Help with Riemann sum: use Midpoint Rule for ∫ cos⁴x dx b=π/2 and a=0 and n=4

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ricecrispie

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Help with Riemann sum: use Midpoint Rule for ∫ cos⁴x dx b=π/2 and a=0 and n=4

Hi all!

I've been trying to work through :

Q: Use the midpoint rule with the given value of n to approximate the integral:

∫ cos⁴x dx b=π/2 and a=0 and n=4

Therefore Δx = 2π

2π∑f(xi) =

2π(f(π/16)+f(3π/16)+f(5π/16)+f(7π/16))
which estimates to about 9.xxxx

however the answer is 0.5890

I don't know where I am going wrong, could someone please help?



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ricecrispie said:
Hi all!

I've been trying to work through :

Q: Use the midpoint rule with the given value of n to approximate the integral:

∫ cos⁴x dx b=π/2 and a=0 and n=4

Therefore Δx = 2π NO! delta x = (b-a)/n =pi/8, so your answer is 16 times as big as it should be

2π∑f(xi) =

2π(f(π/16)+f(3π/16)+f(5π/16)+f(7π/16))
which estimates to about 9.xxxx

however the answer is 0.5890

I don't know where I am going wrong, could someone please help?



Sent from my LG-H840 using Tapatalk
Click to expand...

see comments in red
 
ricecrispie said:
Thank you, I see, but doesn't pi/2/4 = 2pi, I don't understand how it becomes pi/8?
Click to expand...

It sounds like you've already figured it out, but just for future clarification, this is where grouping symbols become very important. The expression exactly as written, pi/2/4, means:

\(\displaystyle \dfrac{\left( \dfrac{\pi}{2} \right) }{4} = \dfrac{\pi}{2} \cdot \dfrac{1}{4} = \dfrac{\pi}{8}\)

By contrast, the expression you're thinking of (that is, indeed, equal to 2pi) would be written as pi/(2/4) which means:

\(\displaystyle \dfrac{\pi}{ \left( \dfrac{2}{4} \right)} = \dfrac{\pi}{ \left( \dfrac{1}{2} \right)} = 2\pi\)

Hope that clears some things up :)
 
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