The area of the region bounded by the curve y = sin x between the ordinates x = 0, x = π/2 and the x-axis is

The correct option is "1 sq, unit".

To find the area of the region bounded by the curve $y=\sin x$, the ordinates $x=0$ and $x=\frac{\pi }{2}$, and the $x$-axis, we use definite integration.

The area $A$ under a curve $y=f\left(x\right)$ from $x=a$ to $x=b$ is given by the integral:
$$A={\int }_a^{b}y dx$$

Here, the curve is $y=\sin x$, and the limits of integration are from $a=0$ to $b=\frac{\pi }{2}$. Since the sine function is non-negative on the interval $[0,\frac{\pi }{2}]$, the area is simply:
$$A={\int }_0^{\frac{\pi }{2}}\sin x dx$$

The antiderivative of $\sin x$ is $-\cos x$. Applying the fundamental theorem of calculus, we evaluate the antiderivative at the upper and lower limits:
$$A={[-\cos x]}_0^{\frac{\pi }{2}}$$

Substituting the limits into the expression:
$$A=-(\cos \frac{\pi }{2}-\cos 0)$$

We know that $\cos \frac{\pi }{2}=0$ and $\cos 0=1$. Substituting these values yields:
$$A=-(0-1)=1\text{sq. unit}$$

Thus, the area of the region is indeed 1 sq. unit.