The area of the region bounded by the curves and is : [2025]

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Q.
The area of the region bounded by the curves $x (1 + y^{2}) = 1$ and $y^{2} = 2 x$ is : [2025]

1 $2 (\frac{π}{2} - \frac{1}{3})$

2 $\frac{π}{4} - \frac{1}{3}$

3 $\frac{π}{2} - \frac{1}{3}$

4 $\frac{1}{2} (\frac{π}{2} - \frac{1}{3})$

Ans.
(3)

We have, $x (1 + y^{2}) = 1$ ... (i)

and $y^{2} = 2 x$ ... (ii)

From equation (i) and (ii), we get

$x (1 + 2 x) = 1 \Rightarrow 2 x^{2} + x - 1 = 0$

$\Rightarrow x = \frac{1}{2}, x = - 1$ (Reject)

From (ii), $y^{2} = 2 (\frac{1}{2}) \Rightarrow y = \pm 1$

$∴$ Required area $= \int_{- 1}^{1} (\frac{1}{1 + y^{2}} - \frac{y^{2}}{2}) d y$

$= (\tan^{- 1} y - \frac{y^{3}}{6}) |_{- 1}^{1} = \frac{π}{2} - \frac{1}{3}$ .

Solution

Q.
The area of the region bounded by the curves $x (1 + y^{2}) = 1$ and $y^{2} = 2 x$ is : [2025]

1 $2 (\frac{π}{2} - \frac{1}{3})$

2 $\frac{π}{4} - \frac{1}{3}$

3 $\frac{π}{2} - \frac{1}{3}$

4 $\frac{1}{2} (\frac{π}{2} - \frac{1}{3})$

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Solution

Q. The area of the region bounded by the curves x(1+y2)=1 and y2=2x is : [2025]

1 2(π2–13)

2 π4–13

3 π2–13

4 12(π2–13)

Ans. (3) We have, x(1+y2)=1 ... (i) and y2=2x ... (ii) From equation (i) and (ii), we get x(1+2x)=1⇒ 2x2+x–1=0 ⇒ x=12, x=–1 (Reject) From (ii), y2=2(12) ⇒ y=±1 ∴ Required area =∫–11(11+y2–y22)dy =(tan–1y-y36)|–11=π2-13.

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Q.
The area of the region bounded by the curves $x (1 + y^{2}) = 1$ and $y^{2} = 2 x$ is : [2025]

1 $2 (\frac{π}{2} - \frac{1}{3})$

2 $\frac{π}{4} - \frac{1}{3}$

3 $\frac{π}{2} - \frac{1}{3}$

4 $\frac{1}{2} (\frac{π}{2} - \frac{1}{3})$