Math 152: int [x=0, 1/2][1/(x^2 - x + 1)]dx = pi/(3sqrt[3])

[huntsiekfind]

Wondering if you can help me with this one. Part (a) asks by completing the square (and then doing a trigonometric substitution),

1/2
?dx/x^2-x+1=?/3?3
0

(b) by factoring x^3 +1 as a sum of cubes, rewrite the integral in part (a). Then express 1/x^3+1 as the sum of a power series and use it to prove the following formula for ?

?
?=3?3/4?(-1)^n/8^n[(2/3n+1) + (1/3n+2)]
N=0

 

[galactus]

Re: Math 152 problem

Wondering if you can help me with this one. Part (a) asks by completing the square (and then doing a trigonometric substitution),

\(\displaystyle \int_{0}^{\frac{1}{2}}\frac{1}{x^{2}-x+1}dx=\frac{\pi}{3\sqrt{3}}\)

complete the square. \(\displaystyle \int\frac{1}{(x-\frac{1}{2})^{2}+\frac{3}{4}}dx\).

Let \(\displaystyle x=\frac{\sqrt{3}}{2}tan(t)+\frac{1}{2}, \;\ dx=\frac{\sqrt{3}}{2}sec^{2}(t)dt\)

It'll whittle down nicely to an easy integral.

(b) by factoring x^3 +1 as a sum of cubes, rewrite the integral in part (a). Then express 1/x^3+1 as the sum of a power series and use it to prove the following formula for \(\displaystyle {\pi}\)

?
?=3?3/4?(-1)”/8”[2/3n=1 + 1/3n=2]
N=0

I am sorry, I can not make out what that last part is. But, the series for \(\displaystyle \frac{1}{1+x^{3}}=\sum_{n=0}^{\infty}(-1)^{n}x^{3n}\)

 

[Deleted member 4993]

Re: Math 152 problem

DUPLICATE POST

http://mathgoodies.com/forums/topic.asp?TOPIC_ID=33884

Wondering if you can help me with this one. Part (a) asks by completing the square (and then doing a trigonometric substitution),

1/2
?dx/x^2-x+1=?/3?3 <<< You are being careless about grouping your operation . What you wrote literally means:
0

\(\displaystyle \int^{\frac{1}{2}}_0\frac{dx}{x^2} \, - \, x \, + \, 1\)

whereas I suspect you meant:

\(\displaystyle \int^{\frac{1}{2}}_0\frac{dx}{x^2\, - \, x \, + \, 1}\)

which should have been written as:

1/2
?dx/(x^2-x+1)
0


You have been taught "completing the square" in high school - prior to taking calculus class that should have been tattooed in your brain. Just to revive those old memory cells:

\(\displaystyle Ax^2 + Bx + C\)

\(\displaystyle = A[(x + \frac{B}{2A})^2 - (\frac{\sqrt{B^2 - 4AC}}{2A})^2]\)

That's it ... very important manipulation - remember it...

(b) by factoring x^3 +1 as a sum of cubes, rewrite the integral in part (a). Then express 1/x^3+1 as the sum of a power series and use it to prove the following formula for ?

That part does not make sense to me... explain further
?
?=3?3/4?(-1)”/8”[2/3n=1 + 1/3n=2]
N=0

Please show your work, indicating exactly where you are stuck - so that we know where to begin to help you.

 

[huntsiekfind]

Re: Math 152 problem

Wondering if you can help me with this one. Part (a) asks by completing the square (and then doing a trigonometric substitution),

\(\displaystyle \int_{0}^{\frac{1}{2}}\frac{1}{x^{2}-x+1}dx=\frac{\pi}{3\sqrt{3}}\)

complete the square. \(\displaystyle \int\frac{1}{(x-\frac{1}{2})^{2}+\frac{3}{4}}dx\).

Let \(\displaystyle x=\frac{\sqrt{3}}{2}tan(t)+\frac{1}{2}, \;\ dx=\frac{\sqrt{3}}{2}sec^{2}(t)dt\)

It'll whittle down nicely to an easy integral.

[quote:2vnc48we](b) by factoring x^3 +1 as a sum of cubes, rewrite the integral in part (a). Then express 1/x^3+1 as the sum of a power series and use it to prove the following formula for \(\displaystyle {\pi}\)

?
?=3?3/4?(-1)”/8”[2/3n=1 + 1/3n=2]
N=0

I am sorry, I can not make out what that last part is. But, the series for \(\displaystyle \frac{1}{1+x^{3}}=\sum_{n=0}^{\infty}(-1)^{n}x^{3n}\)[/quote:2vnc48we]

By factoring x^3 +1 as a sum of cubes, rewrite

1/2
?dx/x^2-x+1=?/3?3
0

Then express 1/x^3+1 as the sum of a power series and use it to prove the following formula for ?

?
?=3?3/4?(-1)^n/8^n[(2/3n+1) + (1/3n+2)]
N=0

 

[Deleted member 4993]

Where is YOUR work?

Where are you stuck?

What did you do with the hints that galactus provided you?

 

[huntsiekfind]

Where is YOUR work?

Where are you stuck?

What did you do with the hints that galactus provided you?

I have the first part done, not sure if I'm on the right track at all on the second part. Scanned my work, but couldn't get it to copy to here, wouldn't upload it as an attachment. Not enough time to type out all of my work, but thank you for your help!

 

[Deleted member 4993]

Hint for second part:

\(\displaystyle \frac{1}{x^3+1} \, = \, \frac{1}{3}\cdot\frac{1}{x+1} \, - \, \frac{1}{6}\cdot\frac{2x-1}{x^2-x+1} \, - \, \frac{1}{2}\cdot\frac{1}{x^2-x+1}\)