check the picture below.
so.. the graph looks like so, since the tangent line is at 5,50, it touches the parabola when y = 50, x = 5.
so, let's get the tangent line at that point,
[tex]\bf y=2x^2\implies \left. \cfrac{dy}{dx}=4x \right|_{x=5}\implies 20\leftarrow m
\\\\\\
y-y_1=m(x-x_1)\implies y-50=20(x-5)\implies y=20x-50\\
\left. \qquad \right. \uparrow\\
\textit{point-slope form}[/tex]
since I don't see an Up/Down function, just a Right/Left one... so, we'll use the function in "y-terms" then, the second one in the graph for each.
now... to get the bounds.... we could make them equal to each other, however, we know that the x-axis( y = 0) is the boundary line at the bottom, and that the tangent line is touching the parabolat a y = 50, thus, our bounds are from 0 to 50.
[tex]\bf \begin{cases}
\cfrac{y+50}{20}=x\impliedby \textit{right function}\\\\
\sqrt{\cfrac{y}{2}}=x\impliedby \textit{left function}
\end{cases}
\\\\\\
\displaystyle \int\limits_{0}^{50}~\left[ \left( \cfrac{y+50}{20} \right)~-~\left( \sqrt{\cfrac{y}{2}}\right) \right]dy
\\\\\\
\displaystyle \cfrac{1}{20}\int\limits_{0}^{50}~y\cdot dy+\int\limits_{0}^{50}~\cfrac{5}{2}\cdot dy-\cfrac{1}{\sqrt{2}}\int\limits_{0}^{50}~y^{\frac{1}{2}}\cdot dy[/tex]
[tex]\bf \left. \cfrac{y^2}{40}+\cfrac{5y}{2}-\cfrac{2\sqrt{y^3}}{3\sqrt{2}} \right]_{0}^{50}\implies \left[ \cfrac{250}{4}+125-\cfrac{500\sqrt{2}}{3\sqrt{2}} \right]-[0]
\\\\\\
\cfrac{375}{2}-\cfrac{500}{3}\implies \cfrac{125}{6}[/tex]
