How do you find the definite integral of # [x / (sqrt(x+8))] dx# from #[0, 6]#?

1 Answer
May 24, 2016

#int_0^6 (x*d x)/(sqrt(x+8) )=4/3(16sqrt2-5sqrt14)#

Explanation:

#int_0^6 (x*d x)/(sqrt(x+8) )=?#

#x+8=u" ; "d x=d u#

#int_0^6 ((u-8)d u)/sqrt u=int _0^6 u^(-1/2)(u-8)d u#

#int_0^6 ((u-8)d u)/sqrt u=int_0^6 (u^(1/2)-8u^(-1/2))*d u#

#int_0^6 (x*d x)/(sqrt(x+8) )=[|2/3u^(3/2)-16* u^(1/2)|_0^6]#

#int_0^6 (x*d x)/(sqrt(x+8) )=[|2/3sqrt((x+8)^3)-16sqrt(x+8)|_0^6]#

#int_0^6 (x*d x)/(sqrt(x+8) )=[(2/3sqrt((6+8)^3)-16sqrt(6+8))-(2/3sqrt((0+8)^3)-16sqrt(0+8))]#

#int_0^6 (x*d x)/(sqrt(x+8) )=[(2/3sqrt(14^3)-16sqrt14)-(2/3sqrt 8^3-16sqrt8)]#

#int_0^6 (x*d x)/(sqrt(x+8) )=[(2/3*14sqrt14-16sqrt14)-(2/3*8sqrt8-16sqrt8)]#

#int_0^6 (x*d x)/(sqrt(x+8) )=[sqrt14(28/3-16)-(sqrt8(16/3-16))]#

#int_0^6 (x*d x)/(sqrt(x+8) )=[-20/3sqrt14-(-32/3sqrt8)]#

#int_0^6 (x*d x)/(sqrt(x+8) )=64/3sqrt2-20/3sqrt14#

#int_0^6 (x*d x)/(sqrt(x+8) )=4/3(16sqrt2-5sqrt14)#