How do you evaluate the definite integral #int 2xsqrt(1-4x^2)# from #[0, 1/2]#?

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How do you evaluate the definite integral #int 2xsqrt(1-4x^2)dx# from #[0, 1/2]#?

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Answer 1 · 2016-09-04T17:09:56

#1/6#.

Let #I=int_0^(1/2) (2xsqrt(1-4x^2))dx#.

We use the Substn. : #1-4x^2=t^2", so that, "-8xdx=2tdt#,

or, #2xdx=-t/2dt#.

Also, #x=0 rArr t=1 ; x=1/2 rArr t=0#.

Hence, #I=int_1^0 t(-t/2)dt=-1/2int_1^0 t^2dt=1/2int_0^1 t^2dt#

#=1/2[t^3/3]_0^1#

#=1/6[1-0]#

#I=1/6#.