The area of the shorter region bounded by |y| | vedclass

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(B) The curves are $|y| = 4 - x^2$ and $|y| = 3x$. Since both are symmetric about the $x$-axis,we consider the region in the first quadrant where $y =

Vedclass · Accepted Answer

$2$

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(B) The curves are $|y| = 4 - x^2$ and $|y| = 3x$. Since both are symmetric about the $x$-axis,we consider the region in the first quadrant where $y = 4 - x^2$ and $y = 3x$ for $x \ge 0$.Intersection point: $4 - x^2 = 3x \implies x^2 + 3x - 4 = 0 \implies (x+4)(x-1) = 0$. Since $x \ge 0$,$x = 1$. At $x=1$,$y=3$.The area in the first quadrant is $\int_0^1 (3x) dx + \int_1^2 (4 - x^2) dx$.$= \left[ \frac{3x^2}{2} ight]_0^1 + \left[ 4x - \frac{x^3}{3} ight]_1^2$$= \frac{3}{2} + \left( (8 - \frac{8}{3}) - (4 - \frac{1}{3}) ight) = \frac{3}{2} + ( \frac{16}{3} - \frac{11}{3} ) = \frac{3}{2} + \frac{5}{3} = \frac{9+10}{6} = \frac{19}{6}$.The total area (shorter region) is $2 	imes \frac{19}{6} = \frac{19}{3} = 6 + \frac{1}{3}$.Given area is $3K + \frac{1}{3} = 6 + \frac{1}{3}$.Comparing,$3K = 6 \implies K = 2$.

The area of the shorter region bounded by $|y| = 4 - x^2$ and $|y| = 3x$ is given by $\left( 3K + \frac{1}{3} \right)$ sq-unit,where $K$ is equal to:

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