Calculus - strange anomaly? can anyone explain

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.
 

suffian

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 alot! Something must be left out of my textbook in that case.

Can you give me more info on this?
 
Last edited:

Hurkyl

Staff Emeritus
Science Advisor
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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.
 

HallsofIvy

Science Advisor
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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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