How do you find the definite integral of #xsqrt(x - 40)dx# from #[40, 41]#?
1 Answer
Dec 4, 2016
Explanation:
#I=int_40^41xsqrt(x-40)color(white).dx#
We should apply the substitution
When doing the substitution, plug the current bounds of
#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
#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#