The area of the region bounded by the y-axis, y = cos x and y = sin x, 0 ≤ x ≤ π2 is

The correct option is (√2 – 1) sq. units.

Let us find the area of the region bounded by the y-axis, $y=\cos x$, and $y=\sin x$ in the interval $0\le x\le \frac{\pi }{2}$ step-by-step.

Step 1: Identify the boundaries and the point of intersection.
The region is bounded by:
1. The y-axis, which corresponds to the line $x=0$.
2. The curve $y=\cos x$.
3. The curve $y=\sin x$.
To find where the two curves intersect in the interval $[0,\frac{\pi }{2}]$, we equate them:
$\sin x=\cos x$
Dividing both sides by $\cos x$ (since $\cos x\ne 0$ in the region of intersection):
$\tan x=1$
In the interval $0\le x\le \frac{\pi }{2}$, this occurs at:
$x=\frac{\pi }{4}$

Step 2: Set up the area integral.
For the interval $0\le x\le \frac{\pi }{4}$, the curve $y=\cos x$ lies above the curve $y=\sin x$ (since $\cos 0=1$ and $\sin 0=0$).
Therefore, the area $A$ of the bounded region is given by the definite integral from the y-axis ($x=0$) to their intersection point ($x=\frac{\pi }{4}$):

$A={\int }_0^{\frac{\pi }{4}}(\cos x-\sin x) dx$

Step 3: Evaluate the integral.
Using standard integration rules, we find the antiderivatives:
$\int \cos x dx=\sin x$
$\int \sin x dx=-\cos x$
Applying these to our definite integral:

$A={[\sin x-(-\cos x\left)\right]}_0^{\frac{\pi }{4}}$

$A={[\sin x+\cos x]}_0^{\frac{\pi }{4}}$

Step 4: Substitute the limits.
Evaluate the expression at the upper limit $x=\frac{\pi }{4}$ and subtract the value at the lower limit $x=0$:

$A=(\sin \frac{\pi }{4}+\cos \frac{\pi }{4})-(\sin 0+\cos 0)$

Substitute the known trigonometric values:
$\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}$, $\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$, $\sin 0=0$, and $\cos 0=1$.

$A=(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})-(0+1)$

$A=\frac{2}{\sqrt{2}}-1$

$A=\sqrt{2}-1$

Thus, the area of the region is $(\sqrt{2}–1)$ sq. units.