mathnewbie1
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I have been asked to finding the area of the graph of the function F(x)=(3x-\pi)\cos\frac{1}{2}x betweenx=-\pi and x=\pi
using integration by parts to integrate the function I get \int 2(3x-\pi)(\sin\frac{1}{2}\pi)+12\cos\frac{x}{2}
when I work out the integral for x=\pi and x =-\pi I get the following
2(3\pi-\pi)(\sin\frac{1}{2}\pi)+12\cos\frac{\pi}{2}-2(3(-\pi)-\pi)(\sin\frac{1}{2}-\pi)+12\cos\frac{-\pi}{2}
when I work this out I get 4\pi-4\pi =0\pi
but the question states give answer to four decimals has anyone any idea of where I have gone wrong
using integration by parts to integrate the function I get \int 2(3x-\pi)(\sin\frac{1}{2}\pi)+12\cos\frac{x}{2}
when I work out the integral for x=\pi and x =-\pi I get the following
2(3\pi-\pi)(\sin\frac{1}{2}\pi)+12\cos\frac{\pi}{2}-2(3(-\pi)-\pi)(\sin\frac{1}{2}-\pi)+12\cos\frac{-\pi}{2}
when I work this out I get 4\pi-4\pi =0\pi
but the question states give answer to four decimals has anyone any idea of where I have gone wrong