Area of the region bounded by the curve $y=\cos x$ and x-axis between $x=0$ and $x=\pi$ is : |
2 sq. units 3 sq. units 4 sq. units 1 sq. unit |
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 |