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

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Answer 1 · 2014-09-22T03:03:50

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

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

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

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

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

$=-[-1/3]$

$=1/3 -> Solution$

How do you find the integral int_0^1x*sqrt(1-x^2)dxint_0^1x*sqrt(1-x^2)dx ?