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 Does anyone know how to do this (the one on the left)? 

 Mar 3, 2019

Best Answer 

 #1
avatar+22184 
+2

Math help

 

 

My attempt:

 

\(\begin{array}{|rcll|} \hline 2A &=& \large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \Big(1-2\sin(\phi)\Big)^2\ d\phi \\ &=& \large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \left( 1-4\sin(\phi)+4\sin^2(\phi) \right) d\phi \\ &=& \large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \left( 1\right)\ d\phi -\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } 4\sin(\phi)\ d\phi +\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } 4\sin^2(\phi) d\phi \\ &=& \large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } d\phi -4\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \sin(\phi)\ d\phi +4\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \sin^2(\phi) d\phi \\ &=& \left[\phi \right]_{\frac{\pi}{6}}^{\frac{5\pi}{6}} -4\left[-\cos(\phi)\right]_{\frac{\pi}{6}}^{\frac{5\pi}{6}} +4\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \sin^2(\phi) d\phi \\ &=& \left[\phi \right]_{\frac{\pi}{6}}^{\frac{5\pi}{6}} +4\left[\cos(\phi)\right]_{\frac{\pi}{6}}^{\frac{5\pi}{6}} +4\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \sin^2(\phi) d\phi \\ &=& \dfrac{5\pi}{6}-\dfrac{\pi}{6} +4\left[ \underbrace{\cos\left(\frac{5\pi}{6}\right)-\cos\left(\frac{\pi}{6}\right) }_{=-\sqrt{3}} \right] +4\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \sin^2(\phi) d\phi \\ &=& \dfrac{2\pi}{3} -4\sqrt{3} +4\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \sin^2(\phi) d\phi \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline \large{ \int \sin^2(\phi) } d\phi \\ &=& -\cos(\phi)\sin(\phi)+ \large{ \int } \cos^2(\phi)\ d\phi \\ &=& -\cos(\phi)\sin(\phi)+ \large{ \int } 1-\sin^2(\phi)\ d\phi \\ &=& -\cos(\phi)\sin(\phi)+ \large{ \int } 1\ d\phi-\large{ \int } \sin^2(\phi)\ d\phi \\ 2\large{ \int \sin^2(\phi) } d\phi &=& -\cos(\phi)\sin(\phi)+ \phi \\ \mathbf{\large{ \int \sin^2(\phi) } d\phi} & \mathbf{=}& \mathbf{\dfrac12\Big(\phi-\cos(\phi)\sin(\phi) \Big) } \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline 2A &=& \dfrac{2\pi}{3} -4\sqrt{3} +4\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \sin^2(\phi) d\phi \\ && \quad \mathbf{\large{ \int \sin^2(\phi) } d\phi=\dfrac12\Big(\phi-\cos(\phi)\sin(\phi) \Big) } \\ &=& \dfrac{2\pi}{3} -4\sqrt{3} +2\Big[ \phi-\cos(\phi)\sin(\phi) \Big]_{\dfrac{\pi}{6}}^{\dfrac{5\pi}{6}} \\ &=& \dfrac{2\pi}{3} -4\sqrt{3} +2\Big[ \phi-\dfrac12\sin(2\phi) \Big]_{\dfrac{\pi}{6}}^{\dfrac{5\pi}{6}} \\ &=& \dfrac{2\pi}{3} -4\sqrt{3} +2\Big[ \dfrac{5\pi}{6}-\dfrac12\sin\left(\dfrac{10\pi}{6}\right)-\left(\dfrac{\pi}{6}-\dfrac12\sin\left(\dfrac{2\pi}{6}\right)\right) \Big] \\ &=& \dfrac{2\pi}{3} -4\sqrt{3} +2\Big[ \dfrac{2\pi}{3}-\dfrac12\sin\left(\dfrac{5\pi}{3}\right)+\dfrac12\sin\left(\dfrac{ \pi}{3}\right) \Big] \\ &=& \dfrac{2\pi}{3} -4\sqrt{3}+2\left( \dfrac{2\pi}{3} + \dfrac{\sqrt{3}}{2} \right) \\ &=& \dfrac{6\pi}{3} -3\sqrt{3} \\ &=& 2\pi -3\sqrt{3} \\ &=& 1.08703288447 \\ A &=& \dfrac{1.08703288447 }{2} \\ \mathbf{A} & \mathbf{=} & \mathbf{0.54351644224} \\ \hline \end{array}\)

 

laugh

 Mar 4, 2019
 #1
avatar+22184 
+2
Best Answer

Math help

 

 

My attempt:

 

\(\begin{array}{|rcll|} \hline 2A &=& \large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \Big(1-2\sin(\phi)\Big)^2\ d\phi \\ &=& \large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \left( 1-4\sin(\phi)+4\sin^2(\phi) \right) d\phi \\ &=& \large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \left( 1\right)\ d\phi -\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } 4\sin(\phi)\ d\phi +\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } 4\sin^2(\phi) d\phi \\ &=& \large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } d\phi -4\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \sin(\phi)\ d\phi +4\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \sin^2(\phi) d\phi \\ &=& \left[\phi \right]_{\frac{\pi}{6}}^{\frac{5\pi}{6}} -4\left[-\cos(\phi)\right]_{\frac{\pi}{6}}^{\frac{5\pi}{6}} +4\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \sin^2(\phi) d\phi \\ &=& \left[\phi \right]_{\frac{\pi}{6}}^{\frac{5\pi}{6}} +4\left[\cos(\phi)\right]_{\frac{\pi}{6}}^{\frac{5\pi}{6}} +4\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \sin^2(\phi) d\phi \\ &=& \dfrac{5\pi}{6}-\dfrac{\pi}{6} +4\left[ \underbrace{\cos\left(\frac{5\pi}{6}\right)-\cos\left(\frac{\pi}{6}\right) }_{=-\sqrt{3}} \right] +4\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \sin^2(\phi) d\phi \\ &=& \dfrac{2\pi}{3} -4\sqrt{3} +4\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \sin^2(\phi) d\phi \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline \large{ \int \sin^2(\phi) } d\phi \\ &=& -\cos(\phi)\sin(\phi)+ \large{ \int } \cos^2(\phi)\ d\phi \\ &=& -\cos(\phi)\sin(\phi)+ \large{ \int } 1-\sin^2(\phi)\ d\phi \\ &=& -\cos(\phi)\sin(\phi)+ \large{ \int } 1\ d\phi-\large{ \int } \sin^2(\phi)\ d\phi \\ 2\large{ \int \sin^2(\phi) } d\phi &=& -\cos(\phi)\sin(\phi)+ \phi \\ \mathbf{\large{ \int \sin^2(\phi) } d\phi} & \mathbf{=}& \mathbf{\dfrac12\Big(\phi-\cos(\phi)\sin(\phi) \Big) } \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline 2A &=& \dfrac{2\pi}{3} -4\sqrt{3} +4\large{ \int \limits_{\frac{\pi}{6}}^{\frac{5\pi}{6}} } \sin^2(\phi) d\phi \\ && \quad \mathbf{\large{ \int \sin^2(\phi) } d\phi=\dfrac12\Big(\phi-\cos(\phi)\sin(\phi) \Big) } \\ &=& \dfrac{2\pi}{3} -4\sqrt{3} +2\Big[ \phi-\cos(\phi)\sin(\phi) \Big]_{\dfrac{\pi}{6}}^{\dfrac{5\pi}{6}} \\ &=& \dfrac{2\pi}{3} -4\sqrt{3} +2\Big[ \phi-\dfrac12\sin(2\phi) \Big]_{\dfrac{\pi}{6}}^{\dfrac{5\pi}{6}} \\ &=& \dfrac{2\pi}{3} -4\sqrt{3} +2\Big[ \dfrac{5\pi}{6}-\dfrac12\sin\left(\dfrac{10\pi}{6}\right)-\left(\dfrac{\pi}{6}-\dfrac12\sin\left(\dfrac{2\pi}{6}\right)\right) \Big] \\ &=& \dfrac{2\pi}{3} -4\sqrt{3} +2\Big[ \dfrac{2\pi}{3}-\dfrac12\sin\left(\dfrac{5\pi}{3}\right)+\dfrac12\sin\left(\dfrac{ \pi}{3}\right) \Big] \\ &=& \dfrac{2\pi}{3} -4\sqrt{3}+2\left( \dfrac{2\pi}{3} + \dfrac{\sqrt{3}}{2} \right) \\ &=& \dfrac{6\pi}{3} -3\sqrt{3} \\ &=& 2\pi -3\sqrt{3} \\ &=& 1.08703288447 \\ A &=& \dfrac{1.08703288447 }{2} \\ \mathbf{A} & \mathbf{=} & \mathbf{0.54351644224} \\ \hline \end{array}\)

 

laugh

heureka Mar 4, 2019

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