Why antiderivative evaluation fails for sqrt(1-x²) from 0 to 1

[AndersHermansson]

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.

 

[Tyger]

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.

 

[AndersHermansson]

Ok thanks a lot! Something must be left out of my textbook in that case.

Can you give me more info on this?

 

[Hurkyl]

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]

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.