Nice idea!
As long as the function $g(x)$ is well-behaved, we have the following very important result. Let
$$G(x)=\int_c^x g(t)dt$$.
Then $G'(x)=g(x)$.
This result (and some related ones) is called the Fundamental Theorem of (Integral) Calculus.
Now let us apply that to your problem. We obtain
$$L'(x)=\sqrt{1+(f'(x))^2}$$
Use the above equation to solve for $f'(x)$ in terms of $L'(x)$. If you take $L(x)$ as known, you have found an explicit formula for $f'(x)$, and all you need to do is to integrate.
Now comes the unfortunate part. For most pleasant functions $L(x)$, the resulting integration problem will be either difficult or more often impossible (in terms of standard functions).
I hope that this gives you something to play with. You will find out why there is such a limited number of different arclength problems in calculus books!
The way to understand this is to step back from thinking about integrals as areas. Fundamentally, integrals are not areas, they are infinite sums of infinitely small values. To understand how it works, let's look at the opposite direction - differentials. Take the equation:
$$ y = x^2 $$
Now, let's differentiate it:
$$ dy = 2x\,dx $$
On the left, you have an infinitely small change in $y$. On the right, you have infinitely small values, which are also equivalent to the same infinitesimal change in $y$.
Now, if you add up all of the infinitesimal changes in $y$ between two $x$ values, you will get the TOTAL change in $y$, correct? Therefore, if you had the original function, you could just subtract the $y$ values at those points to get what the total change is.
A way of stating that mathematically is:
$$ \int_{x_0}^{x^1} dy = y(x_1) - y(x_0) $$
So that's what an integral is. It is the sum of all of the infinitely small changes of an expression. So how does it relate to areas? Let's go back to your problem, where you want the area under the curve $y = x$.
With areas, an area of a rectangle is width times height. So, each area will be $dx$ wide and $y$ high. Therefore, each rectangle will be:
$$ y\,dx $$
Now, we can imagine a function $z$ such that its differential, $dz$, is the same value as this area. So we set:
$$ dz = y\,dx $$
So, using the reasoning we used before, we know that the TOTAL of all of $dz$ is going to be equal to the difference between $z$ evaluated at $x_0$ and $x_1$. In other words:
$$ \int_{x_0}^{x_1} dz = \int_{x_0}^{x_1} y\,dx = \int y\,dx\, \biggr|_{x_0}^{x_1} $$
So, since $z(x_1) - z(x_0)$ is equal to the sum of the areas, so is $ \int y\,dx\, \biggr|_{x_0}^{x_1}$.
Now, we can't integrate $y\,dx$ directly, but, thankfully, we have a formula for $y$, which is $y = x$. Therefore, this becomes:
$$ \int_{x_0}^{x_1} dz = \int_{x_0}^{x_1} y\,dx = \int y\,dx\, \biggr|_{x_0}^{x_1} = \int x\,dx\, \biggr|_{x_0}^{x_1} = \frac{x^2}{2} \biggr|_{x_0}^{x_1} $$
So, what we did was:
- Recognize that the integral is an infinite sum of infinitely small pieces
- Recognize that the difference between two points on a function is equal to the infinite sum of the differentials of that function.
- Because of (2), integrating differentials is the same as performing an infinite sum of them.
- If we can write the formula for an infinitely small area as a differential, we can use (1)-(3) to generate a formula for a function such that the function's change is equal to the sum of the infinitely small values between those points.
Best Answer
The idea is that an integral is a limit of approximations by rectangular strips. So if you split the interval $[0,1]$ into $n$ strips of width $n$, in the strip from say $x=(k-1)/n$ to $x=k/n$, the area under the curve within that strip is between $\frac 1n\sqrt{1-(\frac{k}{n})^2}$ and $\frac 1n\sqrt{1-(\frac{k-1}{n})^2}$. (In general, we have to look at the minimum and maximum values of the function in the strip, but in this case it's easier since the function is decreasing, so the maximum will always be the left end.)
Summing this over all strips gives you bounds $$\sum_{k=1}^n\frac1n\sqrt{1-\Big(\frac{k}{n}\Big)^2}\leq A\leq\sum_{k=1}^n\frac1n\sqrt{1-\Big(\frac{k-1}{n}\Big)^2}.$$ If we increase $n$, we can get better and better bounds. If both bounds approach the same value as $n$ goes to infinity, then we know that value must be equal to the area. (This equality of the bounds happens for the sort of functions you're likely to encounter, although not for every function; in particular it always happens if the function is continuous.)
If you'd started from a simpler function (e.g. the area under $y=x^2$) you could compute these sums exactly and deduce what the limits are. In this case it's rather harder to do that. Most of the time, to actually compute an integral, we don't use the definition but rather the useful fact that integrals are antiderivatives. The intuitive reason for this, writing $F(x)=\int_0^xf(x)\,dx$, is that the difference $F(1+1/n)-F(1)$ is the same as taking one extra strip of width $1/n$ when approximating the area, so $F(1+1/n)-F(1)\approx \frac 1n\times f(1)$ (assuming that $f(x)$ is continuous at $x=1$, so that the height of the strip is close to $f(1)$ throughout). This means if you try to approximate the gradient of $F(x)$ at $x=1$, you get $$F'(1)\approx \frac{F(1+1/n)-F(1)}{1+1/n-1}\approx f(1),$$ and letting $n$ tend to infinity this gives you exactly the right answer.