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How do you evaluate #int sqrt (1-x^2) dx # for [-1, 1]?

Answer 1

#pi/2#

let #x=sintheta# #sqrt(1-x^2)=sqrt(1-sin^2theta)=abs(costheta)# in #-pi/2 to pi/2# #cos theta# is positive so #abscostheta =costheta# changing limits #if x=-1 then theta =-pi/2# #if x=1 then theta = pi/2# #dx=costheta d(theta)# then #int_-1^1sqrt(1-x^2)dx# changes to #int_(-pi/2)^(pi/2)cos^2thetad(theta)# =#int_(-pi/2)^(pi/2)(1+cos2theta)/2d(theta)# =#int_(-pi/2)^(pi/2)d(theta)/2+int_(-pi/2)^(pi/2)(cos2theta)/2d(theta)# =#1/2(pi/2-(-pi/2)+1/2(1/2(sin(pi)-sin(-pi)))# =#pi/2#

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Answer 2

Ezra Adams

To evaluate ( \int \sqrt{1-x^2} , dx ) for ( [-1, 1] ), use the trigonometric substitution ( x = \sin(\theta) ). Then, express ( dx ) in terms of ( d\theta ) and rewrite the integral in terms of ( \theta ). After integrating with respect to ( \theta ), convert back to the variable ( x ) to obtain the result.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.