Answer
$\frac{1}{3}$
Work Step by Step
We have $g(t)=\sin t\cos t$ and $f(t)=\sin t $ for $0\leq t\leq \pi/2$.
For $t=0$, $x=\sin 0=0$
For $t=\pi/2$, $x=\sin\pi/2=1$
Then, area enclosed the curve and the x-axis is given by the integral:
$A=\int_0^{\pi/2}g(t)f'(t)dt$
Evaluate $A$:
$A=\int_0{\pi/2}\sin t\cos t\cdot \cos tdt$
$A=\int_0^{\pi/2}\sin t\cos^2 tdt$
$A=[-\frac{\cos^3t}{3}]_0^{\pi/2}$
$A=(-\frac{\cos^3(\pi/2)}{3})-(-\frac{\cos^30}{3})$
$A=(-0^3/3)-(-1^3/3)$
$A=0+\frac{1}{3}$
$A=\frac{1}{3}$
Thus, the area is $\frac{1}{3}$.