Integral Homework: Solving for 0 to 7 of sqrt((2t-10)^2)

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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
 
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MeMoses said:

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
 
if y= sqrt(x^2)
Can u simplify y?
 
Thats agood question 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?
 
Nvm that last part, I still used sqrt(x^2) = x
 
MeMoses said:
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
 
well one is -10 and the other is 10
 
OK, so what can you learn from that? Do you, or do you not need to use absolute values?

RGV
 
  • #10
i just ended up using x**2 = x*sqrt(x*2), so it keeps its sign and everything works out
 
  • #11
MeMoses said:
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
 
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