Unit Tangent Line Segment

In the above diagram, $\Delta x$ and $\Delta y$ are the distances from $P$ to $Q$ in the x and y directions. Since $PQ=1$ : $\begin{aligned}\sqrt{(\Delta x)^2+(\Delta y)^2}&=1\\ \left(\Delta y\frac{dx}{dy}\right)^2+(\Delta y)^2&=1&\small\text{since }\left|\frac{\Delta y}{\Delta x}\right|=\left|\frac{dy}{dx}\right|\\ y^2\left(\frac{dx}{dy}\right)^2+y^2&=1&\small\text{since }\Delta y=y\text{ at }P\\ \left(\frac{dx}{dy}\right)^2+1=\frac1{y^2}\\ \left(\frac{dx}{dy}\right)^2=\frac{1-y^2}{y^2}\\ \frac{dx}{dy}=\pm\frac{\sqrt{1-y^2}}y\end{aligned}$ As $\frac{dx}{dy}$ is always positive for $y>0$ and so the function is monotonic, and the question states that $f(x)$ maps $[0,\infty)$ to $(0,1]$ , we can infer that when $x=0, y=1$ and as $x$ approaches infinity, $y$ approaches $0$ . A more formal proof of this would involve integrating the formula above with respect to $y$ , which can be done by using the substitution $y=\sin\theta$ .

Once we have found the terminals of the integral in terms of $y$ we can evaluate the integral in the question by substitution. We know the result must be positive, as the question states that the range is $(0,1]$ . $dx=\pm\frac{\sqrt{1-y^2}}ydy$ $\begin{aligned}\therefore I=\int_{x=0}^{x=\infty} y\,dx&=\int_{y=1}^{y=0} y\left(\pm\frac{\sqrt{1-y^2}}y\right)dy\\ &=\pm\int_0^1\sqrt{1-y^2}\\ &=\pm\text{Area of quarter circle}\\ &=\frac{\pi}4\quad\text{where I>0}\\ &=\boxed{0.7854...}\end{aligned}$