MHB Area of graph where have i gone wrong

  • Thread starter Thread starter mathnewbie1
  • Start date Start date
  • Tags Tags
    Area Graph
mathnewbie1
Messages
2
Reaction score
0
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
 
Physics news on Phys.org
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 ??

- - - Updated - - -

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

- - - Updated - - -

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

- - - Updated - - -[/COLOR
 

Similar threads

Back
Top