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[tex]\large\underline{\sf{Solution-}}[/tex]

Given curves are

[tex]\rm \: f(x) = 4 - {x}^{2} \\ \rm \: and \\ \rm \: g(x) = {x}^{2} - 4 \\ [/tex]

So, two curves can be rewritten as

[tex]\rm \: y = 4 - {x}^{2} - - - (1) \\ \rm \: and \\ \rm \: y = {x}^{2} - 4 - - - (2) \\ [/tex]

Step :- 1 Point of intersection of 2 curves

[tex]\rm \: 4 - {x}^{2} = {x}^{2} - 4 \\ [/tex]

[tex]\rm \: 2{x}^{2} = 8 \\ [/tex]

[tex]\rm \: {x}^{2} = 4 \\ [/tex]

[tex]\rm\implies \:x \: = \: \pm \: 2 \\ [/tex]

Hᴇɴᴄᴇ,

➢ Pair of point of intersection of the given curves are shown in the below table.

[tex]\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 2 & \sf 0 \\ \\ \sf - 2 & \sf 0 \end{array}} \\ \end{gathered}[/tex]

Step :- 2 Curve Sketching

[tex]\rm \: y = 4 - {x}^{2} [/tex]

can be rewritten as

[tex]\rm \: {x}^{2} = - (y - 4) \\ [/tex]

represents the downward parabola having vertex at (0, 4) and intersects the x - axis at (0, 2) and (0, - 2)

Now,

[tex]\rm \: y = {x}^{2} - 4 \\ [/tex]

can be rewritten as

[tex]\rm \:{x}^{2} = y + 4 \\ [/tex]

represents the upper parabola having vertex at (0, - 4) and intersects the x - axis at (0, 2) and (0, - 2)

Step :- 3 Required Area

Now, from graph we concluded that the area of the bounded region is symmetrical in all the four quadrants.

So, required area is given by

[tex]\rm \: = \: 4 \times \displaystyle\int_{0}^{2}\rm y_{(downward \: parabola)} \: dx \\ [/tex]

[tex]\rm \: = \: 4 \times \displaystyle\int_{0}^{2}\rm (4 - {x}^{2}) \: dx \\ [/tex]

[tex]\rm \: = \: 4\bigg[4x - \dfrac{ {x}^{3} }{3} \bigg]_{0}^{2} \\ [/tex]

[tex]\rm \: = \: 4\bigg[8 - \dfrac{8}{3} \bigg] \\ [/tex]

[tex]\rm \: = \: 4\bigg[\dfrac{24 - 8}{3} \bigg] \\ [/tex]

[tex]\rm \: = \: 4\bigg[\dfrac{16}{3} \bigg] \\ [/tex]

[tex]\rm \: = \: \dfrac{64}{3} \: square \: units \\ [/tex]

Hence,

[tex]\rm\implies \:\boxed{ \rm{ \:\rm \:Required \: Area = \: \dfrac{64}{3} \: square \: units \: \: }} \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information :-

[tex]\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}[/tex]

Answer:

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