Calculus - strange anomaly? can anyone explain

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Discussion Overview

The discussion revolves around the evaluation of the integral of the function f(x) = (1 - x^2)^(1/2) and the challenges faced in finding its antiderivative. Participants explore the implications of the fundamental theorem of calculus and the correctness of proposed derivatives and antiderivatives.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes that the derivative of f(x) is f'(x) = (1 - x^2)^-1/2 * -2x, as stated in their textbook.
  • Another participant challenges the correctness of the proposed antiderivative F(x), suggesting that it should be F(x) = (1/2)x sqrt(1-x^2) + arcsin(x)/2.
  • There is a suggestion that the evaluation of the integral from 0 to 1 does not yield pi/4 as expected, leading to confusion about the results.
  • A participant mentions the infinite derivative of f(x) at x = ±1, indicating a potential issue with the function's behavior at those points.
  • Another participant advises looking into trigonometric substitution for evaluating the integral correctly.
  • There is a discussion about the fundamental theorem of calculus and its relevance to the claims made regarding derivatives and antiderivatives.

Areas of Agreement / Disagreement

Participants express disagreement regarding the correctness of the proposed antiderivative and the evaluation of the integral. Multiple competing views remain, and the discussion does not reach a consensus on these points.

Contextual Notes

There are limitations regarding the assumptions made about the antiderivative and the evaluation of the integral, as well as the potential impact of the infinite derivative at specific points on the overall analysis.

AndersHermansson
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f(x) = (1 - x^2)^1/2

This all stems from me approximating pi by numerically evaluating the integral S f(x)dx from 0 to 1 and multiply the sum by 4.

Now...

Would you agree that f(x) has a derivative
f'(x) = (1 - x^2)^-1/2 * -2x

?

According to my textbook this is so. Now I can easily find a primary function for f(x).

F(x) = (1 - x^2)^3/2 / -2x

Now it doesn't seem possible to evaluate [ F(x) ] from 0 to 1.
Though it should yeild pi/4, it doesn't.

Doing a riemann sum produces an approximation to pi, while evaluating [ F(x) ] only returns bogus. Since pi is an irrational number I accept that it is impossible to express it exactly. Though, I would like someone to explain why this doesn't work.
 
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If you plot the function

it is of course a circle centered on x and y equals zero, and has an infinite derivative at x=&plusminus;1.
 
Your F(x) is not the antiderivative to f(x) = sqrt(1-x2). It should be F(x) = (1/2)x sqrt(1-x2)+arcsin(x)/2. This new F(x) evaluates the function properly.
 
Ok thanks a lot! Something must be left out of my textbook in that case.

Can you give me more info on this?
 
Last edited:
Try differentiating F(x); you'll see that it does not come out to f(x).

As for evaluating the integral, look through the section on trigonometric substitution.
 
Does your textbook have the "fundamental theorem of calculus"?


If you agree that the derivative of f= (1 - x^2)^1/2 is
f'(x) = (1 - x^2)^-1/2 * -2x then obviously a "primary" function for f' is f itself, not the formula you give.
 

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