+118658 area=
\(area=2\int_{0}^{0.5}\;\;6cos\pi x\;\;dx\;\; +\;\;|\;2\int_{0}^{0.5}\;\;12x^2-3\;\;dx\;|\\ area=2\int_{0}^{0.5}\;\;6cos\pi x\;\;dx \;\;-\;\;\;2\int_{0}^{0.5}\;\;12x^2-3\;\;dx\\ area=2[\;\;\frac{6 sin\pi x}{\pi}\;\;]_{0}^{0.5} \;\;\qquad-\;\;\;2[\;\;4x^3-3x]_{0}^{0.5}\\ area=\;2*\;\frac{6 sin(0.5\pi) - 6 sin(0)}{\pi} \;\;-\;\;\;2[\;\;(4*0.5^3-3*0.5)-(0)]\\ area=\;2*\;\frac{6 - 0 }{\pi} \;\;-\;\;\;2[-1]\\ area=\;\;\frac{12 }{\pi} \;\;+\;\;\;2\\ area=\frac{12+2\pi }{\pi}\qquad units^2 \)
