-2)$
$=1$
