Riemann sum

janeann

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Jun 2, 2011
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Use a riemann sum with 4 rectangles and the midpoint rule to estimate the area under sin(x) on (0,pi). Hint: the double or half angles can be expressed as sin(2theta)=2sin(theta)cos(theta) and sin(theta)=sqrt(1-cos(2theta))/2.

So far i have the change of x being (-pi/4) and xj being -jpi/4. so then i have the integral from 0 to pi sin(x)= the summation n=4 i=1 (-pi/4)(pij/4), then i get pi^2/16 being the area. Im not sure when midpoint would come in.
 
Each subinterval has width π04=π4\displaystyle \frac{{\pi}-0}{4}=\frac{\pi}{4}

Since the midpoint rule is being used, the first point is at π8\displaystyle \frac{\pi}{8}. Half way between 0 and Pi/4.

So, the first one is sin(π8)\displaystyle sin(\frac{\pi}{8}), then pi/4 added thereafter.

second: sin(π8+π4)\displaystyle sin(\frac{\pi}{8}+\frac{\pi}{4})

third: sin(π8+π2)\displaystyle sin(\frac{\pi}{8}+\frac{\pi}{2})

fourth: sin(π8+3π4)\displaystyle sin(\frac{\pi}{8}+\frac{3\pi}{4})

So, we have π4(sin(π8)+sin(π8+π4)+sin(π8+π2)+sin(π8+3π4))\displaystyle \frac{\pi}{4}\left(sin(\frac{\pi}{8})+sin(\frac{\pi}{8}+\frac{\pi}{4})+sin(\frac{\pi}{8}+\frac{\pi}{2})+sin(\frac{\pi}{8}+\frac{3\pi}{4})\right)
 
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