The diagram shows the curve $y = \cos x + \sin x$ for $0 \le x \le \pi$. The curve meets the $y$-axis at the point $A$ and the $x$-axis at the point $B$.

1. Set $y = 0$ to find the $x$-intercept at point $B$.

At point $B$, the curve meets the $x$-axis, so $y = 0$.
$\cos x + \sin x = 0$

2. Rearrange the equation to solve for $x$.

$\sin x = -\cos x$
Dividing both sides by $\cos x$ (since $\cos x \neq 0$ when $\sin x = -\cos x$):
$\tan x = -1$

3. Solve for $x$ in the given domain $0 \le x \le \pi$.

The basic angle is $\alpha = \tan^{-1}(1) = \frac{\pi}{4}$.
Since $\tan x$ is negative and $0 \le x \le \pi$, $x$ must be in the second quadrant.
$x = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$
Thus, the $x$-coordinate of $B$ is $\frac{3\pi}{4}$.

1. Find the coordinates of point $A$.

At point $A$, the curve crosses the $y$-axis, so $x = 0$.
$y = \cos(0) + \sin(0) = 1 + 0 = 1$
So, $A$ is $(0, 1)$.

2. Find the area under the curve from $x = 0$ to $x = \frac{3\pi}{4}$.

Area under curve $= \int_{0}^{\frac{3\pi}{4}} (\cos x + \sin x) \, dx$
$= [\sin x - \cos x]_{0}^{\frac{3\pi}{4}}$

3. Evaluate the definite integral.

$= \left(\sin\left(\frac{3\pi}{4}\right) - \cos\left(\frac{3\pi}{4}\right)\right) - (\sin(0) - \cos(0))$
$= \left(\frac{\sqrt{2}}{2} - \left(-\frac{\sqrt{2}}{2}\right)\right) - (0 - 1)$
$= \sqrt{2} + 1$

4. Find the area of the triangle $OAB$.

The triangle $OAB$ is a right-angled triangle with base $OB = \frac{3\pi}{4}$ and height $OA = 1$.
Area of triangle $= \frac{1}{2} \times \frac{3\pi}{4} \times 1 = \frac{3\pi}{8}$

5. Calculate the area of the shaded region.

Shaded area $=$ Area under curve $-$ Area of triangle
$= (\sqrt{2} + 1) - \frac{3\pi}{8}$
$\approx 2.4142 - 1.1781 = 1.2361$
$\approx 1.24$ (to 3 s.f.)