MHB Area of graph where have i gone wrong

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To find the area under the graph of the function F(x)=(3x−π)cos(1/2)x between x=−π and x=π, it is essential to identify the zeroes of the function within this interval, which occur at specific points including x=π/3 and x=±π/4. The area must be calculated in segments between these zeroes, as the function may change sign, requiring the use of absolute values for each segment's area. Integration by parts was attempted, but the resulting calculations led to an incorrect conclusion of zero area. To accurately compute the total area, evaluate the integral for each segment defined by the zeroes and sum their absolute values.
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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
 
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mathnewbie said:
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

There are zeroes to this function at $\displaystyle \begin{align*} x = \frac{\pi}{3} \end{align*}$ and $\displaystyle \begin{align*} x = \frac{\left( 2n + 1 \right) \, \pi }{4}, n \in \mathbf{Z} \end{align*}$, so in the region $\displaystyle \begin{align*} x \in \left[ -\pi , \pi \right] \end{align*}$ we have zeroes at $\displaystyle \begin{align*} \left\{ -\frac{3\pi}{4}, -\frac{\pi}{4} , \frac{\pi}{4}, \frac{\pi}{3}, \frac{3\pi}{4} \right\} \end{align*}$, so you need to work out the area of each piece between the zeroes (as they may change sign) and add all their absolute values.
 
Thanks , how do I do that , do I pick two values between each interval and solve the normal way ??

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Thanks , how do I do that , do I pick two values between each interval and alive for each value

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mathnewbie said:
Thanks , how do I do that , do I pick two values between each interval and solve the normal way ??

- - - Updated - - -[/COLOR
 

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