Area between two parabolas - 2

Let us first check where the two parabolas intersect. When they intersect, we have:

$\begin{aligned} \frac {(x-3)^2}{16}+5 & = \frac {(x-5)^2}2+1 \\ \frac {x^2-6x+89}{16} & = \frac {x^2-10x+27}2 \\ x^2-6x+89 & = 8x^2-80x+216 \\ 7x^2 - 74x + 127 & = 0 \\ \implies x & = \frac {37 \pm 4\sqrt{30}}7 \end{aligned}$

Then the area bounded by the two parabolas is

$\begin{aligned} A & = \int_{\frac {37-4\sqrt{30}}7}^{\frac {37+4\sqrt{30}}7} \left(\frac {(x-3)^2}{16}+5 - \frac {(x-5)^2}2 - 1 \right) dx \\ & = \frac 1{16} \int_{\frac {37+4\sqrt{30}}7}^{\frac {37-4\sqrt{30}}7} \left(7x^2 - 74x + 127 \right) dx \\ & = \frac 1{16} \left[\frac 73 x^3 - 37x^2 + 127x \right]_{\frac {37+4\sqrt{30}}7}^{\frac {37-4\sqrt{30}}7} \\ & \approx \boxed{17.9} \end{aligned}$