Area of graph where have i gone wrong

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The discussion focuses on calculating the area under the graph of the function F(x) = (3x - π)cos(1/2 x) between x = -π and x = π using integration by parts. The user initially computes the integral but arrives at a result of 0π, which contradicts the requirement for a four-decimal answer. Key insights reveal that the function has zeroes at specific points within the interval, necessitating the calculation of the area between these zeroes to account for sign changes. The correct approach involves integrating over each segment defined by the zeroes and summing 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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