$y=2x^2$ 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.
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:
$$ \int_1^{\frac{5}{2}} f(x)-t_1(x)dx + \int_{\frac{5}{2}}^4 f(x)-t_2(x)dx. $$
Best Answer
You're not going to get anywhere if your approach to problems is a hit-and-miss "can we try vertical integration here?" "okay, what about horizontal integration?"
Instead, you should actually understand what the region looks like, and then use that to set up an integral for the area.
In this case, the parabola and its tangent line look like
and this should tell you why your first integral with respect to $x$ doesn't work: it includes the entire area between the blue and orange curves, which also counts a large triangle below the $x$-axis.
So you have the following approaches: