Simplify definite integral formula

[jvo]

Homework Statement

Working my way through Zienkiewicz's "FEM its basis & fundamentals" I am stuck on the following example:
$- \int_{y_1}^{y_2} t N1 \left\{ \begin{array}{c} \sigma_x \\ \tau_y \end{array} \right\} dy = -\left\{ \begin{array}{c} k_x t (2 y_1 + y_2) (y_2 - y_1) / 6 \\ 0 \end{array} \right\}$

with values:
$N1 = \frac{ y_2 - y }{ y_2 - y_1 }$
$\sigma_x = k_x y$
and $\tau_y = 0$

Homework Equations

n/a

The Attempt at a Solution

My result for the above integral is: $k_x t \frac{ y_2^3 - 3 y_2 y_1^2 + 2 y_1^3 }{6 (y_2 - y_1)}$

Tried to simplify the formula by dividing by (y2 - y1) and (y2-y1)^2 but I don't see how to get to the same result as in the book. What am I doing wrong here?
For more detail, see the attachments with the page from the book and my results.
Thanks in advance.
Jo

 

[Dick]

Homework Statement

Working my way through Zienkiewicz's "FEM its basis & fundamentals" I am stuck on the following example:
$- \int_{y_1}^{y_2} t N1 \left\{ \begin{array}{c} \sigma_x \\ \tau_y \end{array} \right\} dy = -\left\{ \begin{array}{c} k_x t (2 y_1 + y_2) (y_2 - y_1) / 6 \\ 0 \end{array} \right\}$

with values:
$N1 = \frac{ y_2 - y }{ y_2 - y_1 }$
$\sigma_x = k_x y$
and $\tau_y = 0$

Homework Equations

n/a

The Attempt at a Solution

My result for the above integral is: $k_x t \frac{ y_2^3 - 3 y_2 y_1^2 + 2 y_1^3 }{6 (y_2 - y_1)}$

Tried to simplify the formula by dividing by (y2 - y1) and (y2-y1)^2 but I don't see how to get to the same result as in the book. What am I doing wrong here?
For more detail, see the attachments with the page from the book and my results.
Thanks in advance.
Jo

It is the same thing as in the book. $y_2^3 - 3 y_2 y_1^2 + 2 y_1^3=(2 y_1 + y_2) (y_2 - y_1)^2$. You just have to figure out how to factor the numerator.

 

[jvo]

Thanks for your reply but I still don't see how to get there. Can you give a hint on how to approach this? As said in my post I tried long division by $(y_2 - y_1)$ and $(y_2 - y_1)^2$ but don't get even near your factored result.
I know factoring quadratics and checked 'recipes' as given here: http://www.analyzemath.com/polynomials/factor_polynomials.html
It's selfstudy, not homework. I realize missing something essential here, but what is it?
Tia,
Jo

PS this is a scan of some long division trials: http://utopia.knoware.nl/~jovo/images/Scan_img_026.jpg

 

[Dick]

Thanks for your reply but I still don't see how to get there. Can you give a hint on how to approach this? As said in my post I tried long division by $(y_2 - y_1)$ and $(y_2 - y_1)^2$ but don't get even near your factored result.
I know factoring quadratics and checked 'recipes' as given here: http://www.analyzemath.com/polynomials/factor_polynomials.html
It's selfstudy, not homework. I realize missing something essential here, but what is it?
Tia,
Jo

PS this is a scan of some long division trials: http://utopia.knoware.nl/~jovo/images/Scan_img_026.jpg

You started off ok on the first trial. $y_2^2$ is good for the first term. Now keep the rest ordered in decending powers of $y_2$. So what you bring down after the first subtraction is $y_2^2 y_1 - 3 y_2 y_1^2$. Your next term in the quotient should be dividing $y_2$ into $y_2^2 y_1$ (the higher $y_2$ power term). Or $y_2 y_1$. Try it that way. It does divide evenly.

 

[jvo]

I got it! Thank you for your excellent help.
My new year's resolution is to practice factoring more :-)
Regards,
Jo