Determine the area of the region bounded by $y = 2x^2 + 10$ and $y = 4x + 16$
The parabola $y = 2x^2 + 10$ and the line $y = 4x + 16$ intersect at the points $A \left(-1,12\right)$ and $B \left(3,28\right)$.
The required area A is the shaded region ABC.
$\begin{aligned}
\text{Shaded Area ABC} \left(A\right) & = \text{Area BDF } \left(A_1\right) - \left[\text{Area DAE} \left(A_2\right) \right.\\
& \hspace{2em} \left. + \text{Area EACO} \left(A_3\right) + \text{Area OCBF} \left(A_4\right) \right]
\end{aligned}$ $\;\;\; \cdots \; (1)$
Now, $A_1 = \left|I_1\right|$ where
$\begin{aligned}
I_1 & = \int \limits_{-4}^{3} y \; dx \hspace{2em} \text{where } y = 4x + 16 \\\\
& = \int \limits_{-4}^{3} \left(4x + 16\right) \; dx \\\\
& = \dfrac{4}{2} \left[x^2\right]_{-4}^{3} +16 \left[x\right]_{-4}^{3} = 2 \left(9-16\right) + 16 \left(3 + 4\right) = 98
\end{aligned}$
$\therefore$ $\;$ $A_1 = 98$ $\;$ sq units $\;\;\; \cdots \; (2)$
$A_2 = \left|I_2\right|$ where
$\begin{aligned}
I_2 & = \int \limits_{-4}^{-1} y \; dx \hspace{2em} \text{where } y = 4x + 16 \\\\
& = \int \limits_{-4}^{-1} \left(4x + 16\right) \; dx \\\\
& = \dfrac{4}{2} \left[x^2\right]_{-4}^{-1} +16 \left[x\right]_{-4}^{-1} = 2 \left(1-16\right) + 16 \left(-1 + 4\right) = 18
\end{aligned}$
$\therefore$ $\;$ $A_2 = 18$ $\;$ sq units $\;\;\; \cdots \; (3)$
$A_3 = \left|I_3\right|$ where
$\begin{aligned}
I_3 & = \int \limits_{-1}^{0} y \; dx \hspace{2em} \text{where } y = 2x^2 + 10 \\\\
& = \int \limits_{-1}^{0} \left(2x^2 + 10\right) \; dx \\\\
& = \dfrac{2}{3} \left[x^3\right]_{-1}^{0} + 10 \left[x\right]_{-1}^{0} = \dfrac{2}{3} \left[0 - \left(-1\right)^3\right] + 10 \left[0 - \left(-1\right)\right] = \dfrac{32}{3}
\end{aligned}$
$\therefore$ $\;$ $A_3 = \dfrac{32}{3}$ $\;$ sq units $\;\;\; \cdots \; (4)$
$A_4 = \left|I_4\right|$ where
$\begin{aligned}
I_4 & = \int \limits_{0}^{3} y \; dx \hspace{2em} \text{where } y = 2x^2 + 10 \\\\
& = \int \limits_{0}^{3} \left(2x^2 + 10\right) \; dx \\\\
& = \dfrac{2}{3} \left[x^3\right]_{0}^{3} + 10 \left[x\right]_{0}^{3} = \dfrac{2}{3} \times 27 + 10 \times 3 = 48
\end{aligned}$
$\therefore$ $\;$ $A_4 = 48$ $\;$ sq units $\;\;\; \cdots \; (5)$
$\therefore$ $\;$ In view of equations $(2)$, $(3)$, $(4)$ and $(5)$, equation $(1)$ becomes
$A = 98 - \left(18 + \dfrac{32}{3} + 48\right) = \dfrac{64}{3}$ sq units