The area of the figure bounded by $y^{2} = 2 x + 1$ and $x - y - 1 = 0$ is:

2 / 3

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4 / 3

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8 / 3

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11 / 3

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Solution

The correct option is D $11 / 3$
From figure required area
$- \int_{0}^{- \frac{1}{2}} (\sqrt{2 x + 1} - (1 - x)) d x + \int_{- \frac{1}{2}}^{4} (\sqrt{2 x + 1} - (1 - x)) d x$
$= {⎡ ⎢ ⎣ \frac{1}{2} \frac{{(2 x + 1)}^{\frac{3}{2}}}{\frac{3}{2}} - x + \frac{x^{2}}{2} ⎤ ⎥ ⎦}_{0}^{- \frac{1}{2}} + {⎡ ⎢ ⎣ \frac{1}{2} \frac{{(2 x + 1)}^{\frac{3}{2}}}{\frac{3}{2}} - x + \frac{x^{2}}{2} ⎤ ⎥ ⎦}_{- \frac{1}{2}}^{4} = \frac{11}{3}$

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x = - 2 y^{2}, x = 1 - 3 y^{2}

\begin{matrix} L i s t I & L i s t I I P . & A r e a b o u n d e d b y y = x | x |, x - a x i s, & 1. & \frac{10}{3} x = 1 a n d x = - 1 Q . & A r e a b o u n d e d b y x - y + 2, & 2. & \frac{64}{3} x = \sqrt{y} a n d x = 0 R . & A r e a e n c l o s e d b y y^{2} = x & 3. & \frac{2}{3} a n d y = | x | S . & A r e a b o u n d e d b y y^{2} = x & 4. & \frac{1}{6} y = 4 a n d x = 0 \end{matrix}

List-I	List-II
1. Area of region bounded by $y = 2 x - x^{2}$ and $x -$ axis	a. $\frac{1}{3}$
2. Area of the region ${(x, y) : x^{2} \leq y \leq \| x \|}$	b. $\frac{1}{2}$
3. Area bounded by $y = x$ and $y = x^{3}$	c. $\frac{2}{3}$
4. Area bounded by $y = x \| x \|$ , $x$ -axis and $x = - 1, x = 1$	d. $\frac{4}{3}$

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The area bounded by the curve y² = 16x, x = 1, x = 3 and X-axis is:

y = x^{2} - 2 x + 3, y = 3 x - 1

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