Stupid Tabular Method for Integration by parts

[Saladsamurai]

Is this the correct way to use the Tabular Method for
$$\int x^2e^{-5x}dx$$


repeated diff:

$$x^2$$

$$2x$$

$$2$$


repeated Integration:

$$e^{-5x}$$

$$-\frac{1}{5}e^{-5x}$$

$$\frac{1}{25}e^{-5x}$$

$$-\frac{1}{125}e^{-5x}$$


=$$-\frac{x^2}{5}e^{-5x}+\frac{2x}{25}e^{-5x}+\frac{2}{125}e^{-5x}+C$$

I hate this.

Casey

 

[bob1182006]

yes but the signs of the 2nd/3rd terms are wrong.

You should write out the table and include the alternating +/-

writing it out this way might be easier to do since all you do is read across to find the terms

          [tex]e^{-5x}[/tex]
+   [tex]x^2[/tex]   [tex]\frac{-e^{-5x}}{5}[/tex]

-   [tex]2x[/tex]   [tex]\frac{e^{-5x}}{25}[/tex]

 

[Saladsamurai]

Then I clearly have no idea what I am doing. I am looking at the tabular method in my book, and from what i have read I take the diagonal product of x^2*(-1/5)e^{-5x} and then ADD it to the next product of 2x*(1/25)e^{-5x} and now subtract the product of 2*(-1/125)e^{-5x}

Am I off on all of them by a step?

 

[Saladsamurai]

OHHHH...I get it ...form the product then multiply by +1 or -1 THEN add all of the terms...that makes sense. so it is

$$-\frac{x^2}{5}e^{-5x}-\frac{2x}{25}e^{-5x}-\frac{2}{125}e^{-5x}+C$$

Casey