Why do I get -21 instead of 29 for the integral of sqrt((2t-10)^2) from 0 to 7?

[MeMoses]

Homework Statement

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

Homework Equations

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

 

[Ray Vickson]

Homework Statement

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

Homework Equations

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

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

RGV

 

[darkxponent]

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

 

[darkxponent]

Thats agood question Vickson.

 

[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?

 

[MeMoses]

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

 

[Ray Vickson]

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?

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

RGV

 

[MeMoses]

well one is -10 and the other is 10

 

[Ray Vickson]

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

RGV

 

[MeMoses]

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

 

[Ray Vickson]

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

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