Answer
$1$
Work Step by Step
Recall: Let $f(x)\geq 0$ for $a\leq x\leq b$, the area under the graph of $y=f(x)$ and above the $x-$axis between $x=a$ and $x=b$ is represented by $A=\int_a^b f(x)dx$.
Using this knowledge to find the area:
$Area=\int_0^{\pi/2}\sin x dx$
$=[-\cos x]_0^{\pi/2}$
$=(-\cos \frac{\pi}{2})-(-\cos 0)$
$=0-(-1)$
$=1$
Thus, the area is $1$.
