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I just wanted to know how I can find the length of a curve given by $f(x) = x^2$ from $x=0$ to $x=1$.

For appproximation, the length is a bit larger than the hypotenuse of isosceles right triangle with the shorter side being 1 long. It's definitely less than the sum of two shorter sides. Thus, if we represent the length by $L$, the following relationship is expected: $\sqrt 2 < L < 2$

I now regard $L$ as the accumulation of hypotenuses of infinitestimally small right triangles around $f(x)$. Since $f'(x)=2x$, the general right triangle is something like this: If $x$ goes very slightly down the $x$-axis ($\Delta x$), the the $y$ value goes upwards for $2x\Delta x$.

Thus the hypontenuse is the square root of the following: $(\Delta x)^2+(2x\Delta x)^2$. The hypotenuse is: $(\Delta x) {( 4x^2 + 1)^{1/2}}$

Since $L$ has been defined as the accumulation of these hypotenuses, it is: $L = \int_0^1 ( 4x^2 + 1)^{1/2} dx$.

I am stuck just here. Could someone tell me if my chain of thoughts so far is right and how I can go from here? I don't know how to calculate the integral of a function that contains another function in it.

Thanks!!

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For some basic information about writing math at this site see e.g. here, here, here and here. –  Julian Kuelshammer Nov 5 '12 at 12:56
    
Thanks, Julian. I was at a loss how to use LaTeX notations around here. –  amemus Nov 5 '12 at 13:38

3 Answers 3

up vote 1 down vote accepted

Your train of thought is exactly right; you've single-handedly rederived the formula for the length of a curve given by $y=f(x)$ :-) This can be written as

$$ L=\int_a^b\sqrt{1+f'(x)^2}\,\mathrm dx $$

in general. In your case, as you rightly determined, $f'(x)=2x$, and we want the length from $a=0$ to $b=1$, so we have

$$ L=\int_0^1\sqrt{1+4x^2}\,\mathrm dx\;. $$

To solve this integral, you can use the substitution $\sqrt{1+4x^2}=\cosh u$, so $x=\frac12\sinh u$ and $\mathrm dx=\frac12\cosh u\,\mathrm du$, to get

$$ L=\frac12\int_0^{\operatorname{arcosh}\sqrt5}\cosh^2 u\,\mathrm du=\frac14\left[x+\sinh x\cosh x\right]_0^{\operatorname{arcosh}\sqrt5}=\frac14\left(\arccos\sqrt5+2\sqrt5\right)\approx1.479$$

If you don't know how to solve such an integral, you can always ask Wolfram|Alpha; it will usually know the answer and can often tell you the steps to get there if you click on "step-by-step solution"; though the solution will sometimes, as in this case, not be the most simple one.

You might also be interested in the question Intuition behind arc length formula.

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I read the argument by Arturo Magidin over and over. Yes, that expresses what I wanted to do in a far better way. $cosh$ and other hyperbolic functions are still mystery to me. Is there a good starting point in these forums? –  amemus Nov 5 '12 at 14:48
    
@amemus: I don't know about a starting point in this forum. I would try to understand them by exploring the analogies with the trigonometric functions, e.g. $\mathrm e^{\mathrm ix}=\cos x+\mathrm i\sin x$ vs. $\mathrm e^x=\cosh x+\sinh x$ and $\cos^2x+\sin^2x=1$ vs. $\cosh^2x-\sinh^2x=1$ -- you can get these and more like them by viewing the hyperbolic functions as trigonometric functions evaluated at imaginary arguments. –  joriki Nov 5 '12 at 15:12

Use the arc length formula, $$\int_0^1 \sqrt{1+(f')^2}\ dx = \int_0^1 \sqrt{1+(2x)^2}\ dx =\left[\frac 1 2 x\sqrt{4x^2+1} + \frac 1 4 \sinh^{-1}(2x)\right]_0^1 \approx 1.479$$

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In my humble opinion, it's not the most encouraging response when the OP has effectively managed to derive the arc length formula all by her- or himself to just tell them "use the arc length formula". –  joriki Nov 5 '12 at 12:44
    
I'm really sorry, I didn't mean it. –  Tariq Nov 5 '12 at 12:59

height of parabola=1 base=1 hypotenuse length(check1)=1.4142135623730950488016887242097 area between parabola and hypotenuse =(1x1)/6 parabola length= square root ( 1x1+1x1+(1x1)/6) =1.4719601443879744757940071211599.

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