# Help! Improper Integrals Type 2 argg!

1. May 5, 2005

### cmab

I don't understand my Math Homework, here's the number that I don't get.

[int a=1 b=3] 1/(sqrt[abs(x-2)]) dx

sqrt = square root
abs = absolute value
Integral from 1 to 3

Can anyone explain this to me clearly, I'll really appreciate

Thanks

2. May 5, 2005

### cmab

I tried and it keeps giving me 0.

2[sqrt(abs(x-2))] from 1 to t and 2[sqrt(x-2)] from t to 3

(2[sqrt(abs(1-2))])-(2[sqrt(abs(t-2))])
and
(2[sqrt(abs(t-2))])-(2[sqrt(abs(3-2))])

(lim t->2+ (2[sqrt(abs(t-2))])-(2[sqrt(abs(1-2))]) )
+
(lim t->2- (2[sqrt(abs(3-2))])-(2[sqrt(abs(t-2))]) )
=4

in the book it says 3/4

Last edited: May 5, 2005
3. May 5, 2005

### saltydog

Try and remember whenever you have a problem dealing with absolute values, think, "split in two". That is, split it into two parts: one part is when the value between absolute values is positive, and another part when the value is negative. So:

$$|x-2|=x-2 \quad\text{ when }\quad x-2\geq 0$$

$$|x-2|=-(x-2) \quad\text{ when }\quad x-2<0$$

so, split the integral into two, you know, at 2, and then solve it. Remember to replace the absolute value expression in the radical with the appropriate form (from above) for each interval. Can you do this?

4. May 5, 2005

### cmab

Nice, i check it out on gragphmatica, and the graph of both seems to match the original function. Thanks dude, if i have a question i<ll comme back here :tongue:

5. May 5, 2005

### cmab

But the limit for that problem have to approch 2 from the right for the x-2, and from the left for -(x-2). Right ?

6. May 5, 2005

### saltydog

No limit considerations are involved. Just solve two integrals. I'll set up one for you:

$$\int_1^2 \frac{dx}{\sqrt{2-x}}$$

What's the other one?

7. May 6, 2005

### cmab

I did the thing, and it still gives me 4