$A$ : $A$ rea bounded by $x = 4 - y^{2}$ with $Y$ -axis
$B$ : $A$ rea bounded by $x = 6 + 5 y - y^{2}$ with $Y$ - axis
$C$ : $A$ rea bounded by $2 x = y^{2} - 1$ with $Y$ -axis
The ascending order $o f A, B, C$ is

A,B,C

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B,C,A

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A,C,B

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C,A,B

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Solution

The correct option is D C,A,B
A $2 \int - 2 (4 - y^{2}) d y$
$= 4 y - \frac{y^{3}}{3} |_{- 2}^{2}$
$= 4 (4) - [\frac{16}{3}]$
$= \frac{32}{3} s q u n i t s = 10.66 s q u n i t s$

B $6 \int - 1 (6 + 5 y - y^{2}) d y$
$6 (7) + \frac{5}{2} y^{2} |_{- 1}^{6} - y^{3} / 3 |_{- 1}^{6}$
$= 42 + \frac{5}{2} (35) - [\frac{217}{3}]$
$= 57.16 s q u n i t s .$
C $1 \int - 1 \frac{y^{2} - 1}{2} d y = \frac{1}{2} y^{3} / 3 - y |_{- 1}^{1}$
$= ∣ ∣ ∣ 1 / 2 [\frac{2}{3}] - [2] ∣ ∣ ∣ = \frac{5}{3} = 1.6 s q u n i t s .$
ascending order C,A,B.

Suggest Corrections

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x = 3, y^{2} = 3 x

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y = 1 - | x |

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y = \sqrt{3} x

A, B, C

List-I	List-II
1. Area bounded by $y = \| x \|$ and $y = 2$ is	a. 4
2. Area bounded by $\frac{\| x \|}{a} + \frac{\| y \|}{b} = 1$ , when $a, b > 0$ is	b. $\frac{(π - 2) a b}{4}$
3. Area between the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ and the chord $\frac{x}{a} + \frac{y}{b} = 1$	c. 1
4. Area bounded by $y = [x]$ , the $x$ -axis and $x = 1, x = 2$ is	d. 2ab