Question

$\green{\underbrace{ \purple{ \underline{ \large\underline{\sf \: \red{Question:-}}}}}}$



Find the area of ​​the region covered by the parabola f (x) = 4 - x² and g (x) = x²- 4.
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$\underline{ \underline{ \large\rm \text{Answer :-}}} \\ \color{darkcyan}\boxed{ \rm \frac{64}{3} sq.unit}$


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Answer 1

$\large\underline{\sf{Solution-}}$

Given curves are

$\rm \: f(x) = 4 - {x}^{2} \\ \rm \: and \\ \rm \: g(x) = {x}^{2} - 4$

So, two curves can be rewritten as

$\rm \: y = 4 - {x}^{2} - - - (1) \\ \rm \: and \\ \rm \: y = {x}^{2} - 4 - - - (2)$

Step :- 1 Point of intersection of 2 curves

$\rm \: 4 - {x}^{2} = {x}^{2} - 4$

$\rm \: 2{x}^{2} = 8$

$\rm \: {x}^{2} = 4$

$\rm\implies \:x \: = \: \pm \: 2$

Hᴇɴᴄᴇ,

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

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

Step :- 2 Curve Sketching

$\rm \: y = 4 - {x}^{2}$

can be rewritten as

$\rm \: {x}^{2} = - (y - 4)$

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

Now,

$\rm \: y = {x}^{2} - 4$

can be rewritten as

$\rm \:{x}^{2} = y + 4$

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

$\rm \: = \: 4 \times \displaystyle\int_{0}^{2}\rm y_{(downward \: parabola)} \: dx$

$\rm \: = \: 4 \times \displaystyle\int_{0}^{2}\rm (4 - {x}^{2}) \: dx$

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

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

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

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

$\rm \: = \: \dfrac{64}{3} \: square \: units$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \:Required \: Area = \: \dfrac{64}{3} \: square \: units \: \: }}$

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Additional Information :-

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

Answer 2

Answer:

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