# To blow some freshman-minds away....

• MHB
mathbalarka
Problem : Integrate $\displaystyle \int 1 \, \mathrm{d}x$.​

First, let $y = \displaystyle \int 1 \, \mathrm{d}x$. Then $\displaystyle \frac{dy}{dx} = 1$. By the geometric interpretation of differentiation, $\displaystyle \frac{dy}{dx}$ is $\tan(\theta)$, where $\theta$ is the angle made by the tangent at some $y = y(x)$ with the x-axis. In this case, $\tan(\theta) = 1$ is stationary, i.e., invariant under the change of $x$, so the function $y$ must coincide with the tangent made at any point on $y(x)$. Hence $y(x)$ is a linear function. Furthermore, $\tan(\theta) = 1$ is possible iff $\theta = \pi/4$, hence the tangent is the $45^\circ$-cut along the xy-plane, i.e., is the straightline $y' = x$. Then $y = y(x)$ must be parallel to this line, i.e., $y = x + C$ for some arbitrary real $C$. $\blacksquare$