How do you evaluate the definite integral #int (sqrt(4-x^2)) dx# from #[-2,2]#?

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Answer 1 · 2016-11-30T09:12:45

The answer is #=2pi#

We use the substitution

#x=2sintheta#

#dx=2costheta d theta#

#sqrt(4-x^2)=sqrt(4-4sin^2theta)=2costheta#

#intsqrt(4-x^2)dx=int2costheta*2costheta d theta#

#=4intcos^2theta d theta#

We know that #(cos2theta+1)/2=cos^2theta#

So,
#4intcos^2theta d theta=2int(cos2theta +1)d theta#

#=2(sin2theta)/2+ 2theta#

#=2sinthetacostheta+ 2theta#

#=2*x/2*sqrt(4-x^2)/2+2arcsin (x/2)#

#=xsqrt(4-x^2)/2+2arcsin(x/2)+C#

Therefore,

#int_(-2)^2sqrt(4-x^2)dx= [xsqrt(4-x^2)/2+2arcsin(x/2)]_(-2)^2#

#=2*pi/2+2*pi/2=2pi#