Minimising the area between a line and an exponential curve We need to find a suitable k , for which the area between the two curves $y=e^x, y=k(x-1) + \frac{1}{2}(e^2+1)$ is minimum.
The line passes through a fixed point $(\frac{1}{2}(e^2+1))$, and has variable slope. But, I don"t think its possible to determine the intersection points of the two curves , so I had assumed them to be $\alpha$ and $\beta$.
I performed the integral $\int (k(x-1) + 1/2(e^2+1) -e^x)dx$ from $\alpha$ to $\beta$, (the line lies above the curve for this interval) and tried to make use of the fact that : $e^\alpha= k(\alpha-1)+\frac{1}{2}(e^2+1)$ and likewise for $\beta$, but was still unable to get the area explicitly in terms of $k$. 
There might be a geometric argument that minimizes the area but the ways these curves are, I fail to realize it.
Can we have generalized method to minimize/maximize the are enclosed by them? Without explicitly knowing their intersection points?

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 $2$.
