Finding the integral of x^a using a geometric progression

  • Thread starter Thread starter courtrigrad
  • Start date Start date
  • Tags Tags
    Geometric Integral
Click For Summary
SUMMARY

The discussion focuses on finding the integral of the function y = x^a using a geometric progression. The integral is expressed as \(\int^b_a x^a \, dx = \lim_{n \to \infty} \sum_{i=0}^{n} \frac{(b-a)^{i+1}}{i+1}\), derived from the formula for the sum of a geometric series. The method is particularly useful for integrals that are challenging to solve with conventional techniques. Participants emphasized the importance of correctly applying the geometric progression and noted limitations when a = -1.

PREREQUISITES
  • Understanding of integral calculus and Riemann sums
  • Familiarity with geometric progressions and their summation formulas
  • Knowledge of limits in calculus
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the derivation of the geometric series sum formula
  • Explore advanced integration techniques, including improper integrals
  • Learn about the application of Riemann sums in integral calculus
  • Investigate the limitations of integration methods for specific values of a
USEFUL FOR

Mathematicians, calculus students, and educators seeking innovative methods for solving integrals, particularly those involving non-standard functions.

courtrigrad
Messages
1,236
Reaction score
2
Let's say [tex]y = x^a[/tex] and you want to find [tex]\int^b_a x^a[/tex]

How would you find this using a geometric progression?

Thanks
 
Physics news on Phys.org
1.U forgot "dx" in the integral...
2.I'm sure u cannot use geometric progression,if [itex]a=-1[/itex]
3.Do u know how to work with Riemann sums?

Daniel.
 
for sharing this method for finding the integral of x^a using a geometric progression. To find the integral using this method, we can first rewrite the function as y = x^a, which means that y is the sum of a geometric progression with first term 1 and common ratio x.

Next, we can use the formula for the sum of a geometric progression to find the integral:

\int^b_a x^a dx = \lim_{n \to \infty} \sum_{i=0}^{n} \frac{(b-a)^{i+1}}{i+1}

This formula can be derived by using the fact that the sum of a geometric progression with first term a and common ratio r is given by \frac{a(1-r^n)}{1-r}.

Substituting in our values of a=1 and r=x, we get the formula above.

By taking the limit as n approaches infinity, we can find the exact value of the integral. This method can be useful for finding integrals of functions that cannot be easily integrated using traditional methods.

Thanks again for sharing this approach, it can be a valuable tool for solving certain integration problems.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 151 ·
6
Replies
151
Views
6K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
14
Views
3K