Length of a Parabolic Curve 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:
$\sqrt 2 &lt; L &lt; 2$

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!!
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

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

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.
height of parabola=1 

base=1

hypotenuse length (check1) = 1.4142135623730950488016887242097

area between parabola and hypotenuse = $(1 \times 1)/6$

parabola length=$\sqrt(( 1\times1 + 1 \times 1 + 1 \times 1)/6) = 1.4719601443879744757940071211599.$