$y=2x^2$ 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.
In your question, it asks for the area of the shaded region, and area is always positive.
For integration, it is taught as the area between the curve and the $x$-axis. But actually not quite, because in the definition of integration we calculate "Area" as "NET area" (positive if above $x$-axis, and offset by those below $x$-axis).
Note that Area is absolute, but Net Area is relative
Suppose $C$ is the curve $y=f(x)$ with $f$ a twice-differentiable function such that $f''(x) > 0$ for each $x \in [0,a]$, where $a$ is a positive constant. Suppose $T$ is the tangent line to $C$ at a point $P=(r,f(r))$ on $C$ where $r \in (0,a)$. Let $A$ be the area of the plane region bounded by the $y$-axis, the vertical line $x=a$ the curve $C$, and the tangent line $T$. Show that $A$ is minimum if and only if $r=a/2$.
Best Answer
\begin{align} A(r) &= \int_0^af(x)dx-\int_0^a\left(f'(r)(x-r)+f(r)\right)dx \\ &= \int_0^af(x)dx-\int_0^a\left(f'(r)x-rf'(r)+f(r)\right)dx \\ &= F(a)-F(0)-\left[\frac {a^2}2f'(r)-arf'(r)+af(r)\right] \\ &=F(a)-F(0)-\frac {a^2}2f'(r)+arf'(r)-af(r). \end{align}
Then we differentiate to find the minimum of $A(r)$:
\begin{align} A'(r) &= \frac {-a^2}2 f''(r)+af'(r)+arf''(r)-af'(r) \\ &= \frac {-a^2}2 f''(r)+arf''(r) \\ &= af''(r)\left(r-\frac a2\right). \end{align}
Since both $a$ and $f''(r)$ are positive, we only care about $r-a/2$, which is zero at $r=a/2$ and switches from negative to positive at $r=a/2$ meaning that $A'(r)$ switches from negative to positive, meaning that $A(r)$ has a minimum at $r=a/2$. In this case, it is the absolute minimum, since there are no other critical points.