Improper integral + Maclaurin series problem

  • Thread starter mikeXXL
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  • #1
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Can you please offer any hints or suggestions on how to do these two problems:

1) Find the Maclaurin series of (x^2 + 1)/(3x^2 + 2x - 1).

Should I perform long-division first? I can't seem to find any repeating pattern...

2) Evaluate the integral sqrt(12-4x-x^2) from x=2 to x=6.

I realize this is an improper integral since f(x)<0 for x>2 but I am bamboozled as to how to set up and evaluate it as a limit.

Any help is appreciated in advance...
 

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  • #2
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Heh, okay I figured out the first one. I decomposed it using partial fractions to obtain a simpler geometric sum and from then on it was easy.

Now as for the second, when I complete the square I notice that it is a circle centered @ (-2,0) with radius 4 but it asks me to find the area OUTSIDE the circle, namely from (x=2 to x-6)? Does this mean that the integral is divergent?
 
  • #3
lurflurf
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mikeXXL said:
Can you please offer any hints or suggestions on how to do these two problems:

1) Find the Maclaurin series of (x^2 + 1)/(3x^2 + 2x - 1).

Should I perform long-division first? I can't seem to find any repeating pattern...

2) Evaluate the integral sqrt(12-4x-x^2) from x=2 to x=6.

I realize this is an improper integral since f(x)<0 for x>2 but I am bamboozled as to how to set up and evaluate it as a limit.

Any help is appreciated in advance...
for 1)
-factor the denominator
-expand in partial fractions
-expand new denominators in geometric series
-combine
for 2) that is not an improper integral, but an imaginary one
sqrt(12-4x-x^2)=i*sqrt(x^2+4x-12)
then if you like you can find the indefinite integral of that in terms of roots and logs.
 

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