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

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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.
 
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That will give you the area under the curve, not the arc length. Do a bit more searching for the arc length forumal.
 
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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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