How do you evaluate the integral $int xsqrt(x-2)$ ?

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Answer 1 · 2017-01-29T08:46:03

The answer is $=2/15(x-2)^(3/2)(3x+4)+ C$

We need

$intx^ndx=x^(n+1)/(n+1)+C(x!=-1)$

We solve this integral by substitution

Let $u=x-2$

$du=dx$

and $x=u+2$

Therefore,

$intxsqrt(x-2)dx=int(u+2)sqrtu du$

$=int(u^(3/2)+2u^(1/2))du$

$=u^(5/2)/(5/2)+2*u^(3/2)/(3/2)$

$=2/5u^(5/2)+4/3u^(3/2)$

$=2/15u^(3/2)(3u+10)$

$=2/15(x-2)^(3/2)(3x-6+10)+C$

$=2/15(x-2)^(3/2)(3x+4)+ C$

How do you evaluate the integral int xsqrt(x-2)int xsqrt(x-2)?