Area of graph where have i gone wrong

[mathnewbie1]

I have been asked to finding the area of the graph of the function F(x)=$$(3x-\pi)\cos\frac{1}{2}x$$ between$$x=-\pi and x=\pi$$

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

 

[Prove It]

I have been asked to finding the area of the graph of the function F(x)=$$(3x-\pi)\cos\frac{1}{2}x$$ between$$x=-\pi and x=\pi$$

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.

 

[mathnewbie1]

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

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