Improper integral + Maclaurin series problem

In summary, for the first problem, you can find the Maclaurin series by factoring the denominator, expanding in partial fractions, and combining the terms. As for the second problem, it is not an improper integral but an imaginary one, and you can find the indefinite integral in terms of roots and logs.
  • #1
mikeXXL
2
0
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...
 
Physics news on Phys.org
  • #2
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
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.
 

1. What is an improper integral?

An improper integral is an integral where one or both of the limits of integration are infinite or the integrand has an infinite discontinuity within the interval of integration. In other words, it is an integral that does not have a finite value.

2. How do you solve an improper integral?

To solve an improper integral, you must first determine the type of improper integral (infinite limit or infinite discontinuity) and then use appropriate techniques such as limits, substitution, or integration by parts to evaluate the integral. If the integral diverges, it means that it does not have a finite value.

3. What is a Maclaurin series?

A Maclaurin series is an infinite series expansion of a function in powers of x, centered at x=0. It is a special case of a Taylor series, where the center of the series is at x=0. It is used to approximate a function by adding together simpler functions, making it easier to evaluate and manipulate.

4. How do you find the Maclaurin series of a function?

To find the Maclaurin series of a function, you can use the formula for the n-th derivative of a function evaluated at x=0, which is: f(n)(0)/n!. This will give you the coefficients of the series. You can then substitute these coefficients into the general formula for a Maclaurin series to get the final series.

5. How do you solve a problem involving both improper integrals and Maclaurin series?

To solve a problem involving both improper integrals and Maclaurin series, you must first use the appropriate techniques to evaluate the improper integral. Then, substitute the resulting value into the Maclaurin series for the function. If the series converges, you can use it as an approximation for the original function. If it diverges, the series cannot be used as an approximation.

Similar threads

Replies
1
Views
1K
Replies
4
Views
2K
Replies
1
Views
1K
  • Calculus
Replies
1
Views
1K
  • Calculus
Replies
1
Views
3K
Replies
3
Views
1K
  • Calculus
Replies
1
Views
1K
  • Calculus
Replies
2
Views
1K
Replies
1
Views
1K
Back
Top