Length of Curve: Find \int_{0}^{1} (3x^3+11x) dx

[Aerosion]

Homework Statement

To find the length of the curve defined by y=3x^3+11x from the point (0,0) to the point (1,14), you'd have to compute:

\int_{a}^{b} f(x) dx

where a=?, b=? and f(x)=?

Homework Equations

The Attempt at a Solution

So I put a to be 0 and b to be 1, since it's asking for the x axis.

Then I took the equation y=3x^3+11x and got its derivative, 9x^2+11. I subtracted one from the entire equation and square rooted the entire thing, such that it looked like \sqrt{1-9x^2+11}.

Did I do anything wrong here? This is an internet-generated exercise I'm doing.

 

[Integral]

That will give you the area under the curve, not the arc length. Do a bit more searching for the arc length forumal.

 

[Mindscrape]

That will not give you the area under the curve, nor will it give you the arc length!

The arc length would S = \int_a^b \sqrt{1 + (\frac{dy}{dx})^2}

*Note that is (dy/dx)^2, I did something latex didn't like.