What is the net area between #f(x) = sinx - cos^2x# and the x-axis over #x in [0, 3pi ]#?

Question

Isabella Ackerman · Accepted Answer

The net area is #2-(3pi)/2# A small area is #dA=ydx# #=># #dA=(sinx-cos^2x)dx# so the area is #A=int_0^(3pi)(sinx-cos^2x)dx# To integrate #cos^2x#, we use #cos2x=2cos^2x-1# #=>##cos^2x=(1+cos2x)/2# So, #A=int_0^(3pi)(sinx-(1+cos2x)/2)dx# #=[-cosx-x/2-(sin2x)/4 ] _0^(3pi)# #=1-(3pi)/2+1=2-(3pi)/2#

Christopher Agresta · Answer

undefined To find the net area between $ f(x) = \sin(x) - \cos^2(x) $ and the x-axis over $ x $ in the interval $ [0, 3\pi] $, we need to evaluate the definite integral of $ f(x) $ from $ x = 0 $ to $ x = 3\pi $, and take the absolute value of the result since the function dips below the x-axis.

$$ A = \int_{0}^{3\pi} |\sin(x) - \cos^2(x)| \, dx $$

First, we'll find the critical points of $ f(x) $ by setting $ f(x) = 0 $.

$$ \sin(x) - \cos^2(x) = 0 $$
$$ \sin(x) - (1 - \sin^2(x)) = 0 $$
$$ \sin(x) - 1 + \sin^2(x) = 0 $$
$$ \sin^2(x) + \sin(x) - 1 = 0 $$

Using a quadratic formula, let $ u = \sin(x) $:
$$ u^2 + u - 1 = 0 $$

The solutions to this quadratic equation are $ u = \frac{-1 \pm \sqrt{5}}{2} $. Since $ \sin(x) $ is between -1 and 1 for all real $ x $, the only solution in the range is $ u = \frac{-1 + \sqrt{5}}{2} $.

Therefore, $ \sin(x) = \frac{-1 + \sqrt{5}}{2} $ is the only critical point.

Now, we'll evaluate the integral:

$$ A = \int_{0}^{3\pi} |\sin(x) - \cos^2(x)| \, dx = \int_{0}^{3\pi} |\sin(x) - (1 - \sin^2(x))| \, dx $$

$$ = \int_{0}^{3\pi} |\sin^2(x) + \sin(x) - 1| \, dx $$

Since $ \sin^2(x) + \sin(x) - 1 = 0 $ at $ x = \frac{-1 + \sqrt{5}}{2} $, we can split the integral into two parts:

$$ A = \int_{0}^{\frac{-1 + \sqrt{5}}{2}} (\sin^2(x) + \sin(x) - 1) \, dx + \int_{\frac{-1 + \sqrt{5}}{2}}^{3\pi} (\sin(x) - \sin^2(x) - 1) \, dx $$

Now, integrate each part separately and take their absolute values. Finally, sum the absolute values of the two integrals to find the net area.