Home > Calculus > Integration by Parts

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

Answer 1

Isabella Ackerman

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#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Isaiah Allen

To evaluate the integral ( \int x\sqrt{x - 2} ), you can use the substitution method. Let ( u = x - 2 ), then ( du = dx ). Substituting these into the integral:

[ \int x\sqrt{x - 2} , dx = \int (u + 2) \sqrt{u} , du ]

Now distribute and integrate term by term:

[ \int (u + 2) \sqrt{u} , du = \int u\sqrt{u} , du + 2\int \sqrt{u} , du ]

For the first integral, use integration by parts where ( u = u ) and ( dv = \sqrt{u} , du ), and for the second integral, use a simple substitution. After integrating each term, substitute back ( x - 2 ) for ( u ).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.