How do you find the definite integral of #xsqrt(x - 40)dx# from #[40, 41]#?

1 Answer
Dec 4, 2016

#int_40^41xsqrt(x-40)color(white).dx=406/15#

Explanation:

#I=int_40^41xsqrt(x-40)color(white).dx#

We should apply the substitution #u=x-40#. This implies that #x=u+40# and that #du=dx#.

When doing the substitution, plug the current bounds of #x=40# and #x=41# into #u=x-40#. These give bounds of #x=40=>u=40-40=0# and #x=41=>u=41-40=1#. Thus:

#I=int_0^1(u+40)sqrtucolor(white).du#

Expanding this:

#I=int_0^1(u^1+40)u^(1/2)color(white).du#

#I=int_0^1(u^(3/2)+40u^(1/2))color(white).du#

Integrating these using #intu^ncolor(white).du=u^(n+1)/(n+1)+C#, and applying the bounds:

#I=[u^(5/2)/(5/2)+40(u^(3/2)/(3/2))]_0^1#

#I=[2/5u^(5/2)+80/3u^(3/2)]_0^1#

#I=[2/5(1)^(5/2)+80/3(1)^(3/2)]-[2/5(0)+80/3(0)]#

#I=2/5+80/3#

#I=406/15#