Arc Length of Curve: Find Solution (-2,2) to (2,4)

[sepah50]

Homework Statement

Find the arclength of the section y=x² between points (-2,2) and (2,4)

Homework Equations

L = [tex]\int\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}[/tex]

The Attempt at a Solution

So what I did first is find the derivative of y=x² which is y^'=2x

So I put that into the formula and get [tex]\int\sqrt{1+4x^2}[/tex] from limit point 1 to 4

From this point on I attempted to do u-substitution but it didn;t seem to work.. Anyone have any suggestions o how I can get it done?

 

[Char. Limit]

Trig substitution works well here.

 

[sepah50]

Trig substitution works well here.

Yes, some people have told me to use Trigonometric substitution but the thing is that my teacher hasn't taught us that yet!

 

[Dickfore]

Represent it as:

[tex]
\sqrt{1 + 4 x^{2}} = \frac{1 + 4 x^{2}}{\sqrt{1 + 4 x^{2}}} = \frac{1}{\sqrt{1 + 4 x^{2}}} + \frac{4 x^{2}}{\sqrt{1 + 4 x^{2}}}
[/tex]

The integral:

[tex]
\int{\frac{dt}{\sqrt{1 + t^{2}}}} = \sinh^{-1}{(t)} = \ln{(t + \sqrt{1 + t^{2}})}
[/tex]

is sometimes quoted as a table integral. To my knowledge, there is no other way to prove except by those substitutions mentioned earlier, or hyperbolic ones.

The other intergral:

[tex]
4 \, \int{\frac{x^{2} \, dx}{\sqrt{1 + 4 x^{2}}}}
[/tex]

can be integrated out by parts (and using some elementary substitutions). Good luck!

 

[Char. Limit]

Well, basically here's how it works. Let's say that...

[tex]2x=tan(\theta)[/tex]

Then this is also true.

[tex]2 dx = sec^2(\theta) d\theta[/tex]

Just plug those values in for dx and 4x^2=(2x)^2 and then you have a new integral in theta.

But don't forget to change your bounds too.

 

[sepah50]

Well, basically here's how it works. Let's say that...

[tex]2x=tan(\theta)[/tex]

Then this is also true.

[tex]2 dx = sec^2(\theta) d\theta[/tex]

Just plug those values in for dx and 4x^2=(2x)^2 and then you have a new integral in theta.

But don't forget to change your bounds too.

Thanks for the advice, it is much appreciated! What I found out is that I can use the integral table and use this formula: http://www.sosmath.com/tables/integral/integ11/integ11.html" It is #8 of this list. Would that work?

 

[Dickfore]

How about this one:

http://www.wolframalpha.com/input/?i=Integrate[Sqrt[1%2B4*x^2],x]