How do you integrate #int xsqrt(1-x^2)dx# from [0,1]?

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Answer 1 · 2016-12-04T14:59:29

#1/3#

If one recognises that the outside of the sqrt is a function of the sqrt differentiated we can proceed by inspection.

#int_0^1x(1-x^2)^(1/2)dx#

guess#" " y=(1-x^2)^(3/2)#

#d/(dx)((1-x^2)^(3/2))=3/2xx(-2x)(1-2x)^(1/2)#

#=-3(1-2x)^(1/2)#

#:.int_0^1x(1-x^2)^(1/2)dx=-1/3[(1-2x)^(3/2)]_0^1#

#-1/3{[(1-2x)^(3/2)]^1-[(1-2x)^(3/2)]_0}#

#-1/3((0)-1)#

#=1/3#

Answer 2 · 2016-12-04T15:08:47

#int_0^1 xsqrt(1-x^2)dx = 1/3#

Substitute:

#t = 1-x^2#
#dt = -2xdx#

#int_0^1 xsqrt(1-x^2)dx = -1/2int_1^0 sqrt(t) dt = 1/2int_0^1 sqrt(t) dt =#

# = 1/3 t^(3/2) |_(x=0)^(x=1) =1/3#