1. ## defined integral

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

2. ## Re: defined integral

Hint: use the derivative form of the fundamental theorem of calculus.

3. ## Re: defined integral

i dont know how

4. ## Re: defined integral

We are given to verify:

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

The derivative form of the fundamental theorem of calculus is:

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

Use this for the left side, and on the right use the product rule and the fact that we have $0\le x$.

5. ## Re: defined integral

I have tried, but I can't see the relation, excuseme

6. ## Re: 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.

7. ## Re: defined integral

Let's focus on the left side first. What do you get using the theorem I suggested when you differentiate?

8. ## Re: defined integral

|t|, {0,x}=f(x)

9. ## Re: defined integral

Or i don´t know if must be only |x|=f(x)

10. ## Re: defined integral

Yes, you get |x|. Now, what do you get when you differentiate the right side?

edit: Use the definition $|x|\equiv\sqrt{x^2}$.

11. ## Re: defined integral

F'(x)=[(x^2)/2|x|]+(|x|/2)

12. ## Re: defined integral

On the right side, you have:

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

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

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

Thus, we have shown the given result for the definite integral is valid for all real $x$.

13. ## Re: 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. . .

14. ## Re: defined integral

I only sitll have a doubt, what about the condition that x < or = 0?

15. ## Re: defined integral

That condition was an error on my part. x can be any real number, as originally stated in your first post.

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