Minimising the area between a line and an exponential curve
Let $f(k,x)=k(x-1)+\frac12(e^2+1)-e^x$, so $\alpha,\beta$ satisfy $f(k,\alpha)=f(k,\beta)=0$ for some fixed $k$. Minimising the area is equivalent to setting the derivative of the integral to be $0$: $$\frac d{dk}\int_\alpha^\beta f(k,x)\,dx=\int_\alpha^\beta\frac\partial{\partial k}f(k,x)\,dx=\int_\alpha^\beta(x-1)\,dx=(\beta^2-\alpha^2)/2+\alpha-\beta$$ $$=\frac12(\beta-\alpha)(\alpha+\beta-2)=0$$ From this we get $\alpha+\beta=2$ and $$f(k,\alpha)+f(k,\beta)=k(\alpha-1+\beta-1)+e^2+1-e^\alpha-e^{2-\alpha}$$ $$=k(\alpha+\beta-2)+e^2+1-e^\alpha-e^{2-\alpha}$$ $$=-e^\alpha-\frac{e^2}{e^\alpha}+e^2+1=0$$ This last equation rearranges into a quadratic in $e^\alpha$: $$e^{2\alpha}-(e^2+1)e^\alpha+e^2=(e^\alpha-1)(e^\alpha-e^2)=0$$ So we get $\alpha=0$, $\beta=2$ and $k=\frac12(e^2-1)$. The minimum area is $2$.
Perhaps the geometrical trick or insight is as follows.
Suppose the fixed point is P and the line intersects the curve at A and B.
When the enclosed area is minimised, then a small change in the slope $k$ makes no difference to that area. As the line pivots about P through a small angle $\delta \theta$ the triangular area $AP \delta\theta$ removed on one side equals the triangular area $PB\delta\theta$ added on the other side. So $AP=PB$ - ie P is then the midpoint of AB.
(Note : Since this argument does not depend on the form of the curve, it applies for all curves. Also, the minimum area may be a local rather than global minimum. It may even be a local or global maximum since the method only finds turning points.)
The fixed point has co-ordinates $x=1, y=\frac12 (e^2+1)$.
Suppose the co-ordinates of A and B are $(\alpha, e^{\alpha})$ and $(\beta, e^{\beta})$. Then $$\frac12 (\alpha + \beta) = 1$$ $$\frac12 (e^{\alpha}+e^{\beta})=\frac12 (e^2+1)$$ $$k=\frac{e^{\beta}-e^{\alpha}}{\beta-\alpha}$$ from which $$e^{(\alpha+\beta)}=e^{\alpha}e^{\beta}=e^2$$ $$e^{\alpha}(e^{\alpha}+e^{\beta})=(e^{\alpha})^2+e^2=e^{\alpha}(e^2+1)$$ $$(e^{\alpha})^2-(e^2+1)e^{\alpha}+e^2=(e^{\alpha}-e^2)(e^{\alpha}-1)=0$$
If $e^{\alpha}=1$ then $\alpha=0, \beta=2, e^{\beta}=e^2$. Conversely if $e^{\alpha}=e^2$ then $\alpha=2, \beta=0, e^{\beta}=1$.
Therefore $$k=\frac12 (e^2-1)$$
Although this method finds the slope $k$ of the chord through P which minimises (or maximises) area, it does not find the area.