6.1 Areas Between Curves/22

${\begin{aligned}&\color {red}\mathbf {y=\sin({\frac {\pi x}{2}})} &\color {royalblue}\mathbf {y=x} \\\end{aligned}}$

${\begin{aligned}\sin({\frac {x\pi }{2}})&=x\\x&=0\\x&=1\\\end{aligned}}$

$\int _{0}^{1}\left(\sin \left({\frac {x\pi }{2}}\right)-x\right)dx=\int _{0}^{1}\left(\sin \left({\frac {\pi }{2}}\right)\right)-\int _{0}^{1}(x)dx={\frac {2}{\pi }}-{\frac {1}{2}}$

${\begin{aligned}\int _{0}^{1}\left(\sin \left({\frac {x\pi }{2}}\right)\right)dx\\u={\frac {x\pi }{2}}\\du={\frac {\pi }{2}}dx\\{\frac {2}{\pi }}du=dx\\\end{aligned}}$

New upper limit: ${\frac {(0)\pi }{2}}=0$

New lower limit: ${\frac {(1)\pi }{2}}={\frac {\pi }{2}}$

${\begin{aligned}\int _{0}^{1}\left(\sin \left({\frac {x\pi }{2}}\right)\right)dx&={\frac {2}{\pi }}\int _{0}^{\frac {\pi }{2}}\sin(u)du\\&={\frac {2}{\pi }}\left[-\cos(u)\right]{\Bigg |}_{0}^{\frac {\pi }{2}}\\&={\frac {2}{\pi }}\left[-\cos({\frac {\pi }{2}})+\cos(0)\right]\\&={\frac {2}{\pi }}[0+1]={\frac {2}{\pi }}\\\end{aligned}}$

${\begin{aligned}\int _{0}^{1}xdx&=\left[{\frac {x^{2}}{2}}\right]{\Bigg |}_{0}^{1}\\&={\frac {1}{2}}-0={\frac {1}{2}}\\\end{aligned}}$