Area of the region bounded by the curve $y=\cos x$ and x-axis between $x=0$ and $x=\pi$ is :

2 sq. units

y = cos x  

x = 0 to x = π

so

area from 0 to $\frac{\pi}{2}$ is positive

area from $\frac{\pi}{2}$ to $\pi$ is negative

So Area

$=\int\limits_0^{\frac{\pi}{2}} \cos x d x+\left(-\int\limits_{\frac{\pi}{2}}^\pi \cos x d x\right)$   [to make area positive -ve sign added]

⇒ $[\sin x]_0^{\frac{\pi}{2}}-[\sin x]_{\frac{\pi}{2}}^\pi$

= (1 - 0) - (0 - 1)

= 2 sq. units