What is the net area between #f(x)=sinx# in #x in[0,2pi] # and the x-axis?

Question

What is the net area  between #f(x)=sinx# in #x  in[0,2pi]    # and the x-axis?

Emma Aldridge · Accepted Answer

The area is given by the integral #A=int_0^(2pi) sinxdx=[-cosx]_0^(2*pi)=0#

Emily Ammerman · Answer

undefined To find the net area between $ f(x) = \sin(x) $ and the x-axis on the interval $[0, 2\pi]$, you need to compute the definite integral of $ |f(x)| $ over that interval. Since $ \sin(x) $ is positive on the interval $ [0, \pi] $ and negative on the interval $ [\pi, 2\pi] $, you need to split the integral into two parts and take the absolute value of $ \sin(x) $ on the interval where it's negative.

The integral can be split into two parts as follows:

$$
\text{Net area} = \int_{0}^{\pi} \sin(x) \, dx + \int_{\pi}^{2\pi} |\sin(x)| \, dx
$$

After integrating each part separately:

$$
\text{Net area} = \left[-\cos(x)\right]_{0}^{\pi} + \left[-\cos(x)\right]_{\pi}^{2\pi}
$$

$$
= \left[-\cos(\pi) - (-\cos(0))\right] + \left[-\cos(2\pi) - (-\cos(\pi))\right]
$$

$$
= [-(-1) - (-1)] + [-1 - (-1)]
$$

$$
= (1 - (-1)) + (-1 - (-1))
$$

$$
= 2 + 0
$$

$$
= 2
$$

So, the net area between $ f(x) = \sin(x) $ and the x-axis on the interval $ [0, 2\pi] $ is $ 2 $.