# Help me proove absolute value inequalities and Solve an Integral

1. Sep 4, 2014

### RJLiberator

1) True or Fale?
|a-b| $\geq$ ||a|-|b||

My solution: I broke this up into 4 different cases. 1. a> 0 b>0 2. a<0 b<0 and so on...
For each case, I ended up with a more simplified version of the inequality. For instance, in case 1 where I used a<0 and b<0, I ended up with the simplified statement of:
$|-a+b|\geq|-a+b|$ Which is clearly equivalent.

Is this a correct way to prove that this statement is true?

2. With that being out of the way, how can I prove this statement to be true:
$\sqrt{(a+b)^2} = |a+b|$
Would I do the same thing and break it up into 4 cases?

3. This has been my most difficult problem so far.

For any x > 0, consider the function

f(x)=$\int^{\sqrt{x}}_{0}e^{t^{2}}dt$

Computer f'(x)

My work so far: I was told to use the chain rule where the inner function is $\sqrt{x}$ and the outer is 0.

I'm not qutie sure what to do.. I imagine I take the antiderivative of this function using the chain rule?

Last edited: Sep 4, 2014
2. Sep 4, 2014

### HallsofIvy

If you are only considering "1. a> 0, b> 0 2. a< 0 b< 0 and so on" HOW did you get to $|-a+ b|$? You also need to consider, in each of those cases a< b and a> b.

You know that |a|= |b| if and only if $a^2= b^2$, don't you?

This makes no sense. The left side is a function but the right side is a differential. they can't be equal. Did you mean $f(x)= \int e^t dt$? If so finding f' is just the "fundamental theorem of Calculus".

??? There is NO square root in the problem you gave- and an "outer function" of 0 would just make the entire function 0.

You said the problem was to find the derivative not the anti-derivative. What, exactly is the problem.

3. Sep 4, 2014

### RJLiberator

Regarding 1)
To save space I cut my claims short. I did take into account a>b and b>a

2) So if I square both sides, I would receive (a+b)^2 = (a+b)^2 Hm. That makes a lot of sense.

3) It seems you may have quoted my question when I was editing/had a problem with the latex. The proper latex version is up now.
With that being said, I understand why the outer function of 0 would make the chain rule null. Maybe my TA was not looking at the problem carefully enough.

I will need to search through the Fundamental theorem of Calculus then.

Thank you for the input. It is appreciated.

4. Sep 4, 2014

### LCKurtz

Since you apparently solved it using many cases, I will suggest an easier way not needing all those cases. Start with$$|a| = |(a-b)+b| \le |a-b| + |b|$$(I'm assuming you already have the triangle inequality here). So this gives you$$|a|-|b| \le |a-b|$$. Now do you see how this same result with $a$ and $b$ switched gives you your required inequality?

5. Sep 4, 2014

### RJLiberator

Hey guys, I think I figured out the third problem. The integral problem using the fundamental theorem of calculus. Can anyone check my work:

1. denote a new function :
g'(x) = e^(x^2)
f(x)=g(sqrt(x))

Using the chain rule
f'(x) = 1/2(sqrt(x) * e^(sqrt(x)^2)

How does this look?

LCK, I am going to look into what you posted. One moment.

6. Sep 4, 2014

### HallsofIvy

Surely, you can simplify "(sqrt(x))^2".

7. Sep 4, 2014

### RJLiberator

Excellent. e^x.

=)