Evaluate the following integral:

integral (x^3)/square root (x^2 + 16) from 0 to 3

Question

Evaluate the following integral:

integral (x^3)/square root (x^2 + 16) from 0 to 3

hartnn · Answer

did you try u substitution ?

u = x^2+16   [so, x^2 = u-16]
du =... ?

hartnn · Answer

yes/no/trying ?

anonymous · Answer

trying for sure! :) 

Mmm... not quite sure about the du part.. can u explain that to me please

hartnn · Answer

u= x^2+16 
differentiate this w.r.t x
du/ dx =...?

hartnn · Answer

the catch here is to split the numerator's x^3 dx
$x^3dx = x^2 (x dx)$

anonymous · Answer

hmm.. im still a bit confused...

so when i find du i get 2x dx

then im stuck idk where to go from there :(

hartnn · Answer

yes, so 
du = 2x dx, right ?
so, 
x dx = du/2
got this ?

anonymous · Answer

yup

hartnn · Answer

Numerator = x^3 dx  =  x^2 (x dx) =  (u-16) (du/2)
and this ?

hartnn · Answer

any doubts in numerator ?
denominator is just sqrt u, isn't it

anonymous · Answer

ahhh okie dokie got it

hartnn · Answer

good! 
go you can solve further ?

hartnn · Answer

*so

anonymous · Answer

yup i think i'll be ok :) thank you!

If I still get stuck on smthg along the way can i still bug you for help? :P
sorry kinda new to this..

hartnn · Answer

ofcourse! no problem at all :) 
since you're new here, WELCOME to OpenStudy!! :)