# Simple Integral

1. Apr 10, 2012

### MeMoses

1. The problem statement, all variables and given/known data

Ok part of my math homework leads me to do this integral
The integral from 0 to 7 of sqrt((2t-10)^2) dt

2. Relevant equations

3. The attempt at a solution

The answer is 29, but I can only seem to get -21. I have it = integral of (2t-10)dt, so then I get t^2 - 10t from 0 to 7 = 49-70 = -21. Any help is appreciated. Thanks

2. Apr 10, 2012

### Ray Vickson

If f(t) = sqrt(2t-10)^2), what is f(0)? If g(t) = 2t-10, what is g(0)?

RGV

3. Apr 10, 2012

### darkxponent

if y= sqrt(x^2)
Can u simplify y?

4. Apr 10, 2012

### darkxponent

Thats agood question Vickson.

5. Apr 10, 2012

### MeMoses

Do i need an absolute value somewhere? If I don't simplify as I did and solve with substitution I still get (t-5)**2 from 0 to 7 which still yields -21. Where am I going wrong?

6. Apr 10, 2012

### MeMoses

Nvm that last part, I still used sqrt(x^2) = x

7. Apr 10, 2012

### Ray Vickson

In my first reply I asked you a question. You still have not answered it. (BTW: I asked for a very good reason!)

RGV

8. Apr 10, 2012

### MeMoses

well one is -10 and the other is 10

9. Apr 10, 2012

### Ray Vickson

OK, so what can you learn from that? Do you, or do you not need to use absolute values?

RGV

10. Apr 10, 2012

### MeMoses

i just ended up using x**2 = x*sqrt(x*2), so it keeps its sign and everything works out

11. Apr 10, 2012

### Ray Vickson

It sounds like you are trying to avoid the basic issue, which comes up over and over again and so is best dealt with now, once and for all. The point is:
$$\text{for any real number }r, \; \sqrt{r^2} = |r|.$$

RGV