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:

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

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

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

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