Integrating Complicated Functions: (1-a)^s

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In summary, the conversation discusses the integration of a complicated function, ∫₀¹(1-a)^{s}|f′′(x)|^{(1-a)^{s}}|f′′(t)|^{a^{s}}da, with integration from 0 to 1. The speaker mentions that there is no closed form solution for this integral, but it can be solved with numerical techniques by calculating the value of the function for multiple points and averaging them. However, this method may not be accurate and the speaker suggests using more points for better accuracy.
  • #1
adnan jahan
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Hi, i am stuck in integration of a complicated functions which is as,


∫₀¹(1-a)^{s}|f′′(x)|^{(1-a)^{s}}|f′′(t)|^{a^{s}}da

where∫₀¹ is integration from 0 to 1.

waiting for your kind reply.
 
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  • #2
Sorry, but could you please repost this so it can be read more clearly?

That way, we may be able to help you. : ) Also, you must show at least some working.
 
  • #3
i have uploaded the pdf containing the exact form of the integral thanks
 

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  • #4
You integral can be put into the form

[tex]\int^1_0 x^s A^{x^s} B^{ (1-x)^s} dx[/tex] where A and B are constants. No closed form solution for that, though for some particular combinations of s, A and B you might be able to. So unless you can tell us more about f''(a) and f''(t) and s, you are out of luck.
 
  • #5
There is almost certainly no Antiderivative for this function.

But it can be solved easily with numerical techniques.
If accuracy is not too important, just calculate the value of the function
for 5 or 10 values of x between 0 and 1, average those values and
that will be the value of the definite integral.

The accuracy will increase as you increase the number of points used.

Note; ordinarily one would have to multiply the average of f(x) by
the interval, but in this case the interval is = 1 - 0 = 1.
 
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  • #6
paulfr said:
There is almost certainly no Antiderivative for this function.
There certainly is an anti-derivative just because the function is continuous. What you mean is that there is no anti-derivative in terms of elementary functions.

But it can be solved easily with numerical techniques.
If accuracy is not too important, just calculate the value of the function
for 5 or 10 values of x between 0 and 1, average those values and
that will be the value of the integral.

The accuracy will increase as you increase the number of points used.

Note; ordinarily one would have to multiply the average of f(x) by
the interval, but in this case the interval is = 1 - 0 = 1.
 

FAQ: Integrating Complicated Functions: (1-a)^s

1. What does (1-a)^s represent in terms of integration?

(1-a)^s represents a complicated function that involves raising the value of (1-a) to the power of s. This type of function is commonly seen in integration problems and can be solved using various techniques such as substitution, integration by parts, or partial fractions.

2. How do I integrate (1-a)^s?

The process of integrating (1-a)^s can vary depending on the value of s. If s is a positive integer, the function can be expanded using the binomial theorem and then integrated term by term. If s is a negative integer, the function can be rewritten as 1/(1-a)^|s| and integrated using the power rule. For non-integer values of s, techniques like substitution or integration by parts can be used.

3. Are there any special cases when integrating (1-a)^s?

Yes, there are some special cases to consider when integrating (1-a)^s. For example, if s is equal to 1, the integral becomes ln(1-a). If s is equal to 0, the integral becomes -a. Additionally, if s is equal to -1, the integral becomes 1/(1-a), which is a commonly seen integral in calculus problems.

4. Can (1-a)^s be integrated using a substitution?

Yes, (1-a)^s can be integrated using a substitution. This method involves choosing an appropriate substitution, such as u = 1-a, and then rewriting the integral in terms of u. The integral can then be solved using the power rule or other integration techniques.

5. Why is it important to know how to integrate (1-a)^s?

Understanding how to integrate (1-a)^s is important because it is a common type of function that appears in many integration problems. Being familiar with integration techniques for this type of function can help in solving a variety of integration problems and improve overall understanding of calculus concepts.

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