Question

Find area enclosed by $y = \sin x, x-axis, x = 0,x = 2\pi$.

Answer 1

4

Answer 2

The area enclosed by $y = \sin x$, the x-axis, $x=0$, and $x=2\pi$ is calculated by integrating the absolute value of $\sin x$ over the interval $[0, 2\pi]$. Since $\sin x$ is positive for $x \in [0, \pi]$ and negative for $x \in [\pi, 2\pi]$, the integral is split into two parts: $\int_{0}^{\pi} \sin x dx$ and $\int_{\pi}^{2\pi} (-\sin x) dx$.

$\int_{0}^{\pi} \sin x dx = [-\cos x]_{0}^{\pi} = (-\cos \pi) - (-\cos 0) = 1 - (-1) = 2$.

$\int_{\pi}^{2\pi} (-\sin x) dx = [\cos x]_{\pi}^{2\pi} = (\cos 2\pi) - (\cos \pi) = 1 - (-1) = 2$.

The total area is the sum of these two positive values, $2 + 2 = 4$ square units.