October 28th, 2017, 09:33 AM  #1 
Member Joined: Nov 2011 Posts: 71 Thanks: 0  integration
Please integrate the following. the limits are L and 0. X(XB)dx 
October 28th, 2017, 10:47 AM  #2 
Senior Member Joined: Aug 2012 Posts: 1,678 Thanks: 436 
Assuming that by X you mean x, that's a straightforward polynomial.

October 29th, 2017, 04:17 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,919 Thanks: 785 
$x(x B)= x^2 Bx$ Do you know the antiderivative of $x^2$? Bx?

October 29th, 2017, 05:03 AM  #4 
Member Joined: Nov 2011 Posts: 71 Thanks: 0 
in my view the correct answer is ((L^2)/6)*(2L3B) but the book i am referencing shows the answer to be ((L^2)/6)*(2L3B) + B^3 I dont understand where the B^3 came from 
October 29th, 2017, 05:47 AM  #5 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,469 Thanks: 493 Math Focus: Yet to find out. 
$\int_0^L (x^2  Bx) dx = \left[\dfrac{x^3}{3}  \dfrac{Bx^2}{2}\right]_0^L = \dfrac{L^3}{3}  \dfrac{BL^2}{2} = ((L^2)/6)*(2L3B)$ Please post the full question as it appears in the text. 
October 29th, 2017, 08:52 AM  #6 
Member Joined: Nov 2011 Posts: 71 Thanks: 0 
Please see attached full question


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