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How do you find the integral #int_0^1x*sqrt(1-x^2)dx# ?

Answer 1

Charles Anctil

Begin by making a #u#-substitution.

Let #u=1-x^2#

#intxsqrt(u)dx=intu^(1/2)xdx#

#du=-2xdx#

#(du)/(-2)=(-2xdx)/(-2)#

#(-1)/2*du=xdx#

#intu^(1/2)xdx#, Note that #xdx# can be replaced with #(-1)/2*du#

#intu^(1/2)*(-1)/2*du#

#=(-1)/2intu^(1/2)du#

#=(-1)/2[u^(3/2)/(3/2)]#

#=(-1)/2[u^(3/2)*(2/3)]#

#=(-1)/2[(2u^(3/2))/3]#

#=(-1)[(u^(3/2))/3]#

#=-[(u^(3/2))/3]#, now switch from #u# back to #1-x^2#

#=-[((1-x^2)^(3/2))/3]_0^1#, remember to put the original boundaries

#=-[((1-(1)^2)^(3/2))/3-((1-(0)^2)^(3/2))/3]#

#=-[((1-1)^(3/2))/3-((1-0)^(3/2))/3]#

#=-[((0)^(3/2))/3-((1)^(3/2))/3]#

#=-[(0)/3-(1)/3]#

#=-[-1/3]#

#=1/3 -> Solution#

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Answer 2

Christopher Agresta

To find the integral ( \int_{0}^{1} x \sqrt{1 - x^2} , dx ), you can use the substitution method. Let ( u = 1 - x^2 ), then ( du = -2x , dx ). Solving for ( dx ), we get ( dx = -\frac{du}{2x} ). Substituting these into the integral and adjusting the limits accordingly, you can integrate with respect to ( u ).

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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.