To answer this problem, you must derive a system of linear equations to solve for the three unknowns in your parabola $f(x),$ where $$f(x)=ax^2+bx+c. $$ We know $f(2)=1=4a+2b+c$, $f(4)=3=16a+4b+c$, and $f'(\text{B})=-\frac{1}{4}=4a+b$. Thus, the system $$\begin{cases}4a+2b+c=1\\ 16a+4b+c=3\\ 4a+b=-\frac{1}{4} \end{cases} $$ is established. I will leave it up to you to solve for $f(x).$
$y=2x^2$ has derivative $y'=4x$. So the derivative will be positive for all $x>0$. This means the tangent line will intersect the x-axis at some point $x_a<x$ for a given x.
This means the area under the parabola and bounded by the tangent line will have two separate regions, A region with $x<x_a$ which is made up of only area under the parabola, and a region $x>x_a$ where area is from the area between the parabola and the tangent to $(3,18)$.
The equation of a tangent line at $(x_0,y_0)$ is $$\frac{y-f(x_0)}{x-x_0}=f'(x_0)$$
Or : $y=f'(x_0)(x-x_0)+f(x_0)$
The x intercept happens where $y=0$.
Requiring $y=0$ implies an x intercept of $x_c=\frac{-f(x_0)}{f'(x_0)}+x_0$
So from the above arguments with $x_0=3$:
$$A=\int_0^{x_c}2x^2dx+\int_{x_c}^32x^2-(12x-18)dx$$
But this can be simplified. The area under the tangent line is a triangle. So the the parabola can be integrated ignoring the tangent line, and then subtracting the area of the triangle.
From the above, we know $(x_0-x_c)=\frac{f(x_0)}{f'(x_0)}$, the base of the triangle. The height is just $f(x_0)$.
So the area of the triangle is $A_t=\frac{1}{2}\frac{f(x_0)^2}{f'(x_0)}$
So:
$$A=\int_0^{x_0}f(x)dx-\frac{1}{2}\frac{f(x_0)^2}{f'(x_0)}$$
and solve for $x_0=3$.
In this form, an expression can be found for $x_0$ which extremizes the area.
Best Answer
Denote
$$ f(x)\equiv \frac{1}{2}x^{2}-2x+2, \;\;\;\; \text{and} \;\;\;\; f'(x) \equiv x-2$$
The first tangent line, $t_1(x)$, goes through $(1,1/2)$ and has slope $f'(1)=-1$, so:
$$ t_1(x) = \frac{3}{2}-x.$$
The second tangent line, $t_2(x)$, goes through $(4,2)$ and has slope $f'(4)=2$, so:
$$ t_2(x) = -6+2x.$$
The two tangents intersect at
$$ \frac{3}{2}-x = -6+2x \Rightarrow x = \frac{5}{2}. $$
Hence, the integral you are looking for is:
$$ \int_1^{\frac{5}{2}} f(x)-t_1(x)dx + \int_{\frac{5}{2}}^4 f(x)-t_2(x)dx. $$