How to demostrate that Integrate of |t|dt in {0,x} for any x in R is (1/2)x|x|

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- Nov 3rd 2012, 07:10 PMnigromantedefined integral
How to demostrate that Integrate of |t|dt in {0,x} for any x in R is (1/2)x|x|

- Nov 3rd 2012, 07:22 PMMarkFLRe: defined integral
Hint: use the derivative form of the fundamental theorem of calculus.

- Nov 3rd 2012, 07:28 PMnigromanteRe: defined integral
i dont know how

- Nov 3rd 2012, 07:51 PMMarkFLRe: defined integral
We are given to verify:

(1) $\displaystyle \int_0^x |t|\,dt=\frac{1}{2}x|x|$

The derivative form of the fundamental theorem of calculus is:

$\displaystyle \frac{d}{dx}\int_a^x f(t)\,dx=f(x)$

Use this for the left side, and on the right use the product rule and the fact that we have $\displaystyle 0\le x$. - Nov 3rd 2012, 08:10 PMnigromanteRe: defined integral
I have tried, but I can't see the relation, excuseme

- Nov 3rd 2012, 08:16 PMnigromanteRe: defined integral
first I wish to use that t^2/2 is equal to |t|, but im not shure, that it is correct, and in the form that you sugest, I think is better, but sorry, I know that I need more practice, or I have a problem in understand a detail.

- Nov 3rd 2012, 08:24 PMMarkFLRe: defined integral
Let's focus on the left side first. What do you get using the theorem I suggested when you differentiate?

- Nov 3rd 2012, 08:29 PMnigromanteRe: defined integral
|t|, {0,x}=f(x)

- Nov 3rd 2012, 08:32 PMnigromanteRe: defined integral
Or i donīt know if must be only |x|=f(x)

- Nov 3rd 2012, 08:39 PMMarkFLRe: defined integral
Yes, you get |x|. Now, what do you get when you differentiate the right side?

edit: Use the definition $\displaystyle |x|\equiv\sqrt{x^2}$. - Nov 3rd 2012, 08:46 PMnigromanteRe: defined integral
F'(x)=[(x^2)/2|x|]+(|x|/2)

- Nov 3rd 2012, 08:57 PMMarkFLRe: defined integral
On the right side, you have:

$\displaystyle \frac{1}{2}x|x|=\frac{1}{2}x\sqrt{x^2}$

Using the product, power and chain rules, we find:

$\displaystyle \frac{d}{dx}\left(\frac{1}{2}x\sqrt{x^2} \right)=\frac{1}{2}\left(x\frac{x}{\sqrt{x^2}}+ \sqrt{x^2} \right)=\frac{x^2}{\sqrt{x^2}}=\sqrt{x^2}=|x|$

Thus, we have shown the given result for the definite integral is valid for all real $\displaystyle x$. - Nov 3rd 2012, 09:19 PMnigromanteRe: defined integral
OOOOOOh!!!!

It's realy easy, I was near but I didn't know, when I had to use |x| or x^2/2.

I realy thank you for your patience and your time. . . - Nov 3rd 2012, 09:23 PMnigromanteRe: defined integral
I only sitll have a doubt, what about the condition that x < or = 0?

- Nov 3rd 2012, 09:27 PMMarkFLRe: defined integral
That condition was an error on my part.

*x*can be any real number, as originally stated in your first post.