The rectangular grounds at Hillingham University is going to have a new path built which takes the curved shape of $y=2.2x^2$, starting from the south-west corner - (taken as the origin in a graph) over to the point on the path where $x=3.7$. The constructors of this path want to know the path's length.
What is the exact length of this path?
The length of a curve is given by
$$\int_a^b \sqrt{1+(f'(x))^2}\ dx.$$
Let $f(x)= y$, $f'(x) = \frac{dy}{dx}$, $a=0$ and $b=3.7$, and compute the integral in a straightforward manner.
Where does this come from?
Consider estimating the length of a segment of the curve by forming a right triangle whose vertices are $(x,f(x))$, $(x+\Delta x, f(x))$ and $(x+\Delta x, f(x+\Delta x))$.
Let $\Delta y = f(x+\Delta x)-f(x)$.
Then, the length of the hypotenuse, $s$, is given by the Pythagorean Theorem $s = (\Delta x)^2 + (\Delta y)^2$.
Taking the limit as $\Delta x \to 0$, using the notation of infinitesimals, we see that $\Delta y \to dy$ and $\Delta x \to dx$.
To get the total length of the curve, we add up all these infinitesimal lengths:
$$\int_a^b ds.$$
But from the Pythagorean theorem we have that $ds^2 = dx^2+dy^2$, so $\frac{ds^2}{dx^2} = 1+\frac{dy^2}{dx^2} = 1+\left(f'(x)\right)^2$.
Finally, multiplying $dx^2$ back over and taking the square root, we have
$$ds = \sqrt{1+(f'(x))^2}\ dx.$$
The length of the curve is found by adding up all these infinitesimal hypotenuses, so
$$L(f(x)) = \int_a^b ds = \int_a^b \sqrt{1+(f'(x))^2}\ dx.$$