I need help trying to figure out how to start this problem

[Luffy115]

You are given a curve defined by y = 2x + 4. Find the length of the curve via integration between the points (0,4) and (2,8). Use the pythagorean theorem to verify the result.

I need help on starting the integration part. Once I know how to start it, let me try to figure it out myself, so don't show me the answer or how to do it.

 

[chapdawg147]

http://www.math.hmc.edu/calculus/tutorials/arc_length/

scroll down to the second heading called "arc length."(smaller than the first one)

This the the general form of the equation and right below that, there is a sample problem that they run through.

Hope this helps

 

[Luffy115]

Damn, I thought I did something that figured out my mistake, but then I saw I was back to where I was before. With the formula, I'm getting sqrt(10), but when I use the pythagorean theorem I get sqrt(20).

Also, can someone check if this problem is right. I'm supposed to find the general solution to the differential solution of y' = 3x^2. Would the steps then go like:

dy/dx =3x^2
int(dy) = int(3x^2 dx)
y = (3x^3/3) + C

That problem seems easy, but for some reason, I'm not to sure how I'm doing it.[/code][/tex]

 

[galactus]

\(\displaystyle \L\\\int_{a}^{b}\sqrt{1+[f'(x)]^{2}}dx\)


The derivative of 2x+4 is just 2.

\(\displaystyle \L\\\int_{0}^{2}\sqrt{1+4}dx=\int_{0}^{2}\sqrt{5}=\sqrt{5}x\)

\(\displaystyle \L\\\sqrt{5}(2)-sqrt{5}(0)=2\sqrt{5}\)

Pythagoras:

\(\displaystyle \L\\\sqrt{(8-4)^{2}+(2-0)^{2}}=2\sqrt{5}\)

See?.

 

[Luffy115]

Alright, I see what I did wrong, thanks.